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INTRODUCTORY
TOPOLOGY
STEWART SCOTT CAIRNS
PROFESSOR OF MATHEMATICS
UNIVERSITY OF ILLINOIS
THE RONALD PRESS COMPANY · NEW YORK

Copyright © 1961 by
The Ronald Press Company
All Eights Reserved
No part of this book may be reproduced in
any form without permission in writing
from the publisher.
Library of Congress Catalog Card Number: 61-9427
PRINTED IN THE UNITED STATES OF AMERICA

To
My Wife
Kathleen

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Preface
This book is the culmination of repeatedly revised sets of class
notes used by the author in teaching introductory topology courses.
Its purpose is to progress as far as practicable into the fundamental
concepts and the principal results of homology theory, both in their
combinatorial development and in their application to topological
spaces.
First, some of the properties of linear graphs and of surfaces are
presented in such a way as to give an intuitive geometric impression
of the nature of topology. Then enough set-theoretic topology is
given to motivate the subsequent combinatorial theory and to provide
a background for its geometric interpretation. Cohomology groups
are defined and are used in connection with the duality theorems of
Poincare and Lefschetz. Certain aspects of homotopy theory are
treated, and there is a chapter on the fundamental group and covering
complexes. The essential facts from group theory are collected in an
Appendix.
The book can be used as text for a first course in topology either
with or without set-theoretic and group-theoretic prerequisites. It is,
however, advisable to give such a course at the advanced
undergraduate and graduate levels in order to ensure a certain mathematical
maturity and a familiarity with the role of continuity in the calculus.
A class with a background in set theory and group theory might
complete the book in a single semester, using Chapter 3 and the
Appendix for review and reference. For a class without such a
background, portions of the last three chapters could be omitted, or else an
entire academic year could be devoted to the book, perhaps
supplemented with additional topics from set-theoretic topology and group
theory.
The author is deeply indebted to Professor Arthur B. Brown of
Queens College of the City of New York. He read and reread the
manuscript in detail, making numerous valuable suggestions for its
correction and improvement. In several instances, he suggested
alternative arguments or supplied entirely new proofs, leading to a
more efficient and illuminating exposition. In addition, he contributed
a strengthening of Sperner's lemma (Chapter 6, Theorem 4), which
?

permits a substantial simplification of earlier proofs of the topological
invariance of homology properties.
The drawings were executed by Dr. Ali R. Amir-Moez of Purdue
University.
Stewart Scott Cairns
Urbana, Illinois
March, 1961

Contents
CHAPTER PAGE
1 Illustrative Geometric Examples
1-1. The Seven-Bridges Problem ...... 3
1-2. Unicursal Graphs ........ 4
1-3. The Cyclomatic Number ....... 6
1-4. Restrictions on Polyhedra ...... 9
2 Topological Classification of Surfaces
2-1. Polygonal Regions with Matched Edges . . . . 15
2-2. Some Elementary Surfaces ...... 20
2-3. Orientability and Non-Orientability ..... 26
2-4. Standard Form for Spheres with Contours and Handles or
Crosscaps ......... 29
2-5. A Classification Theorem ...... 33
3 Introduction to Set-Theoretic Topology
3-1. Sets and Mappings ........ 41
3-2. Relations, Cartesian Products, Functions . . . . 44
3-3. Continuity for Real Functions of Real Variables . . 48
3-4. Topological Spaces ........ 51
3-5. Homeomorphisms; Definition of Topology ... 54
3-6. Metric Spaces 56
3-7. Compact Spaces ........ 59
3-8. Brouwer Dimension; The Lebesgue Number ... 61
4 Complexes
4-1. Linear and Convex Subspaces of En . . . . . 65
4-2. Dimension Numbers in En 67
4-3. Barycentric Coordinates ....... 70
4-4. Simplexes 71
4-5. Complexes ......... 74
4-6. Polyhedra; Topological Complexes ..... 75
4-7. Abstract and Generalized Complexes .... 77
4-8. Realizations of Abstract Complexes ..... 78
4-9. Isomorphisms and Homeomorphisms .... 80
4-10. Simplicial Mappings 81
4-11. Barycentric Subdivisions. ...... 82
4-12. General Polyhedral Complexes 86
vii

CHAPTER PAGE
5 Homology and Cohomology Groups
5-1. Chains, Cycles, and Bounding Cycles .... 90
5-2. Homology Groups of Finite Simplicial Complexes . . 96
5-3. Some Lower-dimensional Cases ..... 98
5-4. Homology Groups of a Surface ..... 99
5-5. Surface Topology . . . . . . . . 106
5-6. Pseudomanifolds . . . . . . . . 110
5-7. Homology Bases and Incidence Matrices . . . . 113
5-8. Connectivity Groups and Numbers . . . . . 120
5-9. Cohomology Groups . . . . . . . 122
5-10. Dual Bases 125
5-11. Comments on Cohomology Groups . . . . . 126
6 Topological In variance of Homology Properties
6-1. Singular Simplexes . . . . . . . . 128
6-2. Singular ^-Chains and Groups . . . . . . 130
6-3. Sperner's Lemma; Invariance of Dimension . . . 133
6-4. The Brouwer Fixed-Point Theorem 136
6-5. Invariance of Regionality . . . . . . 139
6-6. Singular and Simplicial Groups on a Topological Polyhedron 141
6-7. Simplicial Subsets of Singular Homology Classes . . 142
6-8. Chains on Prism Complexes . . . . . . 144
6-9. Invariance of Homology Properties . . . . . 148
6-10. Classes of Mappings . . . . . . . 151
7 Manifolds
7-1. Some Homology Properties of Pseudomanifolds . . . 153
7-2. Thera-Sphere 154
7-3. Projective m-Space. ....... 155
7-4. Local Homology Groups . . . . . . . 159
7-5. Topological Manifolds and Homology Manifolds. . . 162
7-6. Cell Complexes 164
7-7. Cellular Subdivisions of a Homology Manifold . . . 167
7-8. The Poincare Duality Theorem 171
7-9. Relative Homology . . . . . . . 175
7-10. The Lefschetz Duality Theorem 177
7-11. The Alexander Duality Theorem and Consequences . . 180
8 The Fundamental Group; Covering Surfaces
8-1. Paths and Path Products 185
8-2. The Fundamental Group 187
8-3. Relation Between 0(2) and ^(S) 193

CHAPTER PAGE
8-4. The Fundamental Groups of En and of a Circle . . . 199
8-5. The Fundamental Group of a Surface .... 201
8-6. Covering Complexes ....... 205
8-7. Fundamental Groups and Coverings ..... 208
Bibliography 211
Appendix: Group-Theoretic Background
A-l. Basic Terminology ........ 215
A-2. Homomorphisms and Isomorphisms . . . . . 219
A-3. The Structure of Finitely Generated Abelian Groups . . 221
A-4. Integral Modules, Contravariant and Covariant Components 230
A-5. Dual Bases in a Module 233
Index of Symbols 237
General Index 239

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INTRODUCTORY
TOPOLOGY

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1
Illustrative Geometric Examples
The purpose of this chapter is to present a few examples which
appeal to the intuition and suggest the general nature of topology.
1-1. The Seven-Bridges Problem
In its original form, the problem of the seven Koenigsberg bridges
relates to a situation of the sort shown in Fig. 1-1. The problem is to
Fig. 1-1. The Koenigsberg bridges.
Fig. 1-2. An equivalent linear graph.
devise, if possible, a promenade which crosses each bridge once and
only once. A characteristic of this problem, which makes it topological
Ш nature, is that it depends in no way on the size or shape of the river,
3

4 INTRODUCTORY TOPOLOGY [Ch. I
bridges, and islands, but only on the general manner in which they are
put together. That is, the problem is the same, essentially, as that of
tracing all of Fig. 1-2 without lifting the pencil and without tracing
any arc more than once. One can convince himself experimentally of
the impossibility of doing so in this particular case; not so, however,
with the general problem of unicursal graphs, next to be discussed,
which Leonhard Euler A707-1783), stimulated by the seven-bridges
problem, invented and solved in 1736.
1-2. Unicursal Graphs
A linear graph G will mean a finite set of points, called the vertices
of G, together with a finite set of arcs, called the edges of G, such that
(a) each edge has two distinct vertices for end points, (b) each vertex is
an end point of at least one edge, (c) any two edges are distinct except
that they may have one or two common vertices. The graph may be
thought of as being in 3-dimensional space, where any set of points can
be joined in pairs by non-intersecting arcs. In the plane, it is
impossible, for example, to join five points each to each by non-
intersecting arcs.
By a path on G will be meant a sequence
A-1) ? = (pl9 Ev p2, E2,..., En_v pn)
where A) the p's are vertices, B) the jE"s are edges, and C) Ei has pi
and pi+1 for end points (i = 1, . . . , ? — 1). If p± = pn, the path ?
will be referred to as closed; otherwise ? will be called a path from px
to pn. The graph G is connected if there exists at least one path from
each vertex of G to each other vertex thereof.
The path ? will be described as simple if no two of its vertices pi and
Pi (г 7^ Л are coincident, save perhaps ?? and pn.
Lemma 1. If there exists a path ? on G from a to b, then there
exists a simple path from a to 6, whose edges and vertices are a subset
of those of ?.
Proof of Lemma. If p{ and pj (j > i) coincide in A.1), then
pi9 Ei9 . . . , Ej_1 can be dropped from the sequence and the remaining
terms renumbered, without destroying the defining properties of a
path. This procedure can be repeated until all repetitions are removed
from the sequence.
The graph G will be called unicursal if there exists a sequence A.1)
where Ei has pif pi+1 for end points (i = 1, . . . , ? — 1), and where
each edge of G appears exactly once in the sequence. Intuitively, this
means that it is possible to trace all of G continuously from px to pn
without traversing any edge more than once. It will then be said

Art. 1-2] ILLUSTRATIVE GEOMETRIC EXAMPLES 5
that G is unicursal from px to pn or can be unicursally traced from рл
topn.
The following notation will be useful in certain inductive arguments.
Let ffbea linear graph with at least two edges, and let GE denote the
graph obtained from G by deleting an edge E; it being understood that
if a vertex ? ?? ? belongs to no other edge, then ? is also to be deleted.
A vertex of a linear graph will be called even or odd according as it
belongs to (or is incident with) an even or an odd number of edges.
Lemma 2. A linear graph G has an even number of odd vertices.
Proof of Lemma. If G has one edge, this result is obvious. Assume
it true for all graphs with j edges for some j > 1, and let G be a graph
wiihj + 1 edges. Then, for each edge ? of Cr, GE differs from G in the
number of its odd vertices by 0 or 2. By hypothesis, the lemma holds
for GE. Hence it holds for G.
Lemma 3. If G is connected, then GE either is connected or falls
into two connected linear graphs.
Proof of Lemma. Let a and b denote the end points of E. Let G(a)
and G(b) denote the subgraphs of GE consisting of all vertices, plus
incident edges, which can be joined to a and b respectively by
sequences of the form A.1), hence (see Lemma 1) by simple paths.
Let ? be an arbitrary vertex of G. Since G is connected, ? can be
joined by simple paths on G to a and to b. If each simple path joining
? to b includes the edge E, necessarily as its last edge, then ? is joined
to ? by a path not involving E, and similarly with a and b interchanged.
Hence each vertex can be joined to either a or b on G by a path not
involving E. But this means that G(a) and G(b) exhaust GE, and
Lemma 3 follows.
Theorem 1. The linear graph G is unicursal if and only if it is
connected and either A) there are no odd vertices, in which case the
initial and terminal points of a unicursal tracing must coincide but can
otherwise be arbitrarily selected; or B) there are exactly two odd
vertices, in which case the unicursal tracing must start at one odd
vertex (either will do) and terminate at the other.
Proof. The proof of this theorem will be largely left to the reader.
The following suggestions, however, are offered as a guide to one
particular proof, which is inductive with respect to the number of
edges of G. For the initial case, where G has only one edge, the
theorem is obvious. The inductive hypothesis assumes the theorem
for all graphs with j > 1 edges. The general step then consists in
deducing it for an arbitrary graph G with j + 1 edges. Note that the
inductive hypothesis applies to GE, as defined above, or to its
connected subgraphs, where ? is an edge of G, and that Lemma 3 is

6 INTRODUCTORY TOPOLOGY [Ch. I
useful in the argument. It may prove convenient to divide the general
step into the three cases: where G has no odd vertices, where it has
two such, and where it has more than two.
EXERCISES
1. Prove the following, for a connected graph G: (a) If G has fewer than four
vertices, it is unicursal. (b) If G has exactly four vertices and is not unicursal,
then the graph obtained by adjoining a new edge to G, joining two of its vertices,
is unicursal. (c) If G has exactly nve vertices and is not unicursal, it is possible
to obtain a unicursal graph from it by adjoining an edge connecting two suitably
chosen vertices.
2. (a) Show that the removal of any one of the Koenigsberg bridges would
make the unicursal promenade possible, but that no such promenade would end
where it started, (b) Give an example of a connected graph with four vertices
and with an edge whose removal does not lead to a unicursal graph.
1-3. The Cyclomatic Number
Consider a linear graph G. Let us think of the edges as wires.
Then, from the electrical viewpoint, one might be interested in
knowing whether G is connected and whether or not G contains
circuits, where a circuit means a simple closed path, that is, a sequence
of the form A.1) in which all the vertices (hence all the edges) are
distinct, except that рг = pn. A graph G which is connected and on
which no circuit exists is called a tree.
A terminal vertex of a graph G will mean a vertex incident with only
one edge thereof, and an edge incident with a terminal vertex will be
called a terminal edge. The other edges of G will be called inner edges.
Lemma 4. A tree G is characterized, among connected graphs, by
either of the following properties:
A) For two arbitrary vertices, a and b, of G there exists only one simple
path from a to b.
B) If ? is any inner edge of G, then GE is not connected.
Proof of Lemma. To establish A) as a necessary property, note that
the existence of at least one simple path from a to b follows from the
connectedness of the graph. If more than one should exist, then the
reader can prove that a circuit could be put together from any two of
them, or from selected portions thereof. As for the sufficiency, the
existence of a circuit would imply a violation of condition A), where
a and b are two vertices on such a circuit. The proof of Lemma 3
above can be readily adapted to establish condition B) as a
characteristic property and also to establish the following corollary.

Art. 1-3] ILLUSTRATIVE GEOMETRIC EXAMPLES 7
(A)* If ? is any inner edge of a tree G, then GE consists of two
separate trees.
(B) A tree has at least two terminal vertices. The proof of (B) is
left to the reader.
Given a linear graph G, let a0 denote the number of its vertices and
at the number of its edges. If G is connected, then the number
A.2) ? = ax - a0 + 1
is called the cyclomatic number, also the first Betti number, of G.
Theorem 2. A connected graph G is a tree if and only if ? = 0.
If ? > 0, then it is possible to reduce G to a tree by the removal of ?
suitably selected inner edges. This cannot be done by the removal of
fewer than ? edges, and G is necessarily disconnected by the removal of
? + 1 inner edges.
Proof. Note first that the theorem implies that ? is the number of
wires which must be cut, under the above electrical interpretation, in
order to break every circuit without separating G into distinct
components, f
As in the case of Theorem 1, the proof will be left to the reader,
with the comment that an argument inductive in ax is practicable
(Exercise 3). The initial case, ax = 1, is trivial, and the passage from
?? = j to ax = j + 1 offers no substantial difficulties. The following
auxiliary result, which is readily established, is stated for its intrinsic
interest.
(C) Any connected graph for which ax > 1 can be built up from a
single edge by the successive adjunction of other edges, the graph
being connected at all times.
Like various other aspects of pure mathematics, the study of linear
graphs was partly stimulated by practical problems. In fact, as just
noted, the first important contribution to the theory of such graphs
was made by Kirchhoff in the course of investigating currents in a
connected electrical network. There are two types of equations, all of
the first degree, entering into the determination of the currents, given
an arbitrary distribution of electromotive forces on the wires of the
network, namely:
A) Equations for circuits made up of subsets of the wires, such as the
wires numbered A, 2, 6), A, 2, 4, 5), and D, 5, 6) in Fig. 1-3.
* Letters (A), (JB), (C), . . . , with parentheses, are given for convenience in later
reference.
t See the article by G. Kirchhoff in Poggendorf's Annalen der Physik 72 A847), p. 497.
This article contains "the first important contribution to the theory of linear graphs"
[V, p. 26]. (Here, and throughout this text, such bracketed symbols as [V] refer to books
and articles listed in the bibliography.) The number ?, discovered by Kirchhoff, was
named the cyclomatic number by another physicist, James Clerk Maxwell, A Treatise on
Electricity and Magnetism, (Oxford: Clarendon Press, 1873).

8 INTRODUCTORY TOPOLOGY [Ch. I
B) Equations expressing the fact that the sum of all the currents flowing
into each vertex is zero, provided those flowing out are taken
negatively.
The reader sufficiently familiar with electrical theory should write
the equations and follow through the remainder of this article in detail.
In the notation of linear graphs, there are ?? wires and a0 vertices
in the network. The unknowns of the equations are the currents ih
(h = 1, . . . , ??) in the wires. To
determine them, there must be ax
independent equations in the sets A) and B) just
described; that is, it should be possible
to pick out exactly ax equations from
these two sets, no one of the selected
equations being a consequence of the
others, and to assert that all other
equations of the two sets are consequences of
those selected. It turns out, for example,
that the equations for the three circuits
mentioned under A) are not
independent. Let the wires be oriented by the
arrows in Fig. 1-3, a current being posi-
Fig. 1-3. A network. tive if in the direction of the arrow and
negative if in the opposite direction. As
a consequence of the connectedness of the network, one can show that
there are just a0 — 1 independent equations in set B). All the
equations in set A) are independent of those in set B). Since a total of
ax independent equations is needed to determine the currents in the
wires, there must be ax —- a0 + 1 independent equations in A). This
is the cyclomatic number ? of Eq. A.2) and is the least number of
wires whose cutting will break every circuit. KirchhofF selected a
complete set of independent equations as follows. Let the wires be so
numbered that the cutting of wires A, 2, . . . , ?) breaks every circuit.
Then, in the network left when all but one of these ? wires is removed,
there is just one circuit. If all the wires save wire i are removed, call
the circuit in the remaining network Gi(i = 1, . . . , ?). The equations
for the ? circuits Ci are independent, since each contains a current not
in any of the others. Every equation for a circuit depends on these ?,
since every circuit can be expressed as a combination of the Ct. The
reader should, like KirchhofF, demonstrate this by an argument
recurrent in ?. Thus KirchhofF was led to discover the cyclomatic
number.
EXERCISE
3. Carry out the suggested inductive proof of Theorem 2.

Art. 1-4] ILLUSTRATIVE GEOMETRIC EXAMPLES 9
1-4. Restrictions on Polyhedra*
For the next elementary geometric results, consider the polyhedral
shapes into which one might deform a spherical piece of modeling clay.
It is understood that there is to be no pulling apart or sticking together
of the clay; for example, it is not permissible to punch a hole, like a
tunnel, all the way through it. The surface of the clay is called a
polyhedron ? when it is made up of a finite number of plane polygonal
regions, with their edges and vertices. These plane regions will be
called the faces of ?.
(A) The following properties of 77 offer no difficulties: A) Each face
has at least three edges. B) Each vertex belongs to at least three
edges. C) The vertices on each face can be cyclically ordered so that
each is joined to its successor by an edge. D) The faces having a given
vertex in common can be cyclically ordered so that each has an edge
in common with its successor.
The results of this section are topological in that they apply even if
the surface of the clay is not polyhedral but is divided into regions such
that A) each region has a simple closed boundary with at least three
points, called vertices, on it, B) each vertex belongs to at least three
regions, C) if two regions have in common a boundary point other than
a vertex, then they have in common the entire edgef through that
e number of vertices of the polyhedral surface ?,
e number of edges of ?,
e number of faces of ?.
the number of vertices each belonging to
exactly г edges (i = 3, 4, . . .),
the number of faces possessing exactly г edges
(г = 3,4,...).
The following theorems, presently to be proved, state some of the
possibly surprising restrictions on the structure of ?.
Theorem 3. It is impossible that ? have exactly seven edges,
though it may have six or any number higher than seven. That is,
?? ? 7, but ax = 6, 8, 9, . . . are all possible.
* The material of this article is adapted from the writer's expository paper,
^Peculiarities of polyhedra," American Mathematical Monthly 58 A951), pp. 684-689,
Miere references to sources can be found.
t An edge is here the arc of a boundary between two consecutive vertices.
point.
Let
A3)
Let
A.4)
a0 = th
04 = th
a2 = th
vt =
F< =

10 INTRODUCTORY TOPOLOGY [Ch. I
Theorem 4. There must be at least A) four faces of ? with fewer
than six edges and B) four vertices of ? belonging to fewer than six
edges. This can be formulated thus:
A.5) (a) F3 + F, + F5 > 4, (b) F3 + F4 + V5 > 4.
Theorem 5. It is impossible that, simultaneously, every face have
more than three vertices and every vertex belong to more than three
faces. In fact, the number of triangular faces plus the number of
vertices each belonging to exactly three faces is at least 8; or,
formulated:
A.6) F3+ F3>8.
Theorem 6. It is impossible that ? have either A) an odd number of
odd-sided faces or B) an odd number of vertices each belonging to an
odd number of edges. That is, both (F3 + F5 + F7 + · · ·) and
(^з + У о + У 7 + ' ' ·) must be even.
Theorem 7. A) If each vertex belongs to just three faces and each
face is pentagonal or hexagonal, then exactly twelve faces are
pentagonal. B) If each face is triangular and each vertex belongs to
either five or six faces, then exactly twelve vertices belong to five faces
each. The first part of this theorem is expressible thus:
ao = Уч
A.7)
«2 = F5 + F,\
imply F5 =
The second can be formulated as follows:
*2 = ^3
A.8)
«о = Уъ + У J
imply V5 =
12.
12.
Theorem 8. If all faces have the same number i of edges and all
vertices belong to the same number j of edges, then there are only five
possible topologically* different polyhedral structures, typified by the
tetrahedron, cube, octahedron, dodecahedron, and icosahedron.
In other words, the theorem, known from Grecian antiquity, that
only five regular solids exist is a topological theorem, and the
hypothesis that the faces be congruent in the euclidean sense is
superfluous, since the conclusion of Theorem 8 is a consequence of
an = V. for some г and
A.9)
a2 = Fj for somej.
* The significance of this term is clarified later.

Art. 1-4] ILLUSTRATIVE GEOMETRIC EXAMPLES II
The above six theorems all follow from certain equations and
inequalities among the F's, i^'s, and oc's. In the first place
«2 = -F3 + Fi + · · · + Far.l
since A) each vertex belongs to at least three and at most a2 — 1
edges and B) each face has at least three and at most a0 — 1 edges.
(B) The above theorems illustrate a so-called duality principle; two
statements being dual to one another if each is carried into the other by
the simultaneous interchange of a0 and the F's with a2 and the
respective F's. The property of two faces having a common edge is
described as dual to the property of two vertices being joined by an
edge.
Thus, the two numbered parts of Theorem 4 are dual to one
another, and similarly for Theorems 6 and 7. Theorem 5 is self-dual
and so, in a trivial way, is Theorem 3. Equations A.10) are a dual
pair. As for Theorem 8, the reader should verify A) that the
tetrahedron is self-dual, B) that the cube and octahedron are dual
structures, and C) that the dodecahedron and icosahedron are dual.
The concept of dual maps, explained below, will be helpful.
As the present discussion continues, let it be observed that all the
equations on which the proofs of Theorems 3 to 8 depend either are
self-dual or fall into dual pairs. Hence it will suffice to prove only one
of each pair of dual results in the theorems, since interchanging a0 and
the F's with a2 and the F's will yield a proof of the other.
A geometric insight into the reason for duality can be obtained as
follows. Regard the faces of ? as maps of countries on a polyhedral
earth; or, what amounts to the same thing, let the faces, edges, and
vertices be mapped, preserving their incidence relations, onto a
spherical surface S. The edges need not be arcs of circles, but can be
any simple non-intersecting arcs joining pairs of vertices in accordance
with the structure of the original polyhedron. Interior to each face,
select a vertex. The vertices thus selected will belong to a so-called
dual map and will be called dual vertices. If two faces cpx and ?2 have
in common an edge ?, let the dual vertices v[ and v'% on them be joined
by a new edge ?', crossing ? at a single point and otherwise meeting no
edge or vertex of either the given map or the new dual map (see (B)
above and Fig. 1-4). The possibility of this requires a proof which
would take us too far afield. Accordingly, let it be assumed possible,
or let the edges be restricted to curves on the sphere for which it is
clearly possible. The new edges, which are the edges of the dual map,
subdivide S into new faces, each containing exactly one vertex of the
original map. It is easy to verify that if г; is a vertex of 77 on a given

12 INTRODUCTORY TOPOLOGY [Ch. I
face ?' of the dual map, then ? belongs to as many faces of ? as ?'
has vertices in the dual map. Since the relations between the maps
are symmetric, the duality principle follows.
The same mathematician who solved the problem of unicursal
graphs also established the following identity, known as Euler's
theorem in two dimensions:
A.11) ?0-?! + ?2 = 2.
It holds for a more general subdivision of a spherical surface than
described above, in which there may be as few as two vertices on the
Fig. 1-4. Dual maps.
boundary of a region and, dually, as few as two regions with a given
vertex in common. The simplest such subdivision consists of two
regions, two edges, and two vertices, which obviously satisfies A.11).
Any subdivision can be obtained* from this simplest one by a sequence
of steps, each consisting of either A) the separation of an edge into two
edges by the introduction of a new vertex or B) the separation of a
region into two regions by the introduction of a new edge joining two
vertices. The first type of step increases a0 and ax by one each, leaving
a2 unchanged; the second leaves a0 unchanged and increases a2 and ax
by one each. Since A.11) is therefore preserved by each step, it holds
for all subdivisions of the sphere.
Since each edge of a polyhedron belongs to exactly two faces,
A.12) 2ax = 3FS + 4F, + · · · + (a0 - l)Fao_i;
* This statement requires proof. The reader might consider how he would go about
to develop an inductive demonstration of it. However, certain fine points in the
argument depend on material in later chapters.

Art. 1-4] ILLUSTRATIVE GEOMETRIC EXAMPLES 13
and, dually, since each edge has exactly two vertices,
A.13) 2?? = 3F3 + 4F4 + · · · + (a2 - l)Var_v
Theorem 6 follows from the fact that the right sides of A.12) and
A.13) are equal to even numbers.
From A.10), A.12), and A.13),
A.14) (a) 3a2 < 2al5 (b) 3a0 < 2al5
where the equality sign prevails in A.14a) if and only if all faces are
triangular, and in A.14b) if and only if each vertex belongs to exactly
three edges.
Substituting from A.11) into A.14), one finds
(a) 3?? - 3oc0 + 6 < 2al5
A.15)
(b) 3ax - 3a2 + 6 < 2a!
or
(a) 04 + 6 < 3a0,
A.16)
(b) ax + 6 < 3a2.
The equality sign in A.16a) holds if and only if all faces are triangular;
that in A.16b), if and only if each vertex belongs to exactly three
edges.
From A.16a) and A.14b),
A.17) ??? + 2 < a0 < §ai.
If ax = 7, this would imply the absurdity
A.18) Ц < a0 < 4f.
To complete the proof of Theorem 3, one would have to establish that
ocx = 6, 8, 9, . . . are all possible, for which see Vorlesungen liber die
Theorie der Polyeder by Ernst Steinitz (Springer, 1934), edited by
Hans Rademacher.
Next, let Eqs. A.10) be multiplied on both sides by 6 and, from the
result, let the respective Eqs. A.13) and A.12) be subtracted. The
resulting equations imply
A19) W 6?0-2?1<373 + 274+75,
(b) 6a2 - 2ax < 3i^3 + 2i^4 + F5.
From A.16a), 6a0 - 2ai > 12. Hence, from A.19a),
A.20) 12 < 3F3 + 2F4 + Vb < 3(F3 + F4 + F5),
and A.5b) follows. The other part of Theorem 4 similarly follows with
the aid of A.16b) and A.19b).
In relations A.19), both equality signs prevail if and only if
Vj = Fi = 0 (j > 6). This condition is implied by the hypotheses of

14 INTRODUCTORY TOPOLOGY [Ch. I
the first part of Theorem 7. Since the only non-zero terms on the
right side of A.19) are then Vz and F5, it follows that 6a2 — 2ax = Fъ.
Since a0 = Vz has been seen to imply the equality sign in A.16b),
the latter becomes ax + 6 = 3a2. Accordingly, A.7) follows and, by
a dual argument, so does A.8).
Next, let the quantity 4(oc0 + a2 — ax) be figured by A) multiplying
Eqs. A.10) each by 4 and adding the results, so as to get an expression
for 4(a0 + a2), B) adding Eqs. A.12) and A.13) so as to obtain an
expression for 4al5 and C) subtracting the latter expression from the
former. This gives
(\Щ 4(?0 + ?2 - ??) = (Fs + Vs) - (F5 + Vb) - 2(FQ + Vb) ,
which, with Euler's theorem, implies Theorem 5.
It remains only to establish Theorem 8. From A.9), A.12), and
A.13), 2ax = iV{ = jFj. Hence, again using A.9),
Y i — . '
г
J
and Euler's relation A.11) becomes, in terms of V(,
A.23) (i+!_07( = 2.
Relations A.5) and A.9) yield г = 3, 4, or 5 and j = 3, 4, or 5.
By A.6), either i = 3 or j = 3. If г = j = 3, then, by A.23), F3 = 4
and, dually, Fz = 4, so that the tetrahedral type is implied. If г =3
and j = 4, then, by A.23), Vz = 8 and hence, by A.22), F^ = 6.
Dually, if i = 4 and j = 3, then Fs = 8 and F4 = 6. It is readily
verified that these possibilities correspond to the cubic and octahedral
types. If i = 3 and j = 5, then A.23) and A.22) yield Vs = 20,
Fb = 12. Dually, if i = 5 and j = 3, then Fs = 20, Vb = 12. The
reader should verify that these possibilities imply the dodecahedral
and icosahedral types.
The proofs of Theorems 3 to 8 are now complete. Most of these
results date back to Euler, but the present proofs have been adapted
from more recent sources.
EXERCISES
4. Show that a pyramid with a polygonal base is self-dual.
5. Draw a figure showing (a) a truncated pyramid with a triangular base and
(b) a solid having a dual boundary structure.
A.22)
a2 =

2
Topological Classification
of Surfaces
We continue in the intuitive spirit of Chapter 1, reserving precise
definitions and proofs for later chapters.
2-1. Polygonal Regions with Matched Edges
A surface is an idealization of such an object as a sheet of rubber.
Topology includes the study of properties preserved when a surface is
deformed in various ways, without being torn. A surface may have
edges, like a rubber disk or a disk with a hole in it, in which case the
edges constitute its boundary. If, like a sphere or the surface of a
doughnut, it has no edges, we refer to it as closed.
From the viewpoint of combinatorial topology, surfaces are
investigated by imagining them cut into little patches, "curvilinear
triangles" for example, and considering how these patches are fitted
together.
(A) Accordingly, let us regard a surface ? as a collection of
triangular regions (?1? . . . , ??), with some edges matched (or identified)
in pairs. If an edge A of тг- is identified with an edge of ?,·, we refer to
ri and Tj as adjacent across A. A surface is required to be connected,
in the sense that it is possible to pass from any triangle a of ? to any
other triangle ? by a sequence of triangles (аг = ?, ?2, - - - , 0m = ?,
each belonging to M, where ?{ and ai+1 are adjacent (i = 1, ...,m — 1).
If there are free_edges, that is, edges not matched with other edges,
they constitute the boundary of M. If there are none, ? is closed.
Given (t1? . . . , ??) and a specification of the identifications of
paired edges, we are led to the problem of deforming the regions тг so
as to bring the matched edges into coincidence, thus piecing (т1? . . ., ??)
together into a continuous surface M, as though we had a topological
jigsaw puzzle with deformable pieces. See, for example, Fig. 2-1,
where identified edges are denoted by identical letters and are supplied
with arrows to specify their matching.
15

16
INTRODUCTORY TOPOLOGY
[Ch. 2
When two directed edges are matched, their initial vertices, also
their terminal vertices, are necessarily identified. Thus the twelve
vertices in Fig. 2-la reduce to the four vertices of the tetrahedron in
Fig. 2-lb. Figure 2-2 illustrates a surface with boundary.
a'
b
(b)
Fig. 2-1. A closed surface as a set of matched triangles, (a) The triangles;
(b) the surface assembled into a tetrahedron.
Topologically, the surfaces of a tetrahedron and a sphere are
equivalent. Figure 2-3 shows a dissection of the spherical surface,
topologically equivalent to Fig. 2-lb. Figure 2-4 shows two other
representations of a spherical surface.

Art. 2-1]
CLASSIFICATION OF SURFACES
17
(a)
(b)
Fig.2-2. A surface with boundary, (a) The triangles; (b) the assembled surface.
An edge and its end points are said to be incident with one
another. So are a triangle and its edges or vertices.
Up to a certain point, the piecing
together of (?? ?2, . . . , ??) can be
carried out in the plane as follows.
Let т{ be a triangle having an edge in
common with ?? Then -^and тг can be
brought together and amalgamated
into a quadrilateral region ти along
edge A. See Fig. 2-5, where т± and
т2 of Fig. 2-1 are thus amalgamated,
т2 being first "turned over" and
perhaps changed in size and shape.
Next let rj be a triangle having an
edge В in common with ти. Then ?,·
andrb. can be amalgamated across В Fig 2-3. A representation equi-
to obtain a pentagonal region тш. valent to Fig. 2-lb.

18 INTRODUCTORY TOPOLOGY [Ch. 2
? I I
(b)
Fig. 2-4. Two other representations of the sphere.
Continuing thus, step by step, one can fuse all the regions (rv
?2, . . . , ??) into a single (n + 2)-edged polygonal region, some of
whose outer edges would, in general, have to be fitted together in
pairs to obtain a model of the surface. In the case of Fig. 2-5, where
? = 4, a hexagonal region is obtained. This suggests the following
definition, equivalent to (^4).
(B) A surface or 2-manifold Ж is a plane region bounded by a
single polygon, some of whose edges are identified in pairs according
to prescribed relative directions.
To illustrate (B), it suffices to draw a polygonal region, assign
directions to all its edges, then assign a symbol to each edge, no
symbol being used more than twice. It is easily seen that the same
surface admits many different polygonal representations. For
example, note that the octagonal representation in Fig. 2-6 is
equivalent to the hexagonal representation in Fig. 2-7b, since the paired

Art. 2-1]
CLASSIFICATION OF SURFACES
19
edges ? can be eliminated as suggested by the intermediate Fig.
2-7a.
(C) A convenient symbol for a polygonal representation of a closed
surface is obtained as follows: A) Select a positive sense for the
polygon (clockwise or counterclockwise). B) Select any edge as "first
edge" of the symbol. C) Write the edges in cyclic order, starting with
the first edge and proceeding in the positive sense around the polygon,
c4>;<
r12 shaded
Fig. 2-5. Amalgamation of the triangles of Fig. 2-1 a into a polygonal region.
supplying each edge with the superscript +1 if its sense agrees with
the positive sense of the polygon and — 1 otherwise. The superscript
+1 will generally be omitted in writing
symbols. Figures 2-5 to 2-7 can be thus C^^-*^ A
symbolized as follows:
B.1
(a) GG^F^E^EF
(Fig. 2-5),
(b) АВАС^ВгЮ^ЕЕ-1
(Figs. 2-6 and 2-7a),
(c) АВАС^ВгЮ-1
(Fig. 2-7b).
(D) Any string of letters in which
no letter appears more than twice and
in which the superscripts 1 and —1
are arbitrarily distributed represents a
surface and will be called a polygonal
symbol for it. Each letter in a symbol will be called an edge and will
be described as free or paired according as it appears just once or
twice. Every surface has infinitely many polygonal representations.
Fig. 2-6. An octagonal
polygonal representation.

20
INTRODUCTORY TOPOLOGY
[Ch. 2
(E) Some obvious ways of changing a symbol without changing the
surface represented are as follows:
A) On a free edge, changing the superscript from 1 to —1 or from —1
to 1. On a paired edge, simultaneously changing the superscript on
both appearances of the edge. These changes clearly do not affect
the matching of the edges of the polygon.
B) Writing a symbol in reverse order, changing all or none of the
superscripts.
C) Moving a block of letters from the end of a symbol to the beginning.
A „ ^\ A
Bk
+B
(a)
(b)
Fig. 2-7. Two equivalent representations.
We will admit two-edged regions, as suggested by AA and АА~г
in Figs. 2-8a and 2-8b, and even one-edged regions, as symbolized by
A and shown in Fig. 2-8c.
(F) Making use of Part A) of (E), we will generally write a polygonal
symbol so that the suppressed superscript 1 appears on the first
occurrence of each edge. Thus the symbols B.1) are altered as follows.
(a) C(r1F-1E-1EF -> CCr1FEEr1F-19
B.2) (b) ABAC^B-^^EE-1 -> AВАСВтЮЕЕ^,
(с) А ВАСГ^ЧГ1 -> А ВАСВ-Ю.
EXERCISE
1. Draw eight triangles and indicate, after the fashion of Fig. 2-1, a matching
of their edges corresponding to the surface of an octahedron.
2-2. Some Elementary Surfaces
Some examples of familiar surfaces are now given.
(A) The Disk A: The only one-edged symbol A represents the disk,
which is also represented by any symbol made up entirely of free edges.

Art. 2-2]
CLASSIFICATION OF SURFACES
21
(C)
Fig. 2-8. Two-edged and one-edged polygons.
(B) The Sphere AA~X: Given the 2-edged symbol AA~X, one can
imagine a zipper along the matched edges, whereby the surface can be
closed and then deformed into a spherical shape (Fig. 2-9).
(C) The Projective Plane or Sphere with Grosscap CC: Let the
disk-like surface CG be modified into the form of a sphere with a slit in
it. Then let a part of the sphere containing the slit be blown up into a
sort of blister (Fig. 2-10a). In Fig. 2-10b, the blister is "pinched in"
to bring the matched end points с of the slit close together and to
bring the entire slit close to the vertical line. Finally (Fig. 2-10c) the

(b) (с)
Fig. 2-9. Equivalence of М-1 to a sphere.
edges cpm are brought into coincidence along the vertical line, and so
are the edges cqm. This creates a self-penetration, not regarded as a
self-intersection of the surface. As a horizontal plane moves down
from m, its intersections with the surface, called contour lines, resemble
a figure 8 until с is reached (Fig. 2-10d). Below c, the self-intersection
is lost and the contour lines gradually become circular. The part of
the sphere above К is a crosscap. The equivalence of the sphere with
crosscap to a projective plane is discussed later.
(D) The Moebius Strip K1BK2B or Crosscap KGG: If the crosscap
of Fig. 2-10c is cut off around the circle K, slit open along the segment
C, and flattened out in the plane, the representation suggested by Fig.
2-1 la is obtained. In this figure, К is divided by two points, b and a,
into the arcs K± and K2. Let cuts be made along the dotted lines Bx
and В2, separating the surface into two parts, I and II. In Fig. 2-1 lb,
the separate parts are shown deformed into rectangles. In Fig. 2-1 lc,
rectangle I is turned over to permit a matching with II along С After
this matching, Bx and B2 are amalgamated into a single edge B9
which leads to the symbol K^1BK2B, or, by a change in notation,

Art. 2-2]
CLASSIFICATION OF SURFACES
23
m m
(a)
(b)
(c)
(d)
Fig. 2-10. Representation of CC.

24
INTRODUCTORY TOPOLOGY
[Ch. 2
cq <4 ад
oq
oq
-> ? ? <:
k!\k
г, ^ о /
- °t ? я 4х«
сч .?
0Q ад
ol·1-
*0 (?
-О
0)
о
?
? О
О)
г4

Art. 2-2] CLASSIFICATION OF SURFACES 25
(b)
Fig. 2-12. The torus.
(a) (b)
Fig. 2-13. The handle.
KXBK2B. In Fig. 2-1 Id, the rectangle of Fig. 2-1 lc has its edges В
matched, to produce the Moebius strip, named for the German
mathematician G. F. Moebius A858). To obtain another
representation, let the diagonal cut A (Fig. 2-1 lc) separate the rectangle into
triangular pieces III and IV (Fig. 2-1 le). Then let III be turned over,
so as to permit a matching with IV along B, which leads to the
representation KAA (Fig. 2-1 If). A change of notation yields KCC.
(E) The Torus ABA^B: This, the simplest to visualize of the
closed surfaces requiring a four-edged symbol, is shown in Fig. 2-12a
and, assembled, in 2-12b.
(F) The Handle ????-??~?: This surface (Fig. 2-13) is equivalent
to a torus with a hole in it.

26
INTRODUCTORY TOPOLOGY
[Ch. 2
EXERCISES
2. Give a matching of edges of two triangles such that they fit together to
cover the plane band bounded by two concentric circles.
3. Let (tv . . . , ??) be a set of triangles, some of whose edges are matched in
pairs, and whose vertices are identified only as implied by the matching of edges,
(a) Show that all the triangles incident with a given vertex ? can be arranged in
a cyclic order or a linear order (with a first and a last element) so that consecutive
triangles are adjacent, (b) Show that the linear order results if and only if
? is incident with some free edge; that is, an edge not matched with any
other.
4. State and prove the dual of Exercise 3, insofar as it holds.
5. Under the hypotheses of Exercise 3, with the connectedness condition of
Art. 2-l(A), show that the edges and vertices of (rv . . . , ??), as identified,
constitute a connected linear graph.
6. Show that the boundary of a surface, as defined in Art. 2-1 (A), consists of
a number of circuits, no two of which have a vertex in common. Suggestion:
First do Exercise 3 above.
7. Let the hypotheses of Exercise 3 be modified by making one identification
? = q of vertices not joined by an edge, in addition to the induced identifications.
What statements can be made analogous to (a) and (b) in Exercise 3 with respect
to the resulting vertex? , ^cju^ ьЛ> ла^Аз^™ % ^ ллу+ЛвЛ <ujl^ )
8. Show (a) that each 2-edged symbol/represents a disk or a sphere or a sphere
with erosseap; (b) that each 3-edged symbol\represents a Moebius strip or a
9. Show that each 4-edged symbol is equivalent to one of the following seven
possibilities: (a) the disk, (b) the Moebius strip, (c) the plane band between two
concentric circles, (d) the sphere, (e) the torus, (f) the form CCDD, (g) the sphere
with erosseap. The surface CCDD is considered in more detail in Exercise 18
below.
10. Show that ABA^B^KDK'1 is equivalent to the handle KABA'^Br1.
2-3. Orientability and Non-Orientability
A model of the Moebius strip is made by pasting together the ends
of a narrow strip of paper after putting one twist in it (Figs. 2-1 led
and 2-14). Let a small circle with an assigned positive sense (called an
oriented circle) be slid once around the strip. When it returns to its
original position, its orientation is reversed (see positions 4 and 1 in
Fig. 2-14). The Moebius strip is accordingly described as non-
orientable.
Now imagine a normal vector N from the center of the moving
circle, so directed that the orientation is clockwise as viewed from the
tip of N. When the moving circle returns to its initial position, the
sense of N is reversed. The Moebius strip is accordingly said to be
one-sided (in three-dimensional space).

CLASSIFICATION OF SURFACES
27
Art. 2-3]
(^4) In general, a surface ? is non-orientable or orientable
according as there does or does not exist on it a closed curve G such that
a small oriented circle slid once around G returns to its original
position with orientation reversed. <One-sidedness in relation to
orientability will be discussed later. > ^JU-bJ^ }
(b)
Fig. 2-14. Non-orientability and one-sidedness of the Moebius strip.
Theorem 1. A polygonal symbol represents a non-orientable surface
if it contains two paired edges with equal superscripts.
Proof. A symbol with two paired edges having equal superscripts
can be put in the form
B.3)
·= ...A...A.
Just as in the case of the Moebius strip, a path cutting across the
polygonal region from the midpoint of one copy of A to that of the
other has the non-orientability property of Definition (^4).

(e)
(d)
Fig. 2-15. Reduction of HC) to standard form, (a) HC); (b) HC) as a torus with two handles; (c) HC) as A&A^Bf1
with two handles; (d) separation along A2 with handle shrunk back; (e) separation along B2; (f) preparation to eliminate
th\rd handle.

Art. 2-4]
CLASSIFICATION OF SURFACES
29
2-4. Standard Form for Spheres with Contours and Handles or
Crosscaps
The discussion in Arts. 2-4 and 2-5 is based on some results due to
H. R. Brahana.* His work has been reproduced, with modifications,
several times in the literature ([K2], Ch. II, §§6-7; [S-T], §§37-39,
for example).
From a sphere S, let a number h > 0 of pieces be removed by
circular cuts, and let each of the resulting holes be capped with a
handle (Fig. 2-13b) to obtain a sphere with h handles, which we will
denote with H(h). - Ь е^
Lemma 1. The surface H(h) can be represented by the symbol
B.4) Hft = АхВхА?В?АгВгА?В? . . . ????????.
To establish this result, intuitively, we first interpret H(h) as a
torus with h — 1 handles. Figures 2-15a, b illustrates the case h = 3.
Cuts ?? and Bv analogous to the cuts A and В of Fig. 2-12b, serve to
reduce H(h) to a rectangular region A1B1A{1B^i with h — 1 handles
(Fig. 2-15bc).
The h — 1 handles on the region A1B1A{1B{1 can now be
eliminated one at a time, each by a pair of cuts. Figure 2-15cde shows
the elimination of one of these handles, leading to a representation of
H(h) as an octagonal region A1B1A^1B^1A2B2A21B^1 with h — 2
handles. A repetition of such steps leads eventually to B.4).
Now let q > 0 pieces be removed from the sphere S and let each of
the resulting holes be capped with a crosscap to obtain a sphere with
q crosscaps, C(q) (Figs. 2-16, 2-17).
Lemma 2. The surface C(q) can be represented by the symbol
B-5) Ce = GАВД · · · OqC9.
As in the case of the previous lemma, we outline an intuitive
argument. We commence by showing that the cut Cx indicated in
Fig. 2-16 permits a representation of GA) as G1G1. First make a cut О
through both sheets along the self-penetration cm of the surface; then,
without making the cut Cl5 proceed in reverse order through the stages
illustrated in Fig. 2-10abc. This permits a flattening of the surface
into the disk of Fig. 2-16b, where Сг appears as a diameter, divided by
the points о and с into two segments C[, G'[. In Fig. 2-16c, the cut Gx
has been made, separating the disk into parts I and II. In Fig. 2-16d,
part I has been turned over and the two parts stuck together along C.
* *'Systems of circuits on two-dimensional manifolds," Annals of Mathematics 23
A921), pp. 144-168.

30 INTRODUCTORY TOPOLOGY [Ch. 2
m c
(a)
(b)
c; l c;
?1° ?i'
II
(c)
(d)
(e)
Fig. 2-16. Another reduction of the sphere with a crosscap.

(a)
(b)
С ?
c3l·
^c2
(с)
Fig. 2-17. Reduction ofCC) to standard form. (a)CC); (b) CC) as CXCX with
two crosscaps; (c) CC) as C^C^C^ with one crosscap; (d) CC) = CjCjC^CgC^
31

32 INTRODUCTORY TOPOLOGY [Ch. 2
In Fig. 2-16e, the point с and the line С have been eliminated, and the
arcs G[, G'[ combined into Cv
In the case q > 1, a procedure like the foregoing reduces G(q) to a
disk C1C1 with q — 1 crosscaps. Analogous steps, partly illustrated in
(b)
Fig. 2-18. Surfaces with contours and handles or crosscaps. (a) HB, 3);
(b)CC,2).
Fig. 2-17, eliminate the remaining crosscaps one at a time and lead
to B.5).
Let H(h) be modified by the removal of r disk-like pieces, leading to
H(h, r), a sphere with h handles and r contours (Fig. 2-18). Let G(q)
be similarly modified, leading to C(q, r), a sphere with q crosscaps and
r contours (Fig. 2-18).

Art. 2-5] CLASSIFICATION OF SURFACES 33
Lemma 3. The surfaces H(h, r) and C(q, r) can be represented by
B.6) (a) H(A, r) = НЛКГ, (b) €(q, r) = C,Kr,
where
B.7) Kr = ?,?,?^?,?,?^ . . . KrDrK~\
To verify Lemma 3 for H(h, r), let the procedure of Lemma 1 be
carried through so that the cuts avoid the holes. Thus H(h, r) can be
reduced to the polygonal region symbolized by Hh with r circular
pieces removed. Figure 2-18 suggests a method of making cuts which
lead to B.6).
(^4) The methods used in establishing Lemmas 1, 2, and 3 can be
summarized as follows: A) After 2h + r cuts, two for each handle
and one for each contour (Fig. 2-18a), it is possible to deform H(h, r)
into a polygonal region with matched edges symbolized by НЛКГ.
B) After q + r cuts, one for each crosscap and one for each contour
(Fig. 2-18b), it is possible to deform C(q, r) into a polygonal region
with matched edges symbolized by C0Kr. The cuts just described
afford canonical dissections of H(h, r) and C(q, r).
EXERCISES
11. Make a paper model of a Moebius strip. Cut it in two along the center of
the strip and describe the results in terms of connectedness and orientability of
the resulting model.
-12. Proceed as in Exercise 11, but make the cut along a line one-third the
way across the strip.
2-5. A Classification Theorem
The converse of Theorem 1 is also true; namely, that a polygonal
symbol represents an orientable surface if it contains no paired edges
with equal superscripts. We omit the proof.
Theorem 2. (a) An orientable surface can be represented as a
sphere with h handles and r contours for some h > 0, r > 0. (b) A
non-orientable surface can be represented as a sphere with q crosscaps
and r contours for some q > 0, r > 0.
Proof. (A) By Lemma 3, the theorem can be proved by showing
that a polygonal symbol ? for a surface ? can be reduced to the form
НЛКГ (h > 0, r > 0) if ? is orientable and to the form CtfKr (q > 0,
r > 0) if ? is non-orientable. The reductions will be by a sequence of
steps, each of which converts a symbol into an equivalent symbol.

34
INTRODUCTORY TOPOLOGY
?
[Ch.2
(a)
(b)
a
7 у
/
<
?
^
Hh
(c)
Fig. 2-19. Bringing together the elements of a handle.

Art. 2-5] CLASSIFICATION OF SURFACES 35
(B) An elementary reduction of ? means any one of the following
changes: A) the replacement of a sequence of free edges by a single
free edge; B) if two edges Ex and E2 appear only in the forms EXE2
and E%1E^1, the simultaneous replacement of ???2 by a single edge ?
and E^XE^X by E~x\ C) the elimination of a pair of successive edges
??~?, provided this does not eliminate the entire symbol. An
elementary reduction does not change the surface represented by a symbol.
(C) If 77 contains no paired edges, it represents a disk, which is
equivalent to Я@, 1), the sphere with no handles and one contour.
Hereafter, we assume that ? has at least one pair of identified edges.
By Theorem 1, the orientable case is characterized by the property
that each paired edge in ? appears once with superscript 1 and once
with superscript —1. We follow the convention of Art. 2-1 (F).
(D) The symbol ? will be said to contain a handle if it is of the form
B.8) 77 = oLEfiFyBT^F-h
where the small Greek letters represent sequences of edges. If ?, у, ?
are all vacuous, then 77 contains the handle block EFE^F'1. Thus Нл
in B.4) is a sequence of h handle blocks. The identifications of edges in
Нл induce an identification of the end points of Ai and Вг (г = 1, . . . , h)
into a single vertex 0, to be called the vertex of Нл.
Lemma 4. In the orientable case, there exists a symbol ??, equivalent
to 77, of the form
B.9) 77, = H^ (A > 0)
where the vertex 0 of Нл is not on a free edge and where 77^ contains no
handles.
Proof of Lemma. In the case h = 0, H0 is to be interpreted as
vacuous, but a vertex 0 of 770 will be assigned to it.
Inductive Hypothesis. For some к > 0, there exists a symbol
ттк = Hfc77^ equivalent to 77, where the vertex of H7c is not on a free edge.
If 77 has a vertex belonging to paired edges only, let such a vertex
be regarded as the vertex of the vacuous set of edges H0. If not, let a
paired edge be divided into two edges by the introduction of its
midpoint 0 as a new vertex, and let 770 be the modified symbol, with 0
as vertex of H0. This verifies the hypothesis for к = 0.
If 77^ contains no handle, let к = h and the lemma is proved. If 77^
contains a handle, then nk is of the form
> B·10) 77fc = ?^?????^??^?.
Figure 2-19 shows how to cut and reassemble the polygonal region
bounded by 77? into a region with the symbol

36
INTRODUCTORY TOPOLOGY
[Ch. 2
where n'k+1 = ?????. Note that A = ???2 and ? = ???2, using the
elementary reduction B) of (B) above.
The symbol ък+1 fulfills the hypothesis with к + 1 in place of k,
and nk+1 has four fewer edges than nk. Repetitions of this procedure
must therefore lead to a situation where all the handles are in the
initial sequence of handle blocks, as required by the lemma.
We postpone the discussion of тт'к until after a partial treatment of
the non-orientable case.
ay
(a)
(b)
Fig. 2-20. Conversion of ЬЬосСС? into ???^^???.
Lemma 5. The symbols ???? and ?????? are equivalent, where
a-1 is the block of edges a in reverse order with all superscripts changed.
A proof is suggested by Fig. 2-1 lef, in which Kx and K2 are to be
replaced by the sequences of edges ? and ?. The use of В in both
symbols of the lemma amounts to a reassignment of notation.
Lemma 6. The symbols ??&??? and ????&? are equivalent.
To show this, first convert ?????? into BarW^BfiC as suggested
by Fig. 2-20, with reassignment of notation after the procedure. Then
apply Lemma 5, with a-1C-1 replacing ? and ?? replacing ?, to obtain
??????, equivalent to ?????? by Art. 2-1(^3). Again apply
Lemma 5, with BB playing the role of a, to obtain ССВ^В'1^.
Finally, reassign notation, replacing С by В and В~г by G.
Lemma 7. In the non-orientable case, ? is equivalent to a symbol of
the form
B.12) pQ = Vqp'q for some q > 0
where (a) the vertex of Cq is not on a free edge, (b) p'q does not contain
two identified edges with equal exponents, and (c) pq does not contain
a handle.
Proof of Lemma. The concept of the vertex of Cq is like that for He.

Art. 2-5] CLASSIFICATION OF SURFACES 37
Inductive Hypothesis. For some к > 1, there exists a symbol
pk = Ckpjc equivalent to ?, where the vertex of Ck is not on a free edge.
In the non-orientable case, ? can be put in the form ?&??, with
the aid of Art. 2-1 (E). We verify the hypothesis for the case к = 1
by Lemma 5, where p[ = a-1/? and the cut of Fig. 2-1 le is made from
a vertex not on a free edge. If no such vertex exists in ?, let a paired
edge be subdivided as in the proof of Lemma 4.
If p'k contains paired edges with equal exponents, pk is of the form
B.13) pk = CkyB*Bd.
By Art. 2-1 (E3) and Lemma 5 with ? = 6Cky, рк is equivalent to
СьуВВ&^д. By Lemma 6, with ? playing the role of a, the latter is
equivalent to
B.14) Pk+1 = CkBBYar4 = Ck+lP'k+1
where pk+1 = yoc-1E, which contains two fewer edges than pk, and the
hypothesis is verified with к + 1 replacing k.
If p'k contains a handle, let pk be written in the form
B.15) pk = C^CCaEfiFYE-^F-^.
By Lemma 5, with ?,??? playing the role of a, pk is equivalent to
C^CF-^^E-^CyE^dF-h.
Three additional applications of Lemma 5 lead to the following
sequence of equivalent symbols:
C^CF-^E^E-^C^oidF-h (letting В = Я),
Cl^1CF-1F-1d-1oT1CYEEfie (letting В = i^),
?^??????????? (letting В = С).
Finally, two applications of Lemma 6 lead to
B.16) Pk+2 = C^CCFFEEatfryfie
where pk+2 has four fewer edges than pk.
The final part of the proof of Lemma 7 is like that of Lemma 4.
To complete a proof of Theorem 2, we reduce the ?'? or p'q of Lemma
& or 7 to the form Kr (see B.7)). If Tr'h or p'q is vacuous, we have r = 0
and the proof is complete. We assume the case where n'h is non-
vacuous. The arguments for 77^ and p'q differ only in notation.
Lemma 8. The symbols ?????^? and ????^?? are equivalent.
A proof is suggested by Fig. 2-21, with reassignment of notation
fter the process.

38 INTRODUCTORY TOPOLOGY [Ch. 2
(a) (b)
Fig. 2-21. Shifting the position of a contour in a symbol.
Let all possible elementary reductions be applied. If this eliminates
77д entirely, then r = 0 and we are again through.
Lemma 9. If 7Гд is non-vacuous and admits no elementary reductions,
it contains a sequence of edges of the form KDK-1.
Proof of Lemma. Since тт'к is free from paired edges with equal
exponents, and since the vertex of Нл is not on a free edge, ?'? is of the
form ?????7-1)/. Since no elementary reductions are possible, ? is non-
vacuous. If ? contains a copy G of a paired edge, ? must also contain
С?-1, since 77д contains no handles. Hence ? contains a sequence of
edges of the form 6r/?06r_1, and ?0 consists of fewer edges than ?.
Exercises 15 and 16 below will complete the proofs of Lemma 9 and
Theorem 2.
(E) Later in this book we will show that two surfaces are
topologically equivalent if and only if A) they are both orientable or
both non-orientable and B) their representations as H(h, r) or C(q, r)
agree in contour numbers and handle or crosscap numbers. This means
that a surface is characterized topologically by its orientability class, its
contour number r, and its number of handles or crosscaps, h or q.
EXERCISES
13. List the forms through which the symbol CBB~1ADA~1C~1D~1 would pass
in carrying out the steps in the proof of Theorem 2.
14. Do the same for the symbol ABC В D AC'1.
15. Complete the proof of Lemma 9.
16. With the aid of Lemmas 8 and 9, show that тт'п and p'Q can be reduced to
the forms Kr without affecting Hh or C0, thus completing the proof of Theorem 2.
17. (a) Show that the polygonal symbol FAECGBHDC^JDB-^-1 is
equivalent to GC, 2). (b) Replace the second D by D_1 and find an equivalent
symbol H(h, r) or C(q, r).
18. Figure 2-22 represents a Klein bottle, named after the German
mathematician Felix Klein A849-1925). Show that it can be represented by either

Art. 2-5]
CLASSIFICATION OF SURFACES
39
(b)
Fig. 2-22. The Klein bottle, or sphere with a twisted handle.
Fig. 2-23. A sphere with two twisted handles.

40 INTRODUCTORY TOPOLOGY [Ch. 2
ABA~XB or CCDD. If, in Fig. 2-22a, one of the circles С were reversed, the
resulting surface could be closed to a torus. As matters stand, we can close it to a
Klein bottle if it is made of a topological material which, in addition to deform-
ability, has the property of self-penetrability. One end of Fig. 2-22a is to be
pushed through the wall of the tubular part, suitably blown up, and then brought
along the "interior" to be sealed with its mate according to the specified
orientations.
19. Give a polygonal symbol for the surface of Fig. 2-23. Note that a smart
or lucky insect "trapped inside" a sphere with at least one twisted handle can
escape through the latter. Hence the surface is one-sided. Show that it is also
non-orientable, by considering what would happen to a watch (hence to
"clockwise" orientations) if the insect, assumed strong as well as intelligent or fortunate,
should slide it along from "inside" to "outside" the sphere, returning it to the
point of departure.
20. Show that an inner tube with a hole in it can be "turned inside out," if
made of a deformable but not self-penetrable substance.
21. Show that a Klein bottle is equivalent to two Moebius strips with identified
edges.

3
Introduction to
Set-Theoretic Topology
We next develop general definitions of space, continuity, and other
concepts fundamental to topology.
3-1. Sets and Mappings
We take as basic the idea of a set S of objects, called the elements
of S. Sets are also referred to as aggregates, classes, and collections.
The statement that s is an element of S is symbolized by s e S. To
illustrate a standard notation often used in defining a set, let ? be the
set of all living people, and let ? be a certain room. Then the set of
all people in room ? is denoted by
C.1) S={peP\pisin p}
or merely, if it has been made clear that ? denotes a person, by
C.2) S = {p | ? is in p}.
Such notation as S = {s} is also used, if S has already been defined
or if its defining condition is separately stated, instead of being
written within the braces after a vertical bar.
If the room ? is empty, then C.2) defines the vacuous (or empty,
void, or null) set, which means the set containing no elements,
conventionally denoted by 0.
A slanting line through a symbol for a relation negates that
relation. Thus ? $S means "p does not belong to #."
(A) Any two sets, S and ?7, determine their
union: SuT = {s\seS or seT, or s eS and s ? T}y
intersection: #nT = {s|se# and s e T},
difference: S — ? = {s \ s e S and s ? ?}.
The symbol => stands for implies, <= for is implied by and <=> for
implies and is implied by or if and only if. We will sometimes use o
?? giving definitions as well as in stating results.
41

42
INTRODUCTORY TOPOLOGY
[Ch. 3
(B) We thus define the following terms and symbols:
A) S equals T, S = T: s eSos e T\
B) ? is a subset of T, S с ? or ? з S: seS=>seT;
C) ? is a proper subset of T,S (? ? or ? Ш) S: S с ? and ? - ? ^ 0;
D) ? and ? are disjoint: ? ? ? = 0;
E) i? and S are complements of each other in T: R KJ S = ? and
R nS = 0.
A set consisting of just one element s is called singleton s and is
often denoted by (s). It differs logically from s, just as a person is
different from a committee of one. A set with a finite number of
elements, for example the numbers 1 and 2, will sometimes be denoted
by the elements enclosed in parentheses, thus: S = A,2).
The elements of a set may themselves be sets; for example, the set
consisting of the 4060 possible committees of three from a class of
thirty people, or the set {A} of all subsets of a set S. In the latter
case, if S = A, 2), then
C.3) {A} =@,A), B), A,2)).
(C) The definitions and notation for union and intersection in (A)
above are generalized as follows. If {$} is a collection of sets, then the
union of the sets of {$}, denoted by u{$}, means the set of all objects,
each belonging to at least one of the sets of {$}; and the intersection of
the sets of {$}, denoted by n{$}, means the set of all objects each
belonging to all of the sets of {$}.
Consider two sets, S and ?7, and a subset ? of S. A mapping or
function / from ? to ? (or into T), symbolized by / : ? -> ?, is a
correspondence which associates one and only one element of ? with
each element of M. One writes у = f(x) or / : ? -> у to mean that*
у e ? is thus associated with ? e M, and one calls у the image or map
of ? or the value of / at x. The set ? is the domain of/. Its range is
the set N of all elements of ? each of which is the image of an element
of M. If N = T, f is said to map ? onto T. The preposition into is
consistent with N = ? and with Лт (с Т.
(D) A function^ with domain ?? is an extension of/with domain
? if A) Mx 3 ? and B) ? eM=>f(x) = fx(x). Under these
conditions, / is the restriction of/x to Ж, symbolized by / = /x | M.
(E) Let A be a subset of the domain Ж of a mapping / : ? -> Т.
It is conventional to denote with f{A) the set of all image points of
points of A. Symbolically,
C.4) f(A) = {y\y=f(x) for some ? g A}.
* Неге е means "which is an element of." Similar use will be made of other
relationship symbols.

Art. 3-1] [INTRODUCTION TO SET-THEORETIC TOPOLOGY 43
This notation can be interpreted as representing an extension of/ to
the domain ? KJ {A} where {^4} is the set of all subsets of M.
(F) Given a mapping / with domain A and a mapping g with
domain В = f(A), the product or the composition of/ and g means
the mapping h with domain A and range g(B) defined by
C.5) h(a) = g(f(a)) for each ae A.
The formulations
v Ца) = gf(a)
are also used. Though h represents / followed by g, it is conventional
to write g before/, this being the natural order in C.5).
(G) The mapping / : A -> В is called a one-to-one mapping of A
onto В if each element of В is the image of one and only one element
of A. There then exists an inverse mapping/-1 : В -> A defined by
C.7) f-i(b) = aob=f(a).
(H) A more general concept of an inverse function is defined as
follows: Let/have domain A <= ? and range В <= N. If X <= M, let
C.8) f(X) = {y g N | у = f(x) for some ? e X}.
Note that this extends/even more than does (E), since X need not be
a subset of A. If X <= ? - A, then f(X) = 0. Then, for any
? с ?, let
C.9) ?\?) = {xeM\f(x) = yior some ye ?}.
J iaM Jv^b, ^JtX·,
This defines an inverse mapping /_1/\ \f у ef(A) and у =f(x), then
a; is called an inverse image of y. The set of all inverse images of у is
the image/_1((i/)) of singleton у under/-1. It is frequently denoted
Ьу/-%).
(/) A one-to-one correspondence Г between two sets ? and N is a
collection ? = {(?, у)} of pairs (?, у) of associated elements, where
A) ? g ? and у g N and B) each element of ? appears in exactly one
pair and so does each element of Щ Given ?, let one-to-one mappings
/ : ? -+N and g : ? -> ? be defined by
(ЗЛО) У =/(*)<=> (*, у)еГ,
? = д(у) о (х, у) g Г.
Obviously, / and g are inverse to one another. Conversely, any
one-to-one mapping, or its inverse, defines a one-to-one correspondence.
(J) A set S is finite if it has exactly ? elements for some natural
number* ? e @, 1,2,.. .). Otherwise, it is infinite. It is denumerable
* We presuppose, throughout, a familiarity with certain basic properties of the real
kHumber system.

44
INTRODUCTORY TOPOLOGY
[Ch. 3
or countable if it either is finite or can be put in one-to-one
correspondence with the natural numbers. Its elements can then be
denoted by (sv . . . , sn) or (sl9 s2> . . . , sn, . . .).
EXERCISES
1. Describe geometrically the following sets, given that (x, y) is a rectangular
cartesian coordinate system in the plane: (a) {(x, y) I x2 + y2 < 1} ? {(?, у) |
у > 0}, (b) {(?, у) I ? > 0} ? {(?, у)\у = 0}.
2. An equation у = / (?) defines a mapping / : ? -> ?, where A, on the
a;-axis X, is the domain of definition of/ (x) and where ? is the у-axis. In each
of the following, give the domain and range, and tell whether/ is a mapping onto
? and whether/ is one-to-one. Specify the set/_1A): (а) у =хг,(Ъ)у = V\ -\-?,
(с) у = tan ??.
3. Prove that (a) S = ? <^> S с ? and ? с ?, (b) ? - ? = ? - S ? ? =
(? U ?) - 2\
4. Show that if ? consists of ? elements, then there are exactly 2n elements in
the set of all subsets of S. List them for the case S = (a, b, c, d).
5. With the domain of / extended as in (E) above, show that its range is
N U {B}, where N = f (M) and {B} is the set of all subsets of N. Show that
there is generally more than one subset of ? with a given image B' cz N.
6. Under what conditions are the product mappings gf and fg both defined?
7. Show that the extension of/ in (E) is one-to-one between ? U {A} and
N U {B}, in the notation of Exercise 5, if and only if / : ? -> N is one-to-one,
8. (a) Show that/-1, defined as in (H), is one-to-one between the set {B} of
all subsets of N and/~1({B})A (b) Show that/_1(/ (A)) contains A, generally as
TT~proper subset. (cT WhatTis /_1@)? (d) Show that /-1({iV}) is the set of all
subsets of ? if and only if/ is a one-to-one mapping of ? onto N.
3-2. Relations, Cartesian Products, Functions
We have used several symbols for so-called relations A) between
two sets C)czM))(c)—), B) between an element and a set, ? e M,
C) between two properties (=>, <=,<=>).
A relation ? is a set ?(?) = {(a, b)} of ordered pairs of objects. The
formulations (a, b) e Г(р) and apb are equivalent. Thus ? might stand
for с between sets of elements, or for "is married to" among human
beings.
The domain of ? is the set of all first elements of the pairs ?(?), and
its range is the set of all second elements.
A relation ? is symmetric if apb => bpa, reflexive if ара for each a in
the domain of ? and transitive if apb and bpc together imply ape.
(A) An equivalence relation is one which is reflexive, symmetric,
and transitive. If A is the domain, hence also the range, of such a
relation p, then the equivalence class of an element a e A with respect
to p, or the p-equivalence class of ay means the set [a] = {b e A | bpa].

Art. 3-2] INTRODUCTION TO SET-THEORETIC TOPOLOGY 45
Theorem 1. Two p-equivalence classes are identical if they have an
element in common; that is,
C.11) [?] ? [b] ? 0 => [a] = [b] (Exercise 9).
(B) An equivalence relation p, by Theorem 1, defines an analysis of
its domain A into a collection {a} of subsets, the p-equivalence classes,
such that A) A is the union of the subsets {a} and B) no two of the
subsets {a} intersect. A collection of subsets {a} of a set A with these
two properties is called a partition or decomposition of A.
Theorem 2. A partition {a} of a set A determines an equivalence
relation ? defined by the condition that apb if and only if a and b belong
to the same element of {a}.
Proof. By (Bl), A is the domain of p. The relationship of belonging
to the same class is obviously reflexive, symmetric, and transitive.
(C) Theorems 1 and 2 mean that every equivalence relation defines a
partition of its domain and every partition of a set is defined thus by
some equivalence relation.
(D) The cartesian product S ? ? of two sets S and ? means the
set of all ordered pairs (s, t) such that s e S and t e T. In particular,
8 ? S is the set of all ordered pairs of elements of a set S. The
mappings ?? : S ? ? -> S and ?2 : S ? ? -> ? defined by
C.12) 77^5, t) = S 772E, t) = t
are, respectively, the ^-projection and the T-projection of S ? ?.
A relation ? with domain A <= S and range ? <= ? can be
represented by a unique subset R <^ S ? ?, where (s, t) e R if and only if
spt. Conversely, every subset R c: S ? ? represents the relation spt
defined by the condition that spt if and only if (s, t) e R. We express
the fact that R represents ? by writing R = R(p), which can be
interpreted as defining a mapping of the set of all relations with
domain A <= S and range В c= ? into the set of all subsets of S ? ?.
We call R(p) the graph of the relation p.
(E) The properties of symmetry, reflexivity, and transitivity in a
relation ? are equivalent to the following properties of R = R(p),
where ? has domain A <= S and range В <= ?:
A) ? is reflexive: »-e^4· U В -^ (б, ь) ь R [-^> В =тЦ, 1 SeH -> cs,<0
B) ? is symmetric: E, t) e R => (t, s) e R [=> В = А],
{E, t) g R)
C) ? is transitive: I => (s, и) е R.
\{t, u) g R)
. (F) A function or mapping/ can be interpreted as a special type of
plation, although we write у = f(x) or / : ? -> у rather than xfy as in

46
INTRODUCTORY TOPOLOGY
[Ch. 3
the case of relations. If, for the moment, we define xfy as equivalent
to у = f(x), we can say that a relation ? is a function if and only if the
S-projection of its graph R(p) c: S ? ? is one-to-one. Furthermore,
a relation ? is a one-to-one mapping (function) if and only if both the
S-projection and the T-projection of R(p) are one-to-one.
>S
(a)
?
/?
A,1)
(b)
Fig. 3-1. The sets R for various relations, (a) R(px) :t = s;
To clarify the foregoing remarks, consider the special case where
S = ? is the set of all real numbers. In diagrams, we will represent
S and ? as the usual rectangular coordinate axes for plane analytic
geometry (Fig. 3-1). Then S ? ? is the set of all points (s, t) of the
plane, and a relation ? is represented by a subset R(p) of the plane.

INTRODUCTION TO SET-THEORETIC TOPOLOGY
47
Art. 3-2]
The following relations are illustrated in Fig. 3-1: (a) spj if and only
if t = s (the identity relation), (b) sp2t if and only if s2 < t < Vs,
(c) spzt if and only if t > s, (d) spj; if and only if st > 0.
(c)
?
^s
(d)
(b) R(p2) : s2 < t < Vs; (c) R(ps) :t>s; (d) R(p4) : st > 0.
EXERCISES
9. Prove Theorem 1, first showing that ? ? [a] => [?] = [a].
\ 10. Make diagrams analogous to Fig. 3-1 to illustrate the relations defined
^s follows: (a) s2 + t2 < 1, (b) s2 + s < t, (c) t < Vl - s2.

48 INTRODUCTORY TOPOLOGY [Ch. 3
11. What are the domain and range of the relation "is a daughter of"?
12. (a) Give at least three examples each of symmetric and non-symmetric
human relations, (b) The same with "reflexive" in place of "symmetric."
(c) The same as (a) with "transitive" in place of "symmetric."
13. Discuss the relations between the domain and the range of a relation p
implied by the statement that ? is (a) symmetric, (b) reflexive, (c) transitive.
14. Make a table classifying the six relations (e, з, с, Э, (S, =) with
respect to symmetry, reflexivity, and transitivity.
15. Let AvB if and only if a one-to-one mapping A -> В can be defined.
Show that ? is an equivalence relation.
16. Which of the relations pv p2, /э3, /э4 shown in Fig. 3-1 are (a) functions,
(b) symmetric, (c) reflexive, (d) transitive, (e) equivalence relations?
17. (a) Show that, in diagrams like Fig. 3-1, symmetric relations are
represented by sets symmetric in the line t = s. (b) What geometric property
characterizes sets representing reflexive relations? (c) The same for equivalence
relations.
3-3. Continuity for Real Functions of Real Variables
As a preparation for certain generalizations, we remind the reader
of some of the limiting processes and continuity concepts involved in
the theory of functions of real variables.
The real number system is intuitively interpreted as the set of
points on an axis, in the sense of analytic geometry, and is sometimes
called the arithmetical continuum. A point of the continuum, or axis,
is merely a number. The distance between two points, ? and y, is
generally defined as* \x — y\.
(A) With distance thus defined, the arithmetical continuum becomes
euclidean 1-dimensional space, denoted by E1. The euclidean plane,
or 2-space, is the cartesian product ? ? ? = ?2 of two real number
systems, with the distance between two points (xv yx) and (x2, y2)
defined asf
C.13) p((xv уг), (x2, ya)) = V(*2 - xif + (Уг - Vi?·
A real function of a real variable is a mapping
C.14) f:A-+B i с I, 5 с 7.
An open interval on the x-axis X is a set of real numbers
{? ? ? < ? < b} where a, b are any two numbers with a < b. Its
closure, also called a closed interval or segment, is {x | a < ? < b}.
* The vertical bars here are absolute value signs. Thus \a\ = a if a > 0 and
|a| = — a if a < 0.
f Strictly speaking, 2-dimensional number space is so denned, but it can be proved
to satisfy the postulates of euclidean geometry.

Art. 3-3] INTRODUCTION TO SET-THEORETIC TOPOLOGY 49
The following notation and terminology is often used:
(a) closed interval [a, b]: a < ? <b
[a,b): a < ? <b
(a,b]: a < ? <b
(c) (open*) interval (a, b): a < ? <b.
However, (a, b) will mean a point of E2 unless another meaning is
explicitly specified.
In elementary calculus, it is said that a function / : A —> B,
generally defined in the form у = f(x), approaches the limit с at a (or
as ? approaches a) if (I) f(x) is defined, for some d > 0, at all points ?
such that 0 < \x — a\ < d and B) for any ? > 0, however small,
there exists a ? > 0 so small that
C.16) \f(x) — c\ < ? for all ? such that 0 < \x — a\ < ?.
Note that f(x) may be defined or not when ? = a and, if defined,
may have the value с or any other value.
A function / is said to be continuous at x0 if A) it is defined for
every ? in an interval containing x0 and B) for any ? > 0, there
exists a ? > 0 so small that
C.17) \f(x) — f(xo)\ < ? f°r every x such that \x — x0\ < ?.
In other words, f(x) approaches a limit as ? approaches x0 and that
limit equals f(x0).
(B) An equivalent definition of / being continuous at x0 is the
following: A) f(x) is defined in an interval containing x0 and B) for
every interval Iy on the ?/-axis containing y0 =f(x0) there exists an
interval Ix on the x-axis containing x0 whose image points are all on Iy.
We next give a conventional calculus definition of continuity and
associated concepts for a real function/of two real variables, defined by
C-!8) *=f(*,y) (x,y)eSczE2
where (x, y) is interpreted as a point of a euclidean plane E2 and ? as
a point on a z-axis Z.
A set S of points of E2 is called open if, for each point ? : (?, у)
belonging to S, there exists a ? > 0 so small that all points within
distance ? ?? ? belong to S. The function/ : S -+ ? cz ? or its value
(/(ж, у) is said to approach the limit с at (a, b) if /is defined on some
C.15) (b) half-open intervals I
* The unmodified word interval means an open interval.

50 INTRODUCTORY TOPOLOGY [Ch. 3
open set containing (a, b), save perhaps at (a, b) itself, and if, for each
? > 0, however small, there exists a ? > 0 so small that
C.19) |/(ж, у) - c\ < ? provided 0 < у/(х - aJ + (у - bJ < ?.
(С) The function/ is continuous at (x0, y0) if A) it is defined on an
open set S containing (x0, y0) and B) for each open interval Iz on the
2-axis containing z0 =f(x0,y0) there exists an open set N in E2
containing (x0, y0) such that all the functional values at points of N
lie on the interval Iz.
Euclidean 3-space, Ег, is defined by defining A) a point as an
ordered triple (x, y, z) of real numbers and B) the distance between
two points (x, y, z) and (xf, y', z') as V(x — x'J + (y — y'J + (z — z'J.
Euclidean w-space, En, for each positive integer n, is defined as follows:
A) a point is an ordered set of ? real numbers; B) the distance between
two points ? : {xlf . . . , xn) and q : (ylf . . . , yn) is
C.20) p(p, q) =
? (Xi ~ yf
-1
As in the case of E2, a set S <= En is said to be open if, for each
point ? ? S, there exists a d > 0 so small that p(p, q) < d=> q e S.
A real function of ? real variables with domain А с Еп is a mapping
/ : A —> Y, where ? denotes the real numbers. The range of/ is the
set В = f(A) <= ?. The image of a point ? : (xv . . . , xn) can be
denoted by either f(p) or f(xv . . . , xn). It is said that /, or f(p),
approaches the limit с at a : (a1? . . . , an), or as ? approaches a,
symbolized by
C.21) f(p)~^c as P~+a> or um/(p) = c,
if A) for some d > 0, the domain A of f contains all points within
distance d of a, with the possible exception of a itself, that is,
A 3 {p | о < p(jo, a) < d}, and B) for each ? > 0, there exists a
? > 0 such that
C.22) 0 < p(p, ?)<?=> \f(p) -c\<e.
(D) The function / is continuous at a if A) it is defined at a and
B)f(P) ->/(«) &sp -+a.
Consider two euclidean spaces, an m-space Em and an тг-space Rn.
Let ? : (xv . . . , #w) and q : (з/15 . . . , yn) denote points of Em and Rn,
respectively. The cartesian product Em ? Rn consists of all the ordered
pairs r = (p, q), hence of all the ordered sets (xv . . . , xm, ylt . . . , yn)
of m + ? real numbers. The Em-projection of r : (xv . . ., xm, yv . . . , yn)
is ? : (xv . . . , xw) and its i?w-projection is q : (ylt . . . , ?/w). Let the

Art. 3-4] INTRODUCTION TO SET-THEORETIC TOPOLOGY 51
distance between two points r e Em ? Rn and r' e Em ? Rn, where
r == (pt q) and r' = (p', q'), be defined as
C.23) p(r, r') = y/pl(p,p') + pl(q,q'),
where ?? and p2 are the distance functions in Em and Rn, respectively.
(E) With distance thus defined, Em ? Rn becomes a euclidean
space Em+n (Exercise 22 below).
(F) Consider the euclidean space En+1 = En ? 7, and let/, with
domain A <= En and range В <= У, be interpreted as a relation (Art.
3-2(F)). Its representation R(f) <= 2Jw+i is called the graph or locus
of/ (see Exercise 23 below).
(G) The definitions of limit and continuity above can be extended
to the case that condition A) below C.21) is replaced by the weaker
condition: (?) for every d > 0, the domain A of / contains at least
one point ? satisfying 0 < p(p, a) < d. In this case the condition
? e A is added to the hypotheses of C.22). Furthermore, / is said to
be continuous at each isolated point of A.
EXERCISES
18. (a) Let/ be defined by f (x) = x/\x\ wherever the expression has meaning.
Does/ approach a limit as ? -> 0? Is it continuous there? Justify your answers,
(b) The same if f (x) = x/x. (c) Give an example of a function defined for all ?
which approaches a limit for each point, but is not continuous at ? = 0.
19. Give a formal proof that the function/ of two real variables defined by
/(x, y) = 2x + 3y is continuous at @, 0).
20. Demonstrate the equivalence of Definition (B) to the definition of
continuous formulated in C.17).
21. Prove that Definition (G), Part B), is equivalent to saying that/(x, y)
approaches a limit at (x0, y0) and that the limit equals f (x0, y0).
22. Prove Statement (E).
23. (a) As in Art. 3-2B^), interpret a function / of one real variable as a
relation, and show that its representation R(f ) in the euclidean plane E2 is the
locus or graph у = f (x) of the function, in the sense of analytic geometry. What
geometric restriction on this graph is imposed by the condition that/ be one-to-
one? (b) The same for a function of two variables, represented in Ez.
3*4. Topological Spaces
In later chapters, various properties of euclidean тг-space will be
discussed and used. The definitions in the preceding article were
presented primarily as a basis for defining the more general spaces
'which will concern us.
(A) A topological space (S, {a}) is a set of objects S, to be called
^points, together with a collection {or} of subsets of S, called the open
?ets of S, such that

52
INTRODUCTORY TOPOLOGY
[Ch. 3
Axiom 1. The intersection of any finite collection of open sets is
open.
Axiom 2. The union of any collection of open sets is open.
An example is En, with the open sets defined in Art. 3-3 (Exercise 24
below). A neighborhood of a point ? of S is an open set containing p.
Theorem 3. The null subset 0 of S and the entire set S are open sets.
Proof. The union in S of the null collection ? of open sets is the
null set 0 of S since the union consists of all points each of which
belongs to at least one ? e ?, and there exists no ? e ?. Hence, by
Axiom 2, 0 is an open set. The intersection ??, on the other hand,
is S, since each point of S belongs to all the sets ? e ?, there being no
such sets. Hence, by Axiom 1, S is open.
(B) The points of a topological space might be vectors, mappings,
geometric loci, the states of a mechanical system, and so on. This
fact permits a wide range of applications of topology.
A set S is topologized by the designation of a collection {a} of open
sets satisfying the axioms, and {o*} is then called a topology on S.
There are generally many ways of topologizing a set. The trivial
topology is the one in which only 0 and S are open. The topology in
which every set isVpen is called the discrete topology. For a number of
interesting examples, see [Kx].
A base for a topology {a} on S is a subset 23 of the open sets such
that if ? e a <= S, there exists a ? e 23 such that ? e ? <= o*. In other
words, each open set containing a point ? ? contains a set of the base
which contains p.
(C) For example, A) the set of open intervals is a base for the
topology of E1; B) so is the set of open intervals with rational end
points; C) the set of open circular regions is a base for the topology of
E2; D) so is the set of rectangular regions (a < ? < b, с < у < d)
(Exercise 25).
If © is a base for a topology {a} on S, we will use (S, 23) as equivalent
to (S, {a}), a usage justified by Lemma 1 below. We will also write
merely S for a topological space when it is not important to specify its
topology or when the topology is implied by the context.
Theorem 4. Let S and ? be topological spaces with bases © and (?.
In the cartesian product S ? ?, let © ? (? be the collection of all sets
of the form ? ? ?, where ? ? 93 and ? ? G. Then © ? G is the base
for a topology on S ? ?.
Proof. Let {or} be the open sets of {S, 23}. By definition of base,
each ? ? {?} is the union of the elements of Ж which it contains. By>
Axiom 2 in (A), each union of elements of Ж is an open set. This;
establishes the following result.

Art. 3-4] INTRODUCTION TO SET-THEORETIC TOPOLOGY 53
Lemma 1. Each open set of a topological space with a given base 23
is the union of a collection of elements of 23 and conversely.
It remains to show that the collection {?} of all unions of subsets of
23 X ? satisfies Axioms 1 and 2 of (A). lip e ?? ? ?2, where ?{ e {?}
(г = 1, 2), then, by definition of {?}, there exist sets ?? e © and уг- е (?
such that ? e ?? X yx <= ?? and ? e ?2 xy2c 9V But, writing ? in
the form (s, t) with ? e S and t e T, this implies that s e ?? ?? ?2,
t e ?? ? 72. Since © and (? are bases for # and I7, this implies
sGjSc (/?! П /?2), t ?? <= G? ? 72) for some /? e 23, yeS. Hence
9?? ? 992 is the union of the subsets of 23 X (? which it contains, and
Axiom 1 is fulfilled for the case of two open sets. The remaining
cases are covered in Exercise 26. Axiom 2 offers no difficulty
(Exercise 27).
(D) The space (S ? ?, 23 ? G) will be called the topological
product of (#, 23) and (T, (?). Where confusion with the cartesian
product is unlikely, the simpler notation S ? ? will be used for this
topological product. Such products as Sx X S2 X · · · X Sn are
recursively defined by (Sx X S2 ? · · · X Sj) X Sj+1 = fifx ? · · · ? fifJ+1.
(,?/) Let Л be a subset of aS and let {?} be a topology on S. Let
{a} = {a = ? ? ? | ? e {?}}. Then {a} is a topology on A (Exercise
28). It will be referred to as the topology induced on A by (S, {v})r
(F) A point ? e A c: #, where # is a topological space, is an
inner point of A if there exists an open set a such that pea с Л.
Thus an open set either is vacuous or consists entirely of inner points.
A set consisting entirely of inner points is open (Exercise 29). A
boundary point of a set A c: S is a point ? each neighborhood of
which has a non-vacuous intersection with both A and its complement
S — A. The boundary dA of A means the set of its boundary points.
(G) In a topological space S, a closed set means the complement of
an open set. Thus 0 and S, and possibly other sets, are both open and
closed. The closure of a set A <= S means the set A = А и дА. The
set ^4 is closed (Exercise 30).
A point ? is a limit point (or point of condensation) of A <= S if, for
each neighborhood ? ???, the intersection ? ? (? — ?) is non-vacuous.
(Я) Let Б0 = /S — ^, the complement of the closure of an open
set A cz S. Then the open set A0 = S — B0 is the union of A and
the limit points of A which are not also limit points of B0 (Exercise 32).
The sets A0 and B0 are open sets such that
(a) A0 ? B0 = A0 ? ?0 = 0,
,C-24) (b) 40??0 = ?0??0 = ?,
(с) ЭА0*=дВ0 = А0пВ0.
(I) A set A is closed if and only if dA cz 4. Also ? is the union
of Л and its limit points.

54 INTRODUCTORY TOPOLOGY [Ch. 3
(J) In a topological space, an infinite sequence plt p2, ... of not
necessarily distinct points converges if there exists a point p, each
neighborhood of which contains all but a finite subset of (plt p2, . . .).
In such case, ? is a limit of the sequence.
EXERCISES
24. Show that En is a topological space in the sense of (A), using the definition
of open sets in Art. 3-3. Define an infinite set of open sets whose intersection is
not open.
25. Prove Statement (C).
26. Complete the proof in (C) for the case that the number of elements of
{?} in the collection is (a) zero; (b) one; (c) ? > 2, ? an integer.
27. Verify Axiom 2 for {?}, thus completing the proof of Theorem 4.
28. Prove that {a} is a topology in Statement (E).
29. Show that a set consisting entirely of inner points is open (see (F)).
30. Show that A is closed in Statement (G). (See (F).)
31. Show that regions of the form a{ < xi < b{ (i = 1, . . . , n) form a basis
for the open sets of En, defined in Art. 3-3. Hence show that En = Em ?
En-m = Д1 x . . . ? ?1 (n factors).
32. Prove all parts of Statement (H).
33. The same for Statement (/).
3-5. Homeomorphisms; Definition of Topology
{A) A mapping/ :S -+T, where S = (S, {a}) and ? - (?, {?}) are
topological spaces, is continuous at ? e S if, for each neighborhood ?
off(p), there exists a neighborhood a of ? such that/(o-) <= т. If/ is
continuous at each point of S, it is continuous (on S).
If A is a subset of S, a mapping / : A —> ? is continuous at ? e A
if, for each neighborhood ? of f(p) there exists a neighborhood ? ?? ?
such that/(^ ? ?) <= т.
(Б) A homeomorphism/ : S —> ? is a mapping which A) is one-to-
one, B) is continuous, and C) has a continuous inverse/-1 : ? -+S.
Topology is the study of those properties of spaces which are preserved
by homeomorphisms. A topological property of a space S is a property
shared by each space homeomorphic to S, where two spaces are
homeomorphic if it is possible to define a homeomorphism between
them.
Theorem 5. If f:S -+T is continuous and g :f(S) -+ V is
continuous, then h = gf: S -> V is continuous.
Proof. For any neighborhood ? of h(p) in V, there exists, by the
continuity of g, a neighborhood ? off (?) in ?7, such that <7(? nf(S)) <=
<p. By the continuity of/, there exists a neighborhood ? ?? ? in # such

Art. 3-5] INTRODUCTION TO SET-THEORETIC TOPOLOGY 55
that f(a) с ? so that f(a) с (? П/(?)). But f(a) <= (т n/(S)) and
g(r nf(S)) <= 99 imply gf(a) = h(a) <= 99, and the theorem follows.
Theorem 6. Among topological spaces, the relationship of being
homeomorphic is an equivalence relation.
Proof. The relationship is reflexive, since the identity mapping,
which maps each point onto itself, is a homeomorphism (Exercise 35).
It is symmetric, since the inverse / _1 : ? —> S of a homeomorphism
satisfies the definition of a homeomorphism. The property of being
one-to-one is obviously transitive.
Theorem 5, applied to the product h = gf of two homeomorphisms
and its inverse A-1 = / ~гд~г, completes the proof that the property of
being a homeomorphism is transitive, thus establishing Theorem 6.
A mapping / : S —* ? is called open [closed] if it maps each open
[closed] set of S onto an open [closed] set of T. L *u. $3-U?>]
Theorem 7. A mapping / : 8 —* ? is continuous if and only if its
inverse / _1 is open.
Proof. Since / is not assumed one-to-one, the general definition
C.9) of inverse is needed. Suppose / continuous. Consider /_1(?)
where ? <= ? is open, and let ? ef _1(?). Then ? is a neighborhood of
f(p). Hence there exists a neighborhood a of ? such that/((r) <= ?,
which implies a <= /_1(?). Hence/_1(?) is open. The proof is
completed by showing that if/ _1 is open then/is continuous (Exercise 36). I
Corollary. A homeomorphism is both open and closed (Exercise
36).
(C) In view of Theorem 7, a homeomorphism could be defined as a
one-to-one open mapping with an open inverse.
The above definitions and results are applicable to a topological
space subject only to Axioms 1 and 2 in Definition (A). In most of
our work, other conditions will be satisfied, one of which is stated in
the following definition.
(D) A Hausdorff space is a topological space S in which, for each
pair of different points (p, q), there exist non-intersecting open sets
? and ? containing ? and q respectively.
Exercise 37 gives an important property of Hausdorff spaces.
EXERCISES
34. Show that, for a real function of ? real variables, Definition (A) of
continuity is equivalent to Definition (D) in Art. 3-3.
35. Prove that the identity mapping is a homeomorphism. Show that a
continuous one-to-one mapping need not have a continuous inverse.
36. Complete the proof of Theorem 7 and corollary.

56 INTRODUCTORY TOPOLOGY [Ch. 3
37. Show that a convergent sequence (a) has only one limit in a Hausdorff
space, (b) may have more than one limit in a non-Hausdorff space.
38. Show that S ? ? is homeomorphic to ? ? S for two topological spaces
S and T.
3-6. Metric Spaces
Intuitively, a metric is a measure of distance from one point to
another. Mathematically, for a set of points S, a metric is a function
? of ordered pairs (s, t) of points of S, subject to appropriate axioms.
Since the cartesian product S ? S is the set of ordered pairs of points
of S, it is convenient to define ? as a function on S ? S.
(A) Let S be a set of points and ? the real numbers. A metric ? on
S is a function or mapping ? : S ? S —> R such that
* Axiom 1. p(s, t) > 0
Axiom 2. p(s, t) = Oos = t
Axiom 3. p(s, t) = p(t, s)
Axiom 4 (The Triangle Axiom). p(s, t) + p(t, u) > p(s, u).
Given a set S, a metric p, and a number ? > 0, the (spherical)
^-neighborhood of a point a eS means the set {s eS | p(a, s) < ?}.
Theorem 8. The set © of all spherical neighborhoods of points of a
set S with a metric ? is the base for a topology on S.
Proof. Let {a} be the collection of all subsets of S each of which
is a union of spherical neighborhoods. It is to be shown that {or}
satisfies Axioms 1 and 2 of Art. 3-ЦА). Axiom 2 is obviously satisfied.
To prove Axiom 1, let ? be a point on ?? ? ?*2, where o*l5 o*2 are
members of {a} with a non-vacuous intersection. Then ? e a0(s) ?
arj(t), where a0(s) denotes the ?-neighborhood of a point s and ??(?)
is the ^-neighborhood of a point ?. Let
C.25) ? = min @ - p(s, p), Ч - p(t, p)) > 0
where min (x, y) means the smaller of two numbers ? and y.
Lemma 2. The ?-neighborhood ofp, ??(?),?$ on ??(?) ntan(t)9 hence
ОП ?*! ? 0*2.
Proo/ o/ Lemma. Note first that w e ?*?(?) => p(p, u) < ?. Hence,
using the triangle axiom, p(s, u) < p(s, p) + p{p, u) < p(s, ?) + ?,
and p(s, ?) + ? < p(s, ?) + ? - p(s, p) by C.25). Therefore, p(s, u) <
?. Similarly, p(t, u) < ?, and Lemma 2 follows. But Lemma 2
implies that ?? ? ?2 is the union of the spherical neighborhoods
which it contains, verifying Axiom 1 for the case of two open sets.
The proof for the case of zero, one, or n > 2 members of {a} presents
no difficulty.

Art. 3-6] INTRODUCTION TO SET-THEORETIC TOPOLOGY 57
(B) The space (S, 33) of Theorem 8 is the metric space (S, p) and
its topology {?} is induced by p. Euclidean тг-space is an example.
(C) If (S, {a}) is a topological space, then it may be possible to
define a metric ? on S so that ? induces the topology {a}. If so,
(S, {a}) is metrizable and any metric inducing the topology {?} is a
metric on S.
(D)IfA с (S, p), then the restriction ? \ A, defined in Art. 3-l(Z>),
is a metric on A. The same topology is induced on A by the metric
? | A and by the topology {?} of (S, p) (Exercise 43).
(E) The property of being metrizable is a topological property.
That is, if #x and S2 are homeomorphic spaces and #x is metrizable,
so is S2. Indeed, if ?? is a metric on S± and h : вг —* S2 is a homeo-
morphism, then the function p2 defined on S2 ? #2 by p2(h(s), h(t)) =
рг(8, ?) is a metric on #2 (Exercise 44). The metric p2 is the metric
carried over from (Sv px) to S2 by the homeomorphism h.
(F) If S is metrizable, it is a Hausdorff space (Exercise 45).
(G) Let Л <= $, Б с: S be two non-vacuous subsets of a metric
space # = (S, p). The distance p(A, B) between A and В is defined by
C.26) p(A, B) = g.l.b. p(a, b) aeA,beB
where g.l.b. means greatest lower bound.* Since p(a, b) > 0, p(A, B)
exists. It is either the smallest distance between a point of A and a
point of B, or the largest number less than all such distances.
(Я) The diameter of А с (fif, p) is
C.27) diam A = l.u.b. p(a, b) aeA,beA
provided the least upper bound exists. If it does not exist, A is
unbounded or, symbolically, diam A = oo. When it does exist, A is
bounded.
Theorem 9. Let {/} be the set of all continuous mappings of the
unit segment [0, 1] : 0 < t < 1 into the real numbers, and let
C-28) p(f, 9) = max \f(t) - g(t)\ @ < t < 1).
Then the function defined by ? on {/} is a metric.
Proof. Since/ and g are continuous functions on [0, 1], the same
is true (Exercise 48) of \f(t) — g(t)\, which therefore attains a
maximum. Thus ? is defined for each (/, g) e {/} ? {/}. It obviously is
non-negative, has the property p(f, g) = Oo/ = g, and is symmetric.
* Various properties of real numbers, and various standard concepts are assumed.
The reader unfamiliar with them can readily find them in the literature; for example,
inLLJ and|_T].

58 INTRODUCTORY TOPOLOGY [Ch. 3
This leaves only the triangle axiom to be verified. Given three
elements, (/, g, h) с {/}, let t0, tlt t2 be points such that
l/(*o) - 9(*o)\ = max 1/@ - 9(*)\ = P(f> 9)»
C.29) | g^) - ?(??)[ = max | g(t) - h(t)\ = p(g, A),
1/(У - 4h)\ = max |/(i) - h(t)\ = p(f9 A).
Then
C.30)
\f(k)-g(t2)\ <i/(g-g(gi,
\g(t2) -h(t2)\ <ЫУ-A(OI-
Hence (Exercise 49)
C.31) p(/,A) <P(f,g) + P(g,h).
EXERCISES
39. Test the functions (a) ??(?, у) = (? — уJ and (b) p2(x, у) = ?2 — у2 to
see whether they are metrics on the set R of real numbers.
40. Find p(A, B) and diam A where A and В are the following subsets of the
euclidean (x, y)-plane.
A : 4x2 + $y2 < 36 and В : 9x2 + 4i/2 > 144.
41. Given a point set $, show that the equations p(s, s) = 0, p(s, t) = 1 for
s ? t define a metric on S.
42. (a) Show that the metric of Exercise 41 induces the discrete topology onS.
(b) Show that if S contains more than one point and has the trivial topology, it is
not metrizable.
43. Prove Statement (D).
44. Prove Statement (K).
45. Prove (F) by showing that if ? is a metric on S and (p, q) are two different
points on S, then the ^-neighborhoods of ? and q are disjoint if 26 < p(p, q).
46. In the (x, i/)-plane, let
C.32) Pi@*i> Уг), {x2> 2/2>) = max A^2 - xi\> I2/2 - 2/iD·
Show that C.32) defines a metric px on the (x, y)-plane. Show that ?? induces the
same topology as the euclidean metric.
47. Show that, in a metric space,
(,.)P(A,B)=0oAnB*9, '4i «- ^ - 1>^^J'
C.33) _ - -
(b) p(A, B) = P(A, B) = P(A, B) = P(A, B).
48. Show that p(/, g) in C.28), interpreted as a function of t, is continuous.
49. Deduce inequality C.31) from C.29) and C.30), thus completing the proof
of Theorem 9.

Art. 3-7] INTRODUCTION TO SET-THEORETIC TOPOLOGY 59
3-7. Compact Spaces
Let S be a topological space, and let A <= S. A covering of A means
a collection {a} of subsets of S, such that each point of A is in at least
one set of {a}. If all the sets of {a} are open, then {a} is an open
covering of A. If they are all closed, it is a closed covering.
(A) The set A <= S is called compact in S if and only if each open
covering of A contains a finite covering; that is, given an open
covering {a} of A, there exists a finite subset (?? . . . , ??) of {a} which is
also a covering of A. In the case where A = S, this defines the
concept of a space being compact in itself or, more briefly, compact.
A compact metrizable space is called a compactum.
(?) ?? A is compact in S, then A, with the topology induced by S,
is compact in itself. A compact subspace of a compactum is a
compactum (Exercise 52).
Theorem 10. Let S and ? be two topological spaces, and let A
be compact in S. Then, if / : A —* ? is a continuous mapping, f(A)
is compact in T.
Proof. Let {?} be an open covering off(A) in ?7. Then, if ? e {?},
/_1(?) is an open set in S (Theorem 7). The set {/_1(?) | ? e {?}} is an
open covering of A, and, because Л is compact, it contains a finite
covering /_1(Ti) (* = *> 2> · · · > w)· But (Ti> · · · » О c {T} is tnen a
finite covering of f(A), and the proof is complete.
Corollary. Compactness is a topological property (Exercise 53).
Theorem 11 (Bolzano-Weierstrass). In a compact Hnmrrbirff
space S9 a subset A containing an infinite collection iff r p» pn. .-, .)
of &4$а*@ф4 points possesses a limit point.
Proof. Let ? = (ffl5 ff2, . . . , pn, . . . ), and suppose ? has no limit
point. Let ? be the closure of P. Then #0 = S — ? is open, and so
is the union Sj oiS0 and (pl9 . . . , pj) for any j. By Art. 3-4G), ? = P.
Furthermore, (Sv S2, . . . , Sn, . . . ) is an infinite collection of open
sets covering S, no finite subset of which covers S (Exercise 54),
contrary to the hypothesis that S is compact. Therefore ? has a limit
point. Such limit points need not belong to A.
Theorem 12. If S is metrizable in Theorem 11, then A contains
a convergent sequence of distinct points.
Proof. Let ff be a limit point of P. Let pn be a point other than ?
Jn the spherical (l/w)-neighborhood of ff for each ? = 1, 2, . . . , and
let ffn be taken closer to ? than ffn_x. Then (ffl5 ?2, . . . ) converges to
p, since each neighborhood of ff includes the A/^-neighborhood of ff

60
INTRODUCTORY TOPOLOGY
[Ch. 3
for some ? and hence includes all the points (pv p2, . . . ) except a
subset of (plf . . . , pn).
Corollary. A compact subset A of a metrizable space 8 is closed.
If A is not closed, there is a limit point ? of A not in A (Art.
3-4G)). Let (pvp2, · · · ) be a sequence on A converging to ? and
defined as in the proof of Theorem 12. The sets (#l5 S2, . . . ), defined
exactly as in the proof of Theorem 11, cover A, but no finite subset
of them covers A, contradicting its compactness.
Theorem 13. A compact subset A of a metric space S is bounded.
Proof. Otherwise, for a point ? of S and each ? > 0, there is a
point of A at distance greater than ? from p. Therefore the spherical
^-neighborhoods of ? {? = 1, 2, 3, . . . ) are a covering of A containing
no finite covering, contradicting its compactness.
Theorem 14 (The Heine-Borel Theorem). A subset A of
euclidean w-space En is compact if and only if it is closed and bounded.
This is a standard theorem of analysis, whose proof is readily
accessible in the literature. (See Exercise 66.)
Theorem 15. The topological product Ax ? ?2 of two compact
spaces is compact.
Proof. Let 93л be a base for Ah (h = 1, 2). Then, by Theorem 4,
93x X 93 2 is a base for A1 ? A2. Let {a} be an open covering of A x xA2,
and let {/?} = {/? e 93x x 9321 ? <= <r ? {<r}}. Then {/?} covers Лх ? ?2,
since each ? e {a} is the union of the elements of 93x X 932 which it
contains. Let {??} be the set of ^-projections (see Art. 3-2(Л)) of the
elements of {?} (h = 1, 2). Then {/?л} is an open covering of Ah, and
therefore contains a finite covering {?\, . . . , /?*J. But this implies that
the n1n2 topological products ?] ? /?| (г = 1, . . . , пг\ j = 1, .·. . , ?2)
are a finite subset, which we denote by (/?l5 . . . , ?? ? ), of {?} covering
?? ? A2. Let at be an element of {a} containing ?? (г = 1, . . . , ???2).
Then (?? . . . , ?? ? ) is a finite subset of {?} covering ?? ? ?2, and its
existence means that ?? ? ?2 is compact.
Theorem 16. A continuous function /: A -> R, where A is
compact and R is the real number system, assumes a largest and a
smallest value.
Proof. By Theorem 10, /(Л) is compact. Hence, by Theorem 13
and the corollary to Theorem 12 (interpreting R as E1), f(A) is
bounded and closed. But a closed, bounded set of real numbers
contains a largest and a smallest number, and Theorem 16 follows.
Theorem 17. A compact subset A of a metric space (S, p) is of
finite diameter.
Proof. By Theorem 15, A x A is compact. The metric ? maps

Art. 3-8] INTRODUCTION TO SET-THEORETIC TOPOLOGY 61
A x A continuously into R (Exercise 67), and Theorem 17 follows
from Theorem 16.
Theorem 18. If A and В are compact subsets of a metric space S,
then there exist points a e A and b e В such that p(a, b) = p(A, B).
A proof can be given (Exercise 55) along the lines of the proof of
Theorem 17.
(C) If А с En, В с En are closed and one of them bounded, the
conclusion of Theorem 18 holds.
(D) Let ? = {/} be the set of all continuous mappings of a compact
space A into a metric space (?, ?). Then p(f(p), g(p)) is continuous
in ? on the compact set A. Hence, by Theorem 16, ? assumes a
maximum value. Let
C.34) <p(f, g) = max p(f(p), g(p)).
peA
Theorem 19. The function ? defined by C.34) is a metric on ?.
Proof. This is a generalization of Theorem 9 and its proof. If /
and g are the same mapping, that is f(p) = g(p) for all ? e A, using
the notation of (E), then ?(/, g) = 0 and conversely. The symmetry
of ? follows from that of p. By Theorem 16, there is a point px e A,
such that cp(f, h) = /?(/(??), ???)). For that point, by the triangle
axiom, ptfipj, д{рг)) + р{д{рг), Црг)) > р(/Ы, Црг)) = <p{f, h). But
<p{f, g) > p(/(Pi), д{рг)) and cp(g, h) > р{д(рг)9 Црг)). Hence cp(f, g) +
9%, h) > q>(f, h), completing the proof.
(E) The metric space (?, ?) is called the space of mappings A -> T.
EXERCISES
50. Define an open covering of the set x2 + y2 < 1 in the (x, y)-plane which
contains no finite covering.
51. Define a continuous function on the set x2 + y2 < 1 which has neither an
upper nor a lower bound.
52. Prove Statements (B).
53. Prove the corollary to Theorem 10.
54. In the proof of Theorem 11, why can no finite subset of (SVS2, . . . )
cover St
55. Prove Theorem 18.
56. Prove result (D).
57. Using the example of A : xy > 1 and В : xy < — 1 in E2, show that the
hypothesis that A or В be bounded is necessary m (D).
3-8. Brouwer Dimension; The Lebesgue Number
An ^-covering, or covering of mesh ?, of a compact metric space
(#, p) is a covering {A} of S such that each of the sets of {A} is of
diameter less than ?.

62
INTRODUCTORY TOPOLOGY
[Ch. 3
(^4) For any ?, a compact metric space S admits a finite open [closed']
?-covering. In the open case, a finite subcovering of the covering
by all spherical ^'-regions for any ? < ? is an example. In the closed
case, the set of closures of regions of an open ^-covering is an example.
Let {A} be a covering of a set S. The order of {A} is the largest
natural number v, if such a number exists, with the property that
some collection of ? of the sets {A} has a non-vacuous intersection.
If no such ? exists, {A } is of infinite order. A finite covering is obviously
of finite order, having the number of sets in the covering as a largest
possible value. In general, as ? gets smaller, the number of sets in a
finite ^-covering of a metric space will increase without limit. We will
see, however, that it is possible, for a large class of spaces, to impose
an upper bound on the order, independently of ?.
The closure lofa bounded open region A of euclidean тг-space En
is compact, by Theorem 14. In Art. 6-3 we will show that A) for any
? > 0, there exists a finite closed ^-covering of A of order n + 1, and
B) for ? sufficiently small, there exists no finite closed ^-covering of A
of order less than n + 1.
(B) The property just stated motivates the following definition,
due to L. E. J. Brouwer. * Let S be a compactum, and ? a metric on S.
Then S has (Brouwer) dimension n = dim# if A) for each ? > 0, S
admits a finite closed ^-covering of order n + 1, and B) for ? > 0
sufficiently small, S admits no finite closed ^-covering of order less
than n + 1. If these conditions are not fulfilled for any integer n, the
(Brouwer) dimension of S is infinite: dim S = oo.
Lemma 3. The dimension dim S is independent of the metric used in
the definition.
Proof of Lemma. Let ?? and p2 be two metrics on S. For a number
?1 > 0 and a point ? e S, let аг(р) be the spherical ^-neighborhood
(Art. 3-6) of ? in terms of the metric ?? and let p*(p) be the distance
from ? to S — ?^?) in terms of the metric p2.
(C) Then p*(p) is a positive continuous function on S and therefore
(Exercise 60) has a positive minimum value
C.35) ?2 = min p*(p).
veS
By definition of ?2, ?^?) contains the ^-neighborhood ?2(?) of ? in
terms of the metric p2 for each ? e S.
(D) It is now a simple matter to complete the proof, by showing
that dim S as defined in terms of ?? equals dim S as defined in terms
of p2 (Exercise 61).
Theorem 20. The Brouwer dimension of a compactum 8 is a
topological property.
* "Beweis der Invarianz der Dimensionszahl," Mathematische Annalen, 70 A911), pp.
161-165.

Art. 3-8] INTRODUCTION TO SET-THEORETIC TOPOLOGY 63
Proof. Let h : S —> ? be a homeomorphic mapping of S. Let ? be
a metric on S and p0 the metric carried over to ? by h (Art. 3-6(i?)).
Then ^-neighborhoods on S, and their respective intersections relative
to p, are mapped by h onto ^-neighborhoods on ? and their respective
intersections relative to p0. Theorem 20 now follows readily, with the
aid of Lemma 3.
Theorem 21. A definition of dim# equivalent to Definition (B) is
obtained by replacing "closed" by "open" throughout (B).
Proof. Theorem 21 is equivalent (Exercise 62) to the following
lemma.
Lemma 4. If, for some integer ? > 0 and each number ? > 0, S
admits a closed ^-covering of order v9 then it admits an open ^-covering
of order v; and if it admits an open ^-covering of order v9 then it admits
a closed ^-covering of order v.
Proof of Lemma. Part I. Given the first hypothesis, let ? be a
positive number, let ? = ?/2, and let {A} = (Av . . . , Am) be a closed
?-covering of S of order v. For each г e A, . . . , m), let Bt denote the
union of those sets of the covering {A } which do not intersect A t. If Bt
is vacuous, let pi = I. Otherwise, let pt = p(At, Вг). Let ? =
^ min (?, pl5 . . . , pm), and let a^AJ be the set of all points each at
distance less than ? from At (i = 1, . . . , m).
(E) The regions a^A^ (i = 1, . . . , m) are an open ^-covering of S
of order ? (Exercise 63).
Part II. Given ? > 0, let {?} = (?? . . . , am) be an open ?
covering of S of order v. We will obtain from {a} a closed ^-covering
{f} = (fl5 . . . , fv) possessing useful special properties. The sets f{ will
be defined recurrently.
Step 1. Let ?[ be the union of (?2, . . . , am), let ?? = S — ?[ and let
7i = S — fv Then (?? fx) covers S. Furthermore, ?? is the closure of
the union of (?2, . . . , am), and fx is of diameter less than ?.
Hypothesis. For some j e B, . . . , n), a covering (fl5 . . . , т,_г, у^_х)
of S has been defined, where A) (f1? . . . , т}_г) are the closures of
disjoint open sets (т1? . . . , ?;_?) and each is of diameter less than ?,
B) Yj_x is the complement of the union of (fl5 . . . , f^), and C) y,,^ is
the closure of the union of (oj, . . . , am).
This hypothesis is verified, when j = 2, by Step 1.
Step j. Let yj be the union of (<Ti+1, . . . , ctw), let Tj = ?}_? — yj9
and let ys = S — (fx U · · · U f,). Then (Exercise 64) (fl5 . . . , f„ y,)
satisfy the conditions of the hypothesis with j + 1 in place of j.
Step m, the final step, yields a closed ^-covering (fl9 . . . , fw),
completing the proof.

64 INTRODUCTORY TOPOLOGY [Ch. 3
(F) The closed ^-covering (fl5 . . . , fm) has the following properties:
A) each element f{ is the closure of an open set r{ (i = 1, . . . , m);
B) the open sets (??5 . . . , тт) are disjoint, that is, i ? j => тг П ?, = 0.
A closed covering with properties A) and B) is called a simple covering.
(G) As a corollary to the above proof, a compact metric space of
dimension ? admits, for any ? > 0, a finite closed simple ^-covering of
order n + 1.
Theorem 22 (Lebesgue's Lemma). Let {A} = (Av . . . , Am) be a
closed covering of a compact metric space (S, p). Then, there exists a
number Я > 0 so small that if a set В <= $ with diam ? < ? intersects
each of the sets (Лг , . . . , A{) then Лг ni,· n · - - n A{. ? 0.
Proof. Assume the contrary. Then for each positive integer n
there exists a non-vacuous set Bn <= 8 where A) diam Bn < \\n and
B) the subcollection %n of {A} intersected by Bn has a vacuous
intersection. Since there are only 2W — 1 non-vacuous subcollections of
{A}, some one of them must be repeated infinitely often in the
sequence 2ll9 2I2, . . . . Suppose (^? . . . , Atj) = %^ = %^ = - - - .
Let (Pi>p2i · · ·) be a set of points and k2, &3, &4, . . . a sequence of
integers such that рг е Bni and рг- е Бл^ - (pl5 . . . , ^.J (i = 2, 3,
. . .) (Exercise 65). By Theorem 11, there exists a limit point ? of the
set (pv p2, . . .). Each neighborhood of ? then contains infinitely many
of the sets Вщ , hence a point from each of the sets (Ai, . . . , Ah ).
The latter being closed, ? e A{ ? · · · ? ?{. , contradicting B) above,
and establishing the theorem.
(H) Any number ? satisfying Theorem 22 for given {A} is called a
Lebesgue number of {A}.
EXERCISES
58. Given a positive ? < \, define, for the unit disk x2 + У2 < 1, a finite
closed ^-covering of order 3.
59. Give a positive ? < \, define, for the unit circle x2 + y2 = 1, a finite open
^-covering of order 2.
60. Prove (C).
61. Carry through the argument suggested in (D).
62. Establish that Lemma 4 is equivalent to Theorem 21.
63. Prove (E).
- -^ 64. Verify the conditions of the recurrent hypothesis in Step ,;' of Part II of
the proof of Lemma 4.
65. In the proof of Theorem 22, show that (pv p2, . . . ) and k2, k3, . . . exist
as described.
66. Prove that a subset A of euclidean ?-space En is closed if and only if it
contains its limit points.
67. Prove that if A is a metric space, the metric p maps ? ? ? continuously
into the real number system.

4
Complexes
A topological space was defined in Chapter 3 as a set S of points
together with a collection {?} of open subsets of S, subject to certain
axioms. Thus a foundation was laid for set-theoretic topology.
In this chapter, we similarly lay a foundation for combinatorial
topology, commencing with a definition of a complex as a collection К
of simplexes subject to appropriate structural conditions.
4-1. Linear and Convex Subspaces of En
To avoid undesirable repetition, we remark that all the geometric
objects defined or discussed in this section are understood to be in a
euclidean тг-space ? = En, with a coordinate system (x) = (xv . . . ,
xn). The dimension number ? is merely required to be high enough, in
each case, for a definition or statement to have significance. For
brevity, we will use the term dimension instead of dimension number.
It is to be emphasized, however, that we have not yet established any
relationship between the term dimension, as used in this chapter, and
the Brouwer dimension defined in Art. 3-8.
The simplest general equations defining a line in En are of the form
D.1) x. = ai + bit (i = 1, ... , n)
where the parameter t has the arithmetical continuum, interpreted as a
?-axis, for its domain, and where at least one of the numbers Ъг is not
zero. Equations D.1) map'the ?-axis into E, carrying t = 0 into
(a) = (al9 . . . , an) and t = 1 into (ax + bv . . . , an + bn). The line
PoPn where p0 and px have the coordinates p0 : (a01, . . . , a0n) and
Pi : (#ii> . . . , aln), can therefore be defined by
D.2) Xi = aoi + (au - aoi)t
= aoi(l — 0 + аиг (г = 1, . . . , ?).
Much of the later work is simplified by the use of a parameter
65

66
INTRODUCTORY TOPOLOGY
[Ch. 4
t0 = 1 — t, which, if we write tx for ?, converts D.2) into the following
definition of the line p0pv also to be denoted by A(p0, ??)\
D.3)
MvvViY
(a) xi = a0it0 + autx
1(b) t0 + tx = 1.
(i = 1, . . . , n)
The condition that some bt ? 0 in D.1) is converted into the condition
that p0 and px be distinct points.
(A) A subspace ? of ? is linear if p^ eA (j = 0, 1) implies
MPo>Pi) c ?; that is, ? contains the entire line through any two
of its points.
The segment [????], which we will also denote by ?(?0, ??), is
analytically defined as follows:
D.4)
Г(^о> Pi) :
(a) xi = a0it0 + aj}
(b) t0
L (c) lj > 0
(i = 1, . . . , n)
i1=l
(i = o, i).
(Б) A subset Г of ? is convex if pj e ? (j = 0, 1) implies
?C»?· ?>?) c Г.
In the case of a single point, we make the definitions (Exercise 1)
D.5)
MPo) = Г(й) = Po-
For a subspace aS of jE/, A) the linear space A(S) determined by S
will mean the minimal linear subspace of ? which contains S and B)
the convex hull T(S) ofS will mean the minimal convex set containing
S. In other words, A) A(S) => S and no linear space which is a proper
subset of A(S) contains S and B) T(S) => S and no convex set which is
a proper subset of T(S) contains S.
(C) Equivalent definitions are as follows (Exercise 4): A(S) is the
intersection of all linear spaces containing S, and T(S) is the
intersection of all convex sets containing S.
(D) It is easy to verify that Definitions D.3) and D.4) generalize
as follows to an arbitrary finite subset (p0, . . . , pk) in place of (p0, px):
D.6) A(Po,
D.7) T(Po,
. Pk) ¦
>Pk) ¦
(a) Xi — &o^O + ' ' ' + akfk
1(b) t0 + · · · + tk = 1,
(a) xi = aoit0 ? + akitk
(b) t0 + · · · + tk = 1
L(c) ti > 0 (j = 0, . . . , *).
(г = 1, . . . , ?)
(? = 1, . . . , ?)

Art. 4-2]
COMPLEXES
67
EXERCISES
1. Show that if p0 = pv the defining conditions of A(p0, px) and of T(p0, p±)
both yield the single point p0 = pv
2. Show that (a) in E2 and E3, the definitions of A(p0, ??) and T{p§, ??) agree
with the usual line and line-segment of plane and solid geometry; (b) in En,
linear spaces are convex.
3. Which of the following sets are convex and which linear? Justify your
answers: (a) The set \x{\ < 1 (i = 1, . . . , n). (b) The set \x{\ > 1 (i = 1, . . . , n).
(c) The part of En where all coordinates are positive, (d) The part of En where
exactly one coordinate is negative, (e) The locus ?"=1 #г- = 1. (f) The locus
??-?*? = 1.
4. Show that (a) the intersection of any collection of convex sets is convex,
(b) the intersection of any collection of linear subspaces of ? is linear. Hence
justify Statement (C).
5. Let ? = 3 and let p0, pv and p2 be the points p0 : @, 0, 0), ?? : A, 0, 0),
and p2 : A, 2, 0). Write out D.6) and D.7), then verify that they define the
(xv ?2)-plane and the closure of the triangular region p0p1p2-
6. Discuss D.6) and D.7) for the case ? = 3 and the triple of collinear points
P0 : @, 0, 0),Pl : A, 0, 0), p2 : (-1, 0, 0).
7. Show that the conditions D.6) define a linear space.
8. Show that conditions D.7) define a convex set.
4-2. Dimension Numbers in En
(A) If к + 1 is the smallest number of points whereby a given
linear subspace A ^ En can be determined, then к is the dimension
number or (numerical) dimension of Л. The existence of such a
number к is a consequence of Lemma 2 below. We will call Л а
?-plane (short for ^-dimensional hyperplane). The agreement of
numerical dimension with Brouwer dimension (Art. 3-8) will be
established later (Art. 6-3).
Lemma 1. Euclidean тг-space En is ^-dimensional.
Proof of Lemma. In the first place (Exercise 9), En is the linear
space determined by a particular set of ? + 1 points. It remains to
show that En is not the linear space determined by a set (p0, pv . . . ,
pk) of points where к < n. This amounts to showing (Exercise 10)
that if к < ?, then there exists a point of En not on the locus ? defined
by D.6).
We will assume a number of basic properties of determinants and
matrices which the reader can readily find in the literature. Among
the most fundamental is that a system of equations
D.8) Vi = y?+ ? anxi (< = 1,. . ., л)
/{=1

68
INTRODUCTORY TOPOLOGY
Ch.4
can be solved in the form
D.9) *, = *?+ t^ (j=l,...,n)
г = 1
if and only if the determinant \aH\ is not zero.
It is a straightforward procedure to generalize the basic concepts
and results of plane and solid geometry to higher dimensions. We
present a few essential generalizations now and introduce others as
needed.*
(B) Equations D.8) can be interpreted as a transformation of
coordinates in En, from the given system (x) to (?/), and D.9) is
then the inverse transformation. In general, the euclidean distance
formula
D.10) p2(K, . . . , ??), (bv . . . , bn)) = I (a< - ?,J
г = 1
defined in terms of ^-coordinates, is transformed into a more general
quadratic expression in terms of ^/-coordinates.
Theorem 1. The euclidean distance formula is preserved by the
transformation D.8) if and only if the matrix (aH) is unit-orthogonal,
that is,f
? ?? if г Ф j
D.11) 2 ahfihi = ??= \ . ' (ij = h · · · , n).
/i=i Ц if г = j
We refer to the literature ([S-S], for example) for readers unfamiliar
with this result. Alternatively, we suggest that the reader verify the
result for ? = 2 and 3, then develop his own general proof.
(C) If the matrix (aH) is unit-orthogonal, then (y) is a rectangular
cartesian system. Thus (#), in particular, is such a system. If {aH) is
not unit-orthogonal, (y) is a skew coordinate system. In either case,
D.8) and D.9) are affine transformations. An (affine) coordinate
system in En is a coordinate system obtainable from the original
system (x) by an affine transformation.
(D) According to convenience, we interpret a point ? = (xv . . . , xn)
as an ordered set of ? real numbers or else as a vector ? with components
(xv . . . , xn) and length {x\ -f · · · + xffi. In the latter interpretation,
we use vector addition ? -\- x' = (x^ -j- x^ . . . , xn -j- x'n ). The unit
point (vector) on the a^-axis, in terms of Kronecker deltas, is
D.12) ej = (??? ?2? . . . , dnj) (j = 1, . . . , ?).
* For a more detailed discussion, see [S-S].
f The symbol ог, in D.11) is the Kronecker delta, defined as 1 when the subscripts
are equal and 0 for unequal subscripts.

Art. 4-2] COMPLEXES 69
The (coordinate) (x{ , . . . , #г-л)-р1апе is defined by Xj = 0 (j ? iv . . . ,
ih). It is a coordinate axis if h = 1.
(E) If (y) is a rectangular cartesian system, and ? is the angle
between vectors y, y' of unit length, then cos & = ?*=1 у%у[ (Exercise 13).
(F) To give another useful interpretation to D.8) and D.9), let (x)
be a coordinate system in En and let (y) be a coordinate system in a
second euclidean тг-space E%. Then D.8) defines a linear* homeo-
morphism Я : En -+ Eq and D.9) defines its inverse Я-1. This homeo-
morphism is uniquely determined, among linear homeomorphisms, by
the requirement that the origin, or 0 vector, in the ^-system map onto
y° = (?/0, . . . 5 y®) in the ^/-system and the unit vector e,- of the x-
system map onto y° + ay where a^ = (ajV . . . , ajn) (j = 1, . . . , n).
In En, with the coordinate system (#), let q0 be the origin and let
ql9 . . . , qk be a set of & points (& < n). Let the ^-coordinates of g, be
denoted by FЯ, . . . , bjn) (j = 1, . . . , A).
Lemma 2. The space Л* = Л(д0, . . . , qk) is ^-dimensional if and
only if the п-Ъу-к matrix {bH) is of rank к (Exercise 14).
If (bH) is of rank k, the plane ?? = Л(д0, . . . , qk) can be defined
parametrically by the equations (Exercise 14)
к
D.13) Л* :x{= ? ?;?· (*" = 1,. . . , ?)
J = l
and (wl5 . . . , %) can be interpreted as a coordinate system on ?*, with
origin q0 and with q0 as unit point on the wy-axis.
(G) The coordinate system (uv . . . , uk) on ?* is rectangular
cartesian, in terms of the metric of En, if and only if ?*=1 6Л*6^ = dhj
(h = 1, . . . , k\ j = 1, . . . , k) (Exercise 15). This means that then and
only then will [??=1 « - <J]< equal the distance [??=1 {?[ - x'-Jf
between two points и', и" on ?*.
EXERCISES
9. Let q0 be the origin and qt the unit point on the xraxis (i = 1, . . . , n) in
En. Prove that # = A(q0, qv . . . , gj.
10. Show that A(p0, pl9 . . ., pk) cannot contain the origin and the unit points
on all the axes in En if к < п.
11. Show that the determinant of a unit-orthogonal matrix equals +1 or —1.
12. If (y) is a skew coordinate system, given by D.8), with inverse D.9), what
is the distance formula in terms of (?/)?
13. Establish Statement (E). Obtain an analogous formula for a skew system.
14. (a) Prove Lemma 2. (b) Establish D.13), using D.6).
15. Establish Statement (G).
* That is, a homeomorphism mapping lines onto lines.

70
INTRODUCTORY TOPOLOGY
[Ch.4
4-3. Barycentric Coordinates
An exceptionally useful tool in combinatorial topology is a system
of barycentric coordinates, defined below.
(^4) Suppose the п-Ъу-к matrix (a^) (j = 1, . . . , k; i = 1, . . . , n)
is of rank k. Then, just as in the case of Л* in Lemma 2, the locus Ak
parametrically defined as follows is ^-dimensional.
к
D.14) Ak : x€ = а'ы + ? a'J, (i = 1, . . . , n).
i = l
Also, as above, (tv . . . , tk) can be interpreted as a coordinate system
on Ak. This defines the general &-plane Ak <= En, by contrast with
Л0, which is special in that it contains the origin.
(B) Equations D.14) can also be interpreted as defining a linear
homeomorphism between Ak and a euclidean &-space Ek, in which
(tv . . . , tk) is a coordinate system.
Equations D.14) are simplified by the use of
к
D.15) <0 = 1 - (t, + · · · + tk) => 2 к = 1
г=0
and of the notation
D.16) a0i = a'oi aH = aoi + a'H (i = 1, . . . , n; j = 1, . . . , k).
(C) In terms of (t0, . . . , ik) and the aH, Eqs. D.14) reduce precisely
to Eqs. D.6), so that
D.17) Ak=A(p0>Pv...,Pk).
One calls (t0, . . . , tk) the barycentric coordinates on Ak relative to
(Po> · · · > Pk)· In terms of them, the p's have the coordinates
Pj : (dj0, . . . , djk), using Kronecker deltas (j = 0, . . . , k).
Lemma 3. The set of all ?-planes in En is the set of all images of a
euclidean k-space Ek under linear homeomorphisms mapping Ek
into En.
Proof of Lemma. This follows directly from (B). We note that two
different homeomorphisms may yield the same plane.
(D) An important advantage of the barycentric system (t0, tv . . . , tk)
is that it is invariant under affine transformations of coordinates in the
space En. In other words, the barj^centric coordinates, relative to
(Po> Pi, · · · > Pk), of a point on Ak depend only on (p0, pv . . . , pkI and
are not changed (Exercise 18) if the (x)-system is replaced by the
(i/)-system, in the notation of D.8) and D.9).

Art.4-4] COMPLEXES 71
Since D.14) defines a &-plane if and only if the matrix (a^) is of rank
k, it follows from D.16) that D.6) defines a &-plane if and only if the
matrix
(Ki — ???) · · · ?? — ???)\
(«In — a0n) · · · (akn — aOn)l
is of rank k, hence (Exercise 19) if and only if the matrix
/«oi an · · · aki\
D.19) I
I a0n aln . . . akn I
\l 1 ... ,/
is of rank к + 1.
(?) If Ak = A(p0> pv ... , pk) is a &-plane and if pk+1 e En - Ak,
then A(p0, pl9 . . . , pk+1) is a (k + l)-plane (Exercise 20). Thus the
concept of ?-planes in En could be developed recurrently with respect
to increasing \^alues of k.
(F) If En = A(p0, . . . ,pn), then ?(?0, . . . , pn) contains inner
points (Exercise 21).
EXERCISES
16. What is the locus defined in E3 by the equations ?? = 1 — 4^ — 2t2,
x2 = 10^ + 5t2, x3 = -2tx - t2t
17. Using barycentric coordinate systems, write equations for (a) the line
containing p0 : C, 4) and px : D, —3) in E3, (b) the 2-plane in E* containing
p0 : ( -1, 3, 2, 0), ?? : A, 5, 2, -4), and p2 : D, -5, 4, -3). Verify that these
points determine a 2-plane.
18. Prove Statement (D).
19. Show that the rank of the matrix D.18) is less by 1 than that of D.19).
20. Prove result (E). Also show that if (q0, . . . , qk) is a set of points such that
Mp0> · --,Pk) = M%> --><Ik), and if qk+1 ? A(p0,. . ., pk+1) - A(p0, ...,pk), then
Mq0, · · · , qk+1) = A(p0, . . . , pk+l).
21. Show that if Ak = A(p0, . . . , pk) = A(q0, . . . , qk), then the barycentric
coordinates on Ak relative to (q0, . . . , qk) and to (p0, . . . , pk), respectively, are
expressible linearly and homogeneously in terms of one another.
22. Prove result (F).
4-4. Simplexes
The points p0, pv . . . , pk in En (k < n) are said to be (linearly)
independent if A(p0, pv . . . , pk) is a ?-plane. Even if no two of the

72 INTRODUCTORY TOPOLOGY [Ch. 4
points pt coincide, A(p0, pv . . . , pk) could be a j-plane for any j e
A, 2, . . . , k) (see Exercises 25, 26). If p0, pv . . . , pk are independent,
they determine an (open) ^-simplex, sk, analytically defined by
к
(a) xi = ^aHti (i=l,...,n)
D.20) sk = poPl ...pk
i = 0
к
(b) 2*,= i
i = 0
I (c) t,>0 (j = 0,l, ...,&)
where (ая, . . . , aJn) are the ^-coordinates of pi (j = 0, 1, . . . , k).
Comparing D.7), we see that sk, to be called a closed ^-simplex, is
given by
D.21) Sk=T(p09pv...9pk).
(A) The unmodified term ^-simplex will always mean open simplex.
If к > 0, the points of sk are the inner points of sk, and sk — sk is the
boundary of sk or of sk. It should be noted that sk is not an open subset
of En, unless к = n. However, sk is an open subset of Ak = A(p0,
. . . , pk) and sk — sk is its boundary, relative to Ak, where Ak is
regarded as a subspace of En with the induced topology. The
vertices of sk = p0 . . . p^are (p0, . . . , p^.
(B) For the first few values of k, we have (Exercise 24): A)
s° = Po> a point; B) s1 = p0pv a finite linear interval, that is, a
segment without its end points; C) s2 = ?^???^ & plane triangular
region; D) s* = P0P1P2P& ^ne interior of a tetrahedron. The symbol
s-1 (a ( —l)-simplex) will be interpreted as the vacuous set.
(G) A boundary ^'-simplex of sk = p0 . . . pk means either s-1 or a
simplex sj = qQqx . . . q0 where (g0, ql9 . . . , q}) is a proper subset of
(Po> · · · > Рл) 0' = 0, . . . , A; — 1). A face of sk means a boundary
simplex or sk itself. Each point of sk is on a unique face of sk (Exercise
27). The total number of faces of sk is 2*+1, of which the number of
^'-dimensional faces (to be called j-faces) equals the binomial coefficient
An alternative recurrent definition of sk, equivalent to the above,
is as follows: A 0-simplex is a point, s° = p0> and its boundary simplex
is the vacuous ( —l)-simplex, s_1. Assume, for a non-negative integer
j < n, that a j-simplex, s> = р0рх . . . pj} and its boundary г-simplexes
(i = — 1, 0, . . . , j — 1) have been defined in such a way that A)
sj cz \j cz En for some ^'-plane A\ B) ?; is an open set relative to A\
C) sj — sj is the union of the boundary simplexes ofs3'. Given a point
pj+1 eEn — A\ the (j + l)-simplex
D.22) s}+1 = sjPi+1 = р0рг. . . pj+1
is the set of all points each on a 1-simplex pj+1p as ? ranges over s*.

Art. 4-4] COMPLEXES 73
The boundary simplexes of sj+1 are A) the 0-simplex pj+1, B) sj and its
boundary simplexes, C) the (i + l)-simplexes pj+1sl as si ranges over
the boundary simplexes of sj. The closed simplex sj+1 is the union of
the segments [pj+ip] as ? ranges over sj.
Etymologically, the term barycenter suggests center of gravity. In
En, with the rectangular cartesian coordinate system (xv . . . , xn),
imagine a particle of mass wi at the point pi : (an, . . . , ajn) (j =
0, 1, . . . , k) and let W = w0 + · · · + wk. The formula for the center
of gravity (xv . . . , ?n), generalized, is
? к к
D.23) xt = — У wfla = ? tfla (< = 1, . . . , w)
W j = о j = о
where tj = wJW => ?*=0 tj = 1. This suggests a "physical" reason for
the invariance of barycentric coordinates (Art. 4-3(D)).
(D) Let the vertices (p0, . . . , pm) of an га-simplex sm be
partitioned into two complementary sets (pio> . . . , pik) and (pJQ, . . . ,
Pi ). Let sk = ?,- ... p? and s™-*-1 = ? ... p. . Then sk, is
defined as follows, in terms of the barycentric coordinates (t0, . . . , tm)
relative to p0, . . . , pm:
* i*' = 0 («? = «?0' * * * ^гп-к-l)} k
D.24) **: ? ? =>2^=1·
U<>0 (i = i0,...,ik) J h=o
It follows that (t{, . . . , ^. ) are barycentric coordinates on sk relative to
(Pi0> · · · > РгА)· The faces sk and $w-fc-i are opposite faces of sm, and we
will use such notation as
D.25) sm = Л™-*-1 = ? . . . piksm~k~\ and so on.
B?) Now let (u) = (u0, ...,%) and (г;) = (v0, . . . , vk) be
barycentric coordinates, relative to (p0, . . . , pk) and (g0, . . . , qk)
respectively on two ?-planes A\ = Ak(p0, . . . , pk) and Л* = Ak(q0, . . . , qk).
Let ? : ?* —* ?| be the mapping such that g = ?(?) if and only if the
(^)-coordinates of q equal the (w)-coordinates of p. Then (Exercise 28)
? is a linear homeomorphism uniquely determined, among linear homeo-
morphisms of A\ onto ?*, by the conditions pj -+ qi (j = 0, . . . , k). We
will frequently be interested in the restriction ? | sk where sk = p0... pk.
EXERCISES
23. In E*, let p0,pvp2 be defined as in Exercise 17b above. Find the
«-coordinates of (a) the center of gravity of three particles of equal masses at
p0, pv and p2; (b) the point on s2 = p0p1p2 with barycentric coordinates
D> ?> 0)· Show that these two points are collinear with p2.
24. Prove all parts of (B) for the case En = E3.

74 INTRODUCTORY TOPOLOGY [Ch. 4
25. In the plane E2, define T(S) analytically and describe it geometrically
where S consists of (a) the two points A, i) and (— 1, — 1); (b) the three points
A, 1), B, 2), and C, 3); (c) the points @, 0), A, 0), @, 1); and (d) the points
A, 1), A, -1),(-1, 1), (-1, -1).
26. Show that T(p0, pv p2, ?>3), where the p's are four different points of E3,
is the closure of one and only one of the following: (a) a linear interval, (b) a
plane triangular region, (c) a plane quadrilateral region, (d) the interior of a
tetrahedron.
27. Prove the last two sentences of (C).
28. Establish the italicized part of (E).
4-5. Complexes
Simplexes are the building blocks .for the structures, known as
complexes, with whose topological properties we will be mainly
concerned. The 1-complexes include the linear graphs of Chapter 1, and
the 2-complexes include the surfaces of Chapter 2. In general, a 2-
complex can be described as a collection {s} of simplexes of dimensions
<2 such that A) {s} contains at least one 2-simplex; B) {s} contains
all boundary faces of each of its members; and C) if sj e {s} and
sk e {s}, then the intersection sj ? sk is the closure of a common face
of sj and sk.
Boxlike building blocks, square and cubical for example, might
seem more natural and, indeed, can be advantageously used for many
topological purposes. However, simplexes have many advantages
stemming from the fact that a j-simplex is the convex hull of the
smallest number of points, j + 1, of a euclidean space, whose convex
hull can be j-dimensional.
(A) A formal definition of a finite m-complex Km = {s} in En is as
follows: It is the-tmien л s«.4 of d.^o.^+ «>.™р1«*'*
D.26) {s} = (s'1) U {s0} U {s1} U · · · U {sm}
of singleton is'1) and certain finite collections . ,c
D.27) {**} = (*?,-¦¦,<) (i = 0,l,...,w)
of-disjoint fcrsimpte-xes~Hnutrthat A) each face* of a simplex of the set
{s} is a simplex of {s} and B) y.m > 0. In this definition, m can be any
integer > — 1. If m = — 1, ?~? consists of (s_1) alone.
(B) An w-complex Km contains at least 2w"rl simplexes, since by
B) it contains at least one /^-simplex and all its faces (see Art. 4-4(C)).
Furthermore, a,- ;> Cm_.ltj_v The collection of all the faces of an m-
simplex is the simplest example of an m-complex.
* Singleton (s-1), the vacuous (— 1)-dimensional simplex, is included for the sake of
Property A) and to simplify the statements of other properties and results.

Art. 4-6] COMPLEXES 75
(C) Occasionally we will be concerned with infinite complexes. An
infinite complex Km in En is defined exactly as in (A), except that A)
{sk} = (s*, s\, . . .) is denumerably infinite and B) each point of En has
some neighborhood intersecting only a finite subset of {s}.
(D) We will write sh < sk or sk > sh to symbolize the relationship
of sh being a boundary simplex of sk (implying h < k), and we will then
refer to (sh, sk) as incident simplexes. We will write sh < sk or sk > sh
to signify that sh is a face of sk, which is consistent with sh = sk.
Thus the symbols (<, >, <, >) express relationships among
simplexes, as well as among real numbers.
(E) Consider a set A and a relation <. It is said that < orders A
if A) for any two elements a and a' of A, exactly one of the relations
a < a', a' < a, a = a' holds; B) a < a' and a' < a" imply a < a". It
is said that < partially orders A under the same conditions modified
by changing "exactly one" to "at most one" in A). If < orders
[partially orders] A, it will be said that <, meaning < or =, also
orders [partially orders] A. Equivalent formulations of a < a', a < a'
are a' > a, a' > a, respectively.
EXERCISE
29. Show that the simplexes of a complex {s} are partially ordered by <, but
that they are not ordered by this relation.
4-6. Polyhedra; Topological Complexes
If L is a set of simplexes, then \L\ will denote their point-set union
(the set of all points each on at least one of them), regardless of
whether or not L is a complex.
(A) If К is a complex, \K\ is called a polyhedron, finite or infinite
according as К is finite or infinite, and К is a triangulation of \K\. We
will frequently use such notation as ? = \K\ for a polyhedron. A
given polyhedron ? admits infinitely many triangulations, if its
dimension is positive.
A finite polyhedron is a compactum, since it is a closed bounded
subset of En (Chapter 3, Theorem 14).
(B) It is instructive to interpret К = {s} as a topological space
whose elements are the simplexes {s}. To accomplish this, we define
a subset L c= К of the simplexes {s} as closed if L contains each face of
each of its simplexes; that is, if (s e L, t < s) => t e L. This is
equivalent to the statement that L is a complex. We then call L a
subcomplex of K.
(C) A set L c= К of the simplexes {s} is open if it is the complement
of a closed set. This is equivalent (Exercise 33) to the condition

76
INTRODUCTORY TOPOLOGY
[Ch. 4
(s e L, t > s) => t g L, which says that L contains each simplex which
has a face in L.
Lemma 4. Relative to \K\9 the subset \L\ <= \К\ is closed if and only
if L is a closed subset of К and is open if and only if L is an open subset
of K.
Outline of Proof. The closure of a simplex sk e K, interpreted as a
point set, is sk. Its closure as a subset of К consists of all the faces of
sk, and they have sk for their union. The closure of \L\ is therefore the
point-set union of the closures of the simplexes of L, and the closure of
L as a subset of К consists of all the faces of its simplexes. This leads
to the part of Lemma 4 relating to \L\ and L being closed. The other
part follows because open sets are the complements of closed sets.
(D) If Km is an m-complex, then, for any к e ( —1, 0, ... , m), the
set of all the simplexes of Km of dimensions <k is a complex Kk,
called the ^-skeleton of Km. In the notation of Art. 4-5(Л), Кк =
(s-1) и {s0} и · · · U {sk}. It is a subcomplex of Km.
^> (E)[ln particular, the 1-skeleton of Km is a linear graph K1 = G in
the terminology of Chapter 1. A path on K1 is there defined, as is the
concept of connectedness.) We say that Km is connected if K1 is
connected. If ? e {s0} is a vertex of Km, hence of K1, then the subgraph
of K1 consisting of all vertices, plus incident edges, which can be joined
to ? by paths is a component K\ of K1. The simplexes of Km each
having vertices on a component K\ of K1 constitute a component K%
of Km. A component of Km is a maximal connected subcomplex of Km.
(F) Now suppose Km finite, and let (Kl9. . . , ?? ) be its components.
Then \K{\ is a component of \Km\ and is a connected polyhedron
(Exercise 34), in the sense that each pair of points on \Kf\ can be
joined by a broken line on \K™\. Each component of a complex Km is a
subcomplex and is both open and closed.
(G) Later (Art. 6-9(^)) it will be seen that a polyhedron ? = \Km\
is connected if and only if it has no proper non-vacuous subset which
is both open and closed.
A topological polyhedron means a space ? homeomorphic to a
polyhedron ? = \Km\ in a euclidean space E. A topological ^-simplex
means a space ak homeomorphic to a ^-simplex sk с ?. Consider a
particular homeomorphism h : ? —> ?, and let К = {s} be a trian-
gulation of P, so that ? = \K\. If
D.28) ak = h(sk) where sk e {s},
then ak is a topological ?-simplex on ?. The aggregate Я = {?} of
such topological simplexes is a topological complex, and its point-set
union is |Л| = П. We refer to Я as a triangulation of П.
A simplex sk <= E, as defined in Art. 4-4, will be called a linear
simplex when it is important to distinguish it from a topological

Art. 4-7]
COMPLEXES
77
simplex. Similarly, К will be described as a linear (simplicial)
complex to distinguish it from the topological complex Я, and ? = \K\
will be called a simplicial polyhedron to distinguish it from the
topological polyhedron ? = |Я|. When either the meaning is clear from
the context or a statement applies to both cases, the modifying
adjectives will be omitted.
(Я) We remark, for later reference, that a triangulation Я = {a} of
a topological polyhedron ? is defined by a homeomorphism In : ? —* ?
and a triangulation К of P. The concepts of face, boundary face, and
incidence relation, symbolized by <, are carried over by In from К to Я.
So are the concepts of connectedness, components, and so on.
EXERCISES
30. Let ? be the convex hull in E3 of the six vertices p0 : @, 0, 0), p± : A, 0, 0),
p2 : A, 1, 0), q0 : @, 0, 1), qY : A, 0, 1), g2(l, 1, 1). Show that the 3-simplexes
РоЯоЯ1Я2> PoV\(licl2i Р0Р1Р2Ц2 and their faces are a complex K3 with ? for its point-
set union.
31. In the complex K3 of the preceding exercise give the smallest closed and
open subsets of simplexes (a) containing the 1-simplex p0q2> (b) containing the
2-simplex p0q1q2-
32. Let ? = S\ U S2,, where S2 and S2 are the spheres of radius 1 about the
points B, 0, 0) and ( — 2, 0, 0) in an E3. Show that ? is a topological polyhedron
with S2 and S2 as components and that S2, as a subspace of ?, is both open and
closed.
33. Establish the italicized sentence in (C).
34. Demonstrate result (F).
4-7. Abstract and Generalized Complexes
An abstract ?-simplex is a set of к + 1 points, symbolized thus:
5* = (Po, Pv · · · > Pk) WP0P1 -Pic· The form^o^! . . . pk will be used
only where confusion with a linear simplex is unlikely or immaterial.
A 0-simplex is also called a vertex. A subset of sfc is a face of sfc, and a
proper subset is a proper or boundary face.
Let S be an arbitrary point set. An abstract complex is a collection
Я = {s} of abstract simplexes such that A)(ea,ch O-simplex of ft is a
point of S) and B) each face of an element of Я is an element of Я. We
describe Я as finite if S is finite, and denumerable if S is denumerable.
We describe Я as m-dimensional if it contains an m-simplex but no
(w + l)-simplex, hence no ^-simplex for any ? > m. This does not
define a geometric dimension, since an m-simplex is a finite point set.
To suggest the generality of abstract complexes, note that the
collection of all triples of points, together with all pairs, singletons, and
the vacuous set in E2 or, for that matter, on E1 is an abstract 2-
complex.

78
INTRODUCTORY TOPOLOGY
[Ch. 4
The relations < and < are defined for abstract complexes just as
for linear and topological simplicial complexes, and < partially orders
the elements of an abstract complex.
Abstract complexes originated as abstractions of simplicial
polyhedral complexes. Indeed, if К is a simplicial complex, we obtain from
it an abstract complex Я by letting 8 be the set of all vertices of К and
defining an abstract ^-simplex of Я as the vertices (p0, pv . . . , pk) of
a ^-simplex of K.
(A) We will refer to Я, just defined, as the abstract complex of K,
and we will call K, or \K\, a simplicial, or polyhedral, realization of Я.
We prove, in the next section, that an arbitrary abstract complex Я
admits a simplicial realization in a euclidean space ? if Я is denumer-
able and each of its O-simplexes is incident with only a finite set of
1-simplexes.
(B) Algebraic topology consists of A) an algebraic theory directly
concerned with abstract (or more general) complexes, B) theorems to
the effect that various results of this algebraic theory have topological
significance for realizations of such complexes, and C) the use of
mappings to apply the algebraic theory to topological spaces more
general than topological polyhedra.
(C) A collection A = {a} of objects is called a generalized complex
if it satisfies the following conditions:
A) With each ? e A is associated a non-negative integer k, the dimension
of ? = afc, and afc is called an abstract &-cell.
B) The set A is partially ordered by a relation <, such that aJ < afr =>
j <k.
It is obvious that generalized complexes are more general than
abstract complexes. In order to make effective use of them, one must
impose restrictions in addition to the defining conditions. We have
given Definition (C) to round out our collection of definitions with the
most general one that appears useful.
4-8. Realizations of Abstract Complexes
Theorem 2. Let Я = {s} be a denumerable abstract m-complex,
each of whose vertices is incident with only a finite set of 1-simplexes.
Then Я admits a linear simplicial realization in a euclidean тг-space En,
for ? sufficiently large.
Proof. Case 1 (Я is finite). We treat this case separately, because it
is particularly useful and simple.
Let p0, pv . . . , pN be the vertices of Я. In an EN, consider a linear
JV-simplex ? = qQqx . . . qN. Let {t} be the set of faces of ? defined by
the condition that a face tk = qt q{ ... qik of ? belongs to {t} if and only
if the correspondingly numbered set of p's, namely p{ ... p{ is a

Art.4-8] COMPLEXES 79
simplex of ft. It is a simple matter to verify that К = {t} is a
realization of ft.
(A) The realization К of ft just defined, that is, its realization as a
subcomplex of the faces of an JV-simplex ? where N + 1 is the number
of vertices of ft, will be called a natural realization of ft. It is unique,
given ?, up to a renumbering of the vertices. A barycentric
coordinate system (u) = (u0, uv . . . , us) on ? affords a natural
coordinate system, unique up to a permutation, on K. If ? e tk e К and
tk = qioqi . . . qik, then (ut , ui¦ , . . . , ut) are all positive at ? and are a
barycentric coordinate system on tk. All the other u's are zero on tk.
Case 2 (ft is not necessarily finite). The following argument proves
Theorem 2 in all its generality and, incidentally, yields the following
subsidiary result.
Corollary. The value ? = 2m + 1 is sufficiently large in
Theorem 2.
The following lemma contains the heart of the proof.
Lemma 5. It is possible to define an infinite set of points {q} =
q0, qv q2, ... in a euclidean space En (n > 0) in such a way that
A) Each set of к + 1 of the points {q} determines a linear ^-simplex,
provided к < n.
B) The ^-coordinate of qt equals i (i = 0, 1, 2, . . .).
Proof of Lemma. As the basic step of a recurrent argument, let q0 be
the origin in En. The following hypothesis, read for J = 0, is fulfilled.
Hypothesis. For some non-negative integer j, the points q0, qv
. . . , qj have been so defined in En as to satisfy Conditions A) and B)
of the lemma with [q] replaced by (g0, qv . . . , q^.
Then, as a consequence of the hypothesis, if 1 < к < ? each subset
of (k + 1) of the points (q0, . . . , g,) determines a &-plane.
(B) A &-plane determined by a subset of (q0, . . . , q^ intersects the
(n — l)-plane xx = j + 1 in a (k — l)-plane, and the collection of all
such (k — l)-planes of intersection for ull values of к satisfying
1 < к < ? — 1 does not cover the (n — 1)-plane ?? = j + 1
(Exercise 37).
As a consequence of (B), it is possible to define qj+1 as some point
such that A) xx = j + 1 and B) each subset of к + "L of the points
(#o> · · · > Я.3+1) is linearly independent ?? 1 < к < ? — 1 and hence
so is each subset of k*of these points if 1 < к < п. Thus the
hypothesis is verified with j + 1 in place of n. Lemma 5 follows.
Now let p0, pv p2, . . . ,pN or p0, pv p2, . . . be the vertices of ft
according as ft is finite or infinite. Let ? = 2m + 1, and let {q}
satisfy Lemma 5. Let К denote the set {t} of all &-simplexes tk =
%4ix · · · qik (k = 0,1, ... ,m) such that s* = pioph . . . pik is an

80 INTRODUCTORY TOPOLOGY [Ch. 4
abstract ^-simplex of ft. We will refer to each pair tk and sfc as a pair of
corresponding simplexes. If ft is finite, then qN+v ^?+2, · · · are not
used. C^*>
Lemma 6. (a) The set К = {t} is a niiifllim м ra-complex/\in En.
(b) If ($j, tj) and (sfc, tk) are two pairs of corresponding simplexes then
sy < sfc if and only if V < tk. (c) The simplexes {t} are disjoint.
Proof of Lemma. Let вг = qt ... дг. and s2 = qh . . . qh be two
simplexes of К and let r be the number of distinct vertices in the set
(?v ' ' ' ' 9V ?V ' ' ' ' ?**)· Then r<j+k + 2<2m+2=n + l.
Hence, by Lemma 5, these distinct vertices determine a linear (r — 1)-
simplex cr~1, generally not an element of K, which has sv s2 for faces.
A proof of Lemma 6, hence of Theorem 2 and corollary, in the finite
case is now easy to complete (Exercise 38).
To complete the argument for the infinite case, it suffices to show
that a sufficiently small neighborhood of ? e En intersects only a
finite number of the simplexes of K. But if к <0ВА is an integer
exceeding the a^-coordinate of ? and a is a neighborhood of ? of
diameter less than 1, then a can intersect only those simplexes, finite
in number, each having a vertex in the set (q0, qv . . . , qk). The
simplexes with no vertices in this set are all on the subspace ?? >
к + 1 of En.
(C) Two complexes К = {s} and L = {t}, whether linear,
polyhedral, or abstract (and they need not both be of the same kind), are
isomorphic if there exists a one-to-one/mapping ? : {s} —> {t} which
preserves incidence relations; that is, sj < 8коср(8*) < q)(tk). Such
* a mapping is an isomorphism.
EXERCISES
/ 35. Let Я be the abstract complex of the 1-complex K1 whose vertices are the
vertices of a square and whose edges are the edges and one diagonal of the square.
Draw a figure showing a realization of Я as a subcomplex of the faces of a 3-
simplex.
36. Write out, and illustrate with figures, the proof of Theorem 2, Case 2, for
m = 1, ? = 3, thus showing that an abstract 1-complex has a realization as a
linear graph in E3, with line segments for edges.
37. Prove (B).
„ 38. For the case where Я is finite, complete the proof of Lemma 6 and deduce
Theorem 2, with corollary.
4-9. Isomorphisms and Homeomorphisms
Theorem 3. Let ? : {s} -> {t} be an isomorphism between two
linear simplicial complexes, К = {s} and L = {t}. Then there exists

Art. 4-10] COMPLEXES 81
one and only one linear homeomorphism,* /: \K\ -> \L\ such that
/ maps each s e {s} linearly onto ?(?).
Proof. Let the vertices of К and L be so numbered (pv p2, . . .)
and (qv g2, . . .) that qt = ?(??) (i = 1,2,.. .). If sk = pio . . . ph is
a simplex of K, then tk = qio . . . qik is the simplex tk = q)(sk). By
Art. 4-ЦЕ), there exists a unique linear homeomorphism mapping
sk onto tk so that p{ —> qih (h = 0, . . . , k). It will be described as the
homeomorphism induced by ? on sk. It is easy to verify that sj < sk
implies that the homeomorphism induced by ? on sj is a restriction
of that induced by ? on sk. Since each point ? of \K\ is on a unique
simplex s of {s}, it has a unique image, which we denote by ?(?), on
\L\, where ?(?) is the image of ? under the homeomorphism induced
on s by ?. The mapping ? defined by q = ?(?), ? e \K\, is a
homeomorphism satisfying Theorem 3. Its one-to-one-ness follows from
the fact that its inverse A-1 can be defined just as ? was defined, using
?~? in place of ?. Its continuity on \K\ follows, by a straightforward
though perhaps somewhat tedious argument, from its continuity on
each closed simplex s e {s}.
Theorem 4. Two topological polyhedra Yl1 and ?2 are homeo-
morphic if and only if they admit isomorphic triangulations.
Proof. By Art. 4-6(#), a triangulation ftx of U1 is defined by a
homeomorphism h : ? —> U1 and a triangulation К of P. If ?? : U1 —>
? 2 is a homeomorphism, then In^Jn : ? —> ?2 and i? define a
triangulation ft2 of ? 2 isomorphic to Stv
Now suppose E^ and ?2 admit isomorphic triangulations ftx and ft2.
This means that there exist linear complexes Kx and K2 and homeo-
morphisms h{ : \K{\ -> Пг- (г = 1, 2) which induce isomorphisms
?? : K{ -> Яг. If ? : Ях -> Я2 is an isomorphism, then ?^??? is an
isomorphism ^х -> i^2. Hence, by Theorem 3, there exists a
homeomorphism ? : \Кг\ -> |??2|. But А2ЯА1-1 is then a homeomorphism
U! -> ?2 and Theorem 4 is proved.
4-10. Simplicial Mappings
Let Я = {s} and ? = {t} be two abstract complexes and let Я0 and
fi° be their 0-skeletons; that is, their respective sets of vertices.
Consider a mapping ?0 : Я0 -> ?° of the vertices of Я into those of
fi. It is not required that ?0be either one-to-one or onto.
The mapping ?0 : Я0 -> ?° is simplicial if the image ?0(?0, . . . , ^,)
of a i-simplex sk = (p0, . . . , pk) of Я is a simplex t' = (g0> · · · , ?/)
of ?. The notation does not imply that qt is the image of piy but
Such a homeomorphism is frequently described as piecewise linear.

82
INTRODUCTORY TOPOLOGY
[Ch.4
(<7o> · · · > Qj) *s merely the set of distinct points among the images
<Po(Pi) (i = 0, . . . , A).
Theorem 5. Suppose Я and ? satisfy the hypotheses of Theorem
2, so that they have linear realizations К and L. Let ? : Я -> ? be a
simplicial mapping. Then there exists a unique linear mapping
? : |??| -> |i| which maps the realization of sk e Я onto the realization
of <p(sfc) ? ?.
Proof. We will interpret Я and ? as the abstract complexes of К
and L, so that the same symbols can be used for the vertices of К and
Я and L and ?. We will use the same symbol ? to denote the mapping
{s} -> {t} determined by ? : Я -> ?.
Lemma 7. Let sk = p0 . . . pk be a simplex of K9 and let ?J* =
q0 . . . qi be its image, ? = q)(sk). For each vertex qh of ? let 7гл =
<p_1(gj, the subset of (p0, . . . , ??) mapping onto qh. There exists a
unique linear mapping ? : sk -> ?5 such that ЯGгл) = qh (h = 0, . . . , j).
Proof of Lemma. Let (w0, . . . , uk) and (v0, . . . , Vj) be the bary-
centric coordinates on sk and V relative to (p0, . . . , pk) and (g0, . . . , дД
respectively.
(Л) A mapping ? satisfying the lemma is defined as follows
(Exercise 40). We will say that the coordinate ut is associated with p{
(i = 0, . . . , k). Then ? maps a point (u0, . . . , uk) onto the point
(v0, . . . , Vj) where vh is the sum of the w's associated with the vertices
in the set ?? = (p^iq^; that is,
D.29) vh= 2 Щ (* = 0,...,j).
We will refer to ? as the mapping induced by ?.
The remainder of the proof of Theorem 5 follows along the same
lines as that of Theorem 3. In fact, Theorem 3 deals with the special
case of Theorem 5 where the simplicial mapping ? is an isomorphism.
(B) It follows that a simplicial mapping ?0 : Я0 -> ?° has a natural
extension into a mapping ? : Я —> ?. We will also describe ? as
simplicial.
EXERCISES
39. Discuss the mapping ? of Statement (A) geometrically in the case к = 3,
3 =2.
40. Prove (A) and (B).
4-11. Barycentric Subdivisions
Combinatorial topology is concerned with abstract complexes (or
with generalized complexes) rather than with their realizations. It

Art. 4-11] COMPLEXES 83
is through such realizations or, more generally, through mappings
into topological spaces, that combinatorial topology is applied to
set-theoretic problems.
For the most satisfactory interplay between combinatorial and
set-theoretic methods, it would be desirable to define combinatorial
equivalence so that abstract complexes with realizations (Theorem 2)
are combinatorially equivalent if and only if their realizations are
homeomorphic. However, no combinatorial criterion* has been
discovered*!* whereby we can determine, in general, whether the
realizations of two such abstract complexes are homeomorphic. The
methods actually used to apply combinatorial results to topological
spaces depend on subdivisions of triangulations, especially barycentric
subdivisions, which we proceed to define and discuss. (See Art.
4-12F) below.)
(A) Let Я and Ях be two triangulations of the same topological
polyhedron, so that ? = |Я| = IftJ. Then Ях is a subdivision of Я
if and only if each simplex of Ях is on a simplex of Я. It follows
that each simplex a e Я is the union of the simplexes of Ях which
intersect a.
The only subdivision of a 0-complex K° is K° itself, which is
accordingly defined as the first barycentric subdivision of K°. Now
let m be a positive integer, let Km = {s} be a linear simplicial m-
complex, and suppose the first barycentric subdivision ?™~? defined
for the (m - l)-skeleton Km~x of Km. The boundary B™-1 of an
m-simplex sm e Km is then covered by a subcomplex ?™~? of Km~1.
Let ? be the barycenter of sm; that is, the point all of whose
barycentric coordinates relative to the vertices of sm equal l/(m + 1).
The first barycentric subdivision of sm then consists of A) the vertex p,
B) the complex B™~x, and C) the simplexes pP (see D.25) for notation)
for ? ? .BJ1-1. The first barycentric subdivision Kf of Km is the union
of the first barycentric subdivisions of its closed m-simplexes. It is
easy to verify that ?? is a linear simplicial m-complex and is a
subdivision of Km. (See Fig. 4-1.) This completes a recurrent definition
of the first barycentric subdivision of a linear simplicial m-complex
Km, with respect to increasing values of m.
The vth barycentric subdivision of Km is the first barycentric
subdivision of its (v — l)th barycentric subdivision, a definition recurrent
in ? = 1, 2, 3, . . . . The 0th barycentric subdivision of Km is Km. The
term barycentric subdivision without the numerical adjective vth will
mean the first barycentric subdivision.
* Such a criterion would have to be expressible in terms of the incidence relations
among the simplexes of a complex.
t (Added in proof.) While this book was in preparation, A. A. Markov showed (''The
Unsolvability of the Problem of Homeomorphism," Doklady Akad. Nauk SSSR 121
A958), pp. 218-220) that no such criterion can be discovered.

84 INTRODUCTORY TOPOLOGY [Ch. 4
Fig. 4-1. The first two barycentric subdivisions of s2. (a) K2 = the faces of s2;
(B) Let sm = р0рг . . . pm and let pit { be the barycenter of the
face piQph . . .Pi. of sm (Fig. 4-lb). Note'that pio...{, =Pr0...r, if
(r0, . . . , Tj) is a permutation of (i0, . . . , i}). A typical m-simplex of
the barycentric subdivision of sm is then
D·30) ST0... im = PioPiohPiohh · · · Pi0 . . . im
where (i0, . . . , im) is a permutation of @, 1, . . . , m). The barycentric
subdivision of sm is the set of all the faces of the (m + 1)! m-simplexes
г0г1 · · · гт'
Theorem 6. On the m-plane Am ofsm = р0рг... pm, let (w0,..., ww)

Art. 4-11] COMPLEXES 85
be the barycentric coordinate system relative to (p0, . . . , pm).
Then sf im is defined by the following inequalities:
D.31) <...im:^>\>--->V
Proof. The following lemma will be used.
Lemma 8. The barycentric coordinates (v0, . . . , vj on Am relative
to pio,Pioii> · · · >Pi0...im are elated to (u0, . . . , uj as follows:
D.32) utj = J —*- C = 0,..., m).
h = j h + 1
To establish this lemma, it is sufficient to make the simple
verification that Eqs. D.32) yield the correct ^-coordinates for the point
Pi ...i > h e @, . . . , m), when we substitute vh = 1 and let all the
other v's be zero. For A) we know that the u's are expressible linearly
in terms of the v's (Exercise 21), and B) the coefficients in D.32) must
be correct for each h in order to give correct results when vh = 1 and
vk = 0(k^h).
To complete the proof of Theorem 6 it is sufficient to show (Exercise
43) that the defining conditions Vj > 0 (j = 0, 1, . . . , m) of sf{ _im
are equivalent to ui. > и{._ (j = 1, . . . , m).
Corollary to Theorem 6. Each ^-simplex tk of the barycentric
subdivision of sm is definable by a set of conditions of the form D.31)
with all but к of the symbols > replaced by =, and tk is the common
face of all the m-simplexes D.31) from which its defining conditions
can be thus obtained.
Theorem 7. If each m-simplex of Km is of diameter less than d9
then each simplex of its vth barycentric subdivision Kv is of diameter
less than \mj{m + l)]vd.
Proof. The following two lemmas will be used.
Lemma 9. Let q e~sk, where sk is a linear ^-simplex, and let ? be a
point on sk such that p(q, p) = max p(q, r). Then ? is a vertex of sk.
resk
Proof of Lemma. Let sj be the face of sk containing p. If j > 0,
let p'p" be a segment on sj through p, with its end points on sj — sj.
Then (Exercise 45), either p(q, p') > p(q, p) or p(q, p") > p(q, p). This
contradiction of the definition of ? and q implies j = 0, which means
that ? is a vertex.
Corollary. Suppose p(p, q) = diam sk in Lemma 9. Then pq is
an edge of sk.
Lemma 10. Let d = diam sm and let ? = p(q, r), where qr is a
1-simplex of the barycentric subdivision S™ of sm. Thenx <dm/(m + 1),
with equality only if m = 1.

86 INTRODUCTORY TOPOLOGY [Ch. 4
Proof of Lemma. For m = 1, ? = d/2, verifying the lemma.
Assume the lemma with m — 1 in place of m. Then each edge ? of Sf
not incident with the barycenter q0 of sm is an edge to which the
inductive hypothesis applies for some face of sm. Hence ? <
d(m — \)jm < dm\(m + 1). Consider, finally, an edge pq0 of S™,
x = p(p} qQ). By Lemma 9, if such an edge maximizes ?, ? is a vertex
?j of sm. The barycentric coordinate ui on sm associated with pi equals
1 at ? j, l/(m + 1) at q0, and 0 at the barycenter qs of the face of sm
opposite pj. Since pj} q0, qs are collinear (Exercise 46), and since ui is
proportional to the distance from gy along q^p^ it follows that
ptioiPi) = mp(q3>Pi)Km + !)· But pfaotPi) = *\ and PteuPs) < d,
by the corollary to Lemma 9. This implies Lemma 10.
Theorem 7 now follows (Exercise 47·).
(G) As a consequence of Theorem 7, there exists, for each ? > 0,
an integer ? so large that each simplex of Kv is of diameter less than ?.
We will then say that Kv is of mesh ?.
EXERCISES
41. On a finite m-complex Km with vertices (pl9 . . . , pN) let (ul9 . . . , uN) be
the natural coordinate system as defined in Art. 4t-8(A). Show that щ =
max (uv . . . , uN) defines the union of the closed simplexes incident with p}- on the
barycentric subdivision K™ of Km.
42. Carry through the suggested proof of Lemma 8.
</ 43. Carry through the suggested completion of the proof of Theorem 6.
44. With the aid of the corollary to Theorem 6, show that each point of sm
is on one and only one simplex of the barycentric subdivision of sm.
45. In the proof of Lemma 9, show that p(q, pf) > p(q, p) or p(q, p") > p(q, p).
46. Show that q0, qjy and Pj are collinear in the proof of Lemma 10.
47. Deduce Theorem 7 from Lemmas 10 and 9, and the corollary to Lemma 9.
4-12. General Polyhedral Complexes
It is frequently convenient to consider polyhedral complexes which
are structures of c'cells" that are not simplexes; for example, the
surface of a cube with square faces, or of a dodecahedron with
pentagonal faces.
In En, with a coordinate system (x) = (xv . . . , xn), an open half-
space and a closed half-space are defined by inequalities, thus:
?
Open half-space: ^ aixi + a>o> 0,
Closed half-space: 2 aixi + ao ^ 0>
i = l
where at least one of the numbers (av . . . , an) is not zero.

Art. 4-12]
COMPLEXES
87
(A) A convex open polyhedral ?-cell cn in En is a non-vacuous
bounded set which is the intersection of a finite collection of open half-
spaces, and cn is a convex closed polyhedral w-cell.
Consider the intersection
D.34)
с : 2>гА + a0j > 0 (j=l,...,m)
of a finite set of closed half-spaces, and let Ek = Л(с). If с is vacuous,
we will denote Л(с) = 0 by 2?_1. If с is a single point, к = 0.
(В) If к > 0, let (г/) be a coordinate system in En such that i?fc is the
(Уг> · · · > 2/fc)_plane· The transformation from (x) to (y) carries D.34)
into a system
к
? ЪаУг
D.35)
such that
D.36)
г = 1
-г^г ? &(W > 0
r*·^
г=1
&oi > 0
1УЛ = 0
(j=l,...,»0
(А = ifc + 1, . . . , ?)
(j=l,...,W)
(/?, = ^ + 1, . . . , ?)
is non-vacuous. If ck is a bounded set, it is a convex open polyhedral
?-cell, and ck is a convex closed polyhedral ?-cell, to be referred to, for
brevity, as merely a cell ck or ck, respectively.
(C) The boundary of ck is ck — ck. It falls naturally into convex
J-cells (j < k), called faces of ck, defined by replacing some of the signs
> in the defining relations of ck by =. If this be done for all possible
subsets of the defining relations, then all the faces are obtained,
including the vacuous c_1, some of them possibly in several different
ways. Thus ck is reckoned as one of its own faces. See Exercise 49 for
the simplicial case.
While the replacement of > by = leads to a system formally
different from D.34), note that ? = 0 is equivalent to the pair of
relations ? > 0 and ? < 0.
A finite polyhedral m-complex Cm in En means a finite set {c} of
convex polyhedral j-cells (j < m), including at least one m-cell,
where A) every face of a member of {c} is a member of {c} and B) any
two members of {c} intersect in a common face, which of course may
be vacuous.
If each cell of Cm is a simplex, Cm is a simplicial complex.
(D) Let Gm and Cf be two finite polyhedral m-complexes which
coincide as point sets, \Gm\ = \??\. Then ?? is a subdivision of Cm if

88 INTRODUCTORY TOPOLOGY [Ch. 4
and only if each cell of Cf is on a cell of Gm. A subdivision each of
whose cells is a simplex is called a simplicial subdivision.
Theorem 8. A finite polyhedral m-complex Gm admits a simplicial
subdivision.
Proof. We prove Theorem 8 by giving a specific subdivision which
will be useful in Chapter 6.
The ^-skeleton of Cm means the polyhedral subcomplex Ck
consisting of the cells of Gm of dimensions < k.
Hypothesis. For some positive к < ??, Gk has a simplicial
subdivision G[, which contains as a subcomplex all the simplicial cells of
Ck (compare Exercise 49).
For к = 1, this hypothesis is fulfilled, and G\ = C1, since all
polyhedral 1-complexes are simplicial.
If all cells of Cfc+1 are simplicial, let G\+1 = Ck+1. If not, let c^1 be
an arbitrary non-simplicial cell of Gk+1. The boundary ck+1 — ck+1 is
covered by a subcomplex B\ of G\. Let ? be the barycenter* of cfc+1,
and let {pB\} be the set of all the simplexes psj for sj e B\.
(E) The set of all the faces of the simplexes {pB\} is a simplicial
subdivision of c\+1. If each non-simplicial cell of Ck+1 is thus
subdivided, a simplicial subdivision С\+1 of Gk+1 is obtained, which
satisfies the hypothesis with к + 1 in place of k.
(F) The above process, repeated for к = 1, . . . , m — 1, leads to a
simplicial subdivision C™ of Cm, which we will call its minimal central
subdivision, it being minimal in the sense that only the non-simplicial
cells are subdivided in passing from Cm to Cf.
(G) Two complexes are combinatorially equivalent if they can be
subdivided into isomorphic complexes.
Theorem 9. A topological polyhedron ? is a compact space if and
only if it admits a finite triangulation. If it admits a finite triangulation,
it does not admit an infinite triangulation.
Proof. Let Я = {?} be a triangulation of ?. The star ?,- of a vertex
of is the union of all the (topological) simplexes of Я incident with of.
The set {?} = (?13 ?2, . . .) of all such stars is an open covering of ?.
If Я is infinite, {?} is an infinite covering admitting no finite subcover-
ing, so that ? is not compact.
Now suppose Я finite. Then ? is homeomorphic to ? = \K\ where
К is a linear simplicial realization of Я in some euclidean space E. As
a closed bounded subset of ?, ? is compact (Chapter 3, Theorem 14);
hence (Chapter 3, Theorem 10) so is ?. The last conclusion of
Theorem 9 now becomes obvious.
* That is, the point xt = A/?) ??=? ?*,· (j = 1, . . . , ?), where qt : (au, . . . , ain)
(i = 1, . . . , h) are the vertices of ck+1.

Art. 4-12]
COMPLEXES
89
EXERCISES
48. Draw diagrams illustrating the minimal central subdivision (a) where
m = 2 and C2 consists of the faces of a square 2-cell с2 = Р±Р2РзРа anc* a 2-simplex
s2 = ?0???2^ (b) where m = 3 and C3 is the set of all faces of a cube.
49. Show that (a) a cell cn in En is 'an ?-simplex if it is definable by ? + 1
inequalities ?"=1 aijxi + a0j > 0, and (b) that the faces of cn, as defined in (C)
above, are the same as its faces in terms of the definition in Art. 4-4@).
50. Prove Statement (E).
51. Let ? be the unit segment 0 < ? < 1 and let it be subdivided using the
points ? = Oand 1/n (n = 1, 2, . . .) as O-simplexes and the intervals l/(n + 1) <
? < 1/n (n = 1, 2, . . .) as 1-simplexes. This yields an infinite subdivision of
the compact space ?. Why does it not contradict Theorem 9?

5
Homology and Cohomology Groups
Algebraic topology, to which we now turn, has the homology theory
of complexes as its most important and fundamental part.
5-1. Chains, Cycles, and Bounding Cycles
Throughout this chapter, attention is confined to finite simplicial
complexes, which may be linear, topological, or abstract. In
statements applicable to all of these three types, we will use the terms
simplex and complex without modifier.
The algebraic part of combinatorial topology employs oriented
simplexes and complexes. The distinction between a ^-simplex and an
oriented ^-simplex generalizes that between a segment p0px and a
directed segment p0px = —PiP^· Indeed, we orient a 1-simplex s1 =
PoPi by associating +1 with one of the permutations of р0рг and — 1
with the other; for example:
(+l)si = +si = +PoPv
E.1)
(-l)s1 = -s1 = -?0?? = +РгР0.
We refer to p0 and px as the initial and terminal points respectively of
+p0Px and represent +S1 by the vector ?0?? (Fig. 5-lb).
(A) The arrangements of the к + 1 vertices of a ^-simplex sk =
PoPi - - - Pk (Art. 3-2(B)) can be separated into the even permutations
and the odd permutations of the arrangement (p0, pv . . . , pk). The
association of sk with either of these two classes of permutations is an
oriented ^-simplex and may be denoted by sk, in which case the
association of sk with the other class is denoted by — sk.
In the case к = 2, for example, we use the notation
E.2) +52 = p^Pz = ???2?? = ??????
and
E.3) -s2 = ргр0р2 = p0p2Pi = P2P1P0 = ~PoPiP2> etc·
90

Art. 5-1] HOMOLOGY AND COHOMOLOGY GROUPS 91
(C) (d) ?2
Fig.5—1. Oriented j-simplexes 0 = 0, 1,2,3). (a) + s°= + p0; (ty+s^
PoPv (c) +s2 = P0P1P2 = P1P2P0 = P2P0P1'. (d) +s3 = P0P1P2P3 = P0P2P3PP etc·
Note that the even permutations assign one sense of description to the
boundary of s2, and the odd ones assign to it the opposite sense (Fig.
5-lc). Thus, intuitively, one may think of a 2-simplex as being
oriented either clockwise or counterclockwise.
In three dimensions also, a fairly simple intuitive interpretation of
orientation is possible. Suppose the "oriented 3-simplex"
E·4) +s3 = +PoP!P2Ps
is such that a right-handed screw with head at p0 would be driven into
the face PiP2Ps by a rotation in the sense of the cyclic permutation
PiP2Ps (Fig. 5-ld). It is easy to verify that if Pi Pi Pi Pi is an even
permutation of ?0???2?3, then a right-handed screw with head at pt
would be driven into the opposite face by a rotation in the sense of
РггРг2Рг - For an odd permutation, representing —ss, a left-handed
screw would have this property. Thus oriented 3-simplexes can be
interpreted as either "right-handed" or "left-handed." Otherwise
expressed, a 3-simplex +s3 with a right-handed [left-handed]
orientation has the property that if pt pt pt pt = +ss then the 2-simplex
+s2 = PiPijPi appears to be oriented clockwise [counterclockwise]
when viewed from p{ .
Since there is a unique permutation of a single point, we orient a
" 0-simplex s° = p by writing either +s° = +p and —s° = —p or
+s° = — ? and —s° = +p. We generally let +s° = +p.

92
INTRODUCTORY TOPOLOGY
[Ch. 5
A simplicial complex К is oriented when an orientation is assigned
to each of its simplexes. Since each ^-simplex sk e К admits two
different orientations, there are 2a ways of orienting К where ? =
?™=0 ai5 and аг is the number of its г-simplexes. In spite of the use of
the term oriented complex, one should bear in mind that the 2a
orientations of К are not different complexes but different orientations of
the same complex. Orientations are a tool for discovering properties
of complexes, much as a coordinate system in analytic geometry is a
device to facilitate the analysis of geometric properties.
The positive simplexes of an oriented m-complex will conventionally
be denoted as follows, with sk instead of -\-sk when it is clear that
oriented simplexes are meant:
m
E.?) ? = {s} = U {sk}
fc=0
where {sk} = (s^, . . . , skk) (the positive ^-simplexes of K).
(B) In the terminology of the Appendix (Art. A-4), let {sk} be
regarded as the initial basis for the integral module [sk] = [s\, . . . , s* ],
which will be called the (integral) &-chain group (?k. An element of
(?л is then a linear form
<*k
E.6) Ck = 2аг5? (аг an integer),
i = l
and is called an (integral) /c-chain. If к = — 1 or if к > ??, (?fc reduces
to a null set 0. If Dk = ?«*=? Ь$, then Gk + Dk = ?^=1 {a, + b^.
We next define, for each &-chain Ck, a (k — l)-chain called the
boundary chain, or simply the boundary, of Gk, and denoted by dCk.
Consider an oriented ^-simplex
E.7) tk = +ад1 . . . qk.
By omitting a vertex qt from the symbol д0дг . . . qk we obtain a
symbol, which we denote by q0 . . . $г- . . . qk, for the face of tk opposite
qt. It will prove convenient to orient that face thus
E.8) tf1=(-l)iq0---4i---q*·
In the case к = 2, for example,
t2 = Mife *o = Wi?2 = Ыъ
E.9)
*? = ?0$L?2 = —???2» 1\ = Wl?2 = №v
and the edges form a clockwise or counterclockwise circuit,
corresponding to the orientation of t2 (Fig. 5-2b).
(C) We define the boundary (chain) of tk as the (k — l)-chain
E.10) &* = ??-\
г=0

Art. 5-1] HOMOLOGY AND COHOMOLOGY GROUPS 93
(C)
Fig. 5-2. Oriented simplexes tk and boundary chains dtk. (a) dt1 = qx — q0;
(b) dt2 = tl + t\ + tj; (c) at3 = ??=01? (looking down on vertex q0).
a definition justified by the following result. See Fig. 5-2 for the first
few dimensions.
Lemma 1. The chain dtk depends only on the orientation of tk, not
on the particular permutation in E.7).
Proof of Lemma. A transposition of q5 and qJ+1 has the following
effects: A) It changes the sign of tk. B) It changes the sign of ?'1
(* ? h 3 + !)· C) ?* replaces ? and ?+? by —?+\ and —^respectively
(Exercise 1). It thus changes the sign of dtk. Since an even
permutation is the product of an even number of transpositions, the lemma
follows. As a corollary,
E.11) d(-tk) = -dtk.
(D) Let tk~x be an oriented face of an oriented ^-simplex tk. Then
tk~x has the orientation induced by tk, and tk has the orientation
induced by ifc_1 if tk-x appears with the coefficient +1 in dtk. The
- (Jc _ i)-faces of tk are coherently oriented if each has the orientation
induced by tk or else each has the opposite orientation, induced by — tk.

94 INTRODUCTORY TOPOLOGY [Ch. 5
An orientation assigned to a (k — l)-face of tk uniquely
determines coherent orientations for all the (k — l)-faces of tk (Exercise 2).
The boundary of an integral &-chain expressed in the form E.6) is
E.12) дС* = д 2 a fit = ? «, Эв? = ? V?.
г'¦ = 1 г = 1 j = ?
where the coefficient 6, of sj"-1 is ??1? a^ with eu = 0, +1, or —1
according as s*_1 and s\ are not incident, incident with induced
orientations, or incident with opposite orientations.
(E) Since, as a result of E.12), d(Ck + Dk) = BCk + dDk, the
boundary operator has the important interpretation as a homomorphism
(Art. A-2)
E.13) Э :(?,->?,_!
of the k-chain group into the (k — l)-chain group. The kernel (Art.
A-2@)) of д consists of those ^-chains whose boundaries are null.
They are called ^-cycles:
О
E.14) дСк=%о Ск is a &-cycle,
and the subgroup
E-15) Зк с (?fc
consisting of them is the &-cycle group. For к = 0, we define dCk as 0.
(F) To motivate the above definitions geometrically, consider first
a path 77 = q0q1 . . . qv on a complex K, where t] = ?г-_!?г-(г = 1, ... , ?)
is an edge of K, with the indicated orientation. By the chain of ? we
mean the 1-chain C(n) = ??? = 1t]. Since dt] = qt — дг_1? it follows that
V V
E.16) ЭО(тт) = Э ?&-1& = 2 fef - Vi-i) = iv - ??»
г=1 г=1
which is the terminal point of the path minus the initial point. If the
path is closed, qv = g0, then дС(тт) = 0, so that the chain of a closed
path is a cycle. (See Exercise 3.)
Theorem 1. A bounding chain dCk+1 is a &-cycle. That is,
О
E.17) ddCk+1 = 0, the null (k - l)-cycle.
Proof. Since д is a linear* operator, it is sufficient to establish
E.17) for Ck+1 = sk+1. In particular, it suffices to show that an
arbitrary face sfc_1 of sk+1 does not appear in ddsk+1. Let sk+1 be
written in the form sk+1 = PoPi5*-1· The 0П^У terms in dsk+1 giving
rise to is*-1 in ddsk+1 are +p1sk~1 and — p0sk~1, which lead to
+ e*-i — e*-i = 0.
* This means that d(Ck + Dk) = dCk + Э2Х

Art. 5-1] HOMOLOGY AND COHOMOLOGY GROUPS 95
Lemma 2. The bounding ^-cycles constitute a subgroup gfc of the
cycle group 3k.
Proof of Lemma. The lemma follows without difficulty from the
fact that G\ = дС\+1 (i = 1, 2) implies C\±C\ = д{С\+1 ± C|+1),
so that the sum and the difference of two bounding cycles is a
bounding cycle.
(G) The work of this article generalizes easily from integral cycles
to cycles of the form
a*
E.18) <?* = !>?
i = l
where g{ e ©, an arbitrary abelian group. This leads to the concepts
of A) the &-chain group (?fc(©), B) the &-cycle group 3u(®)> and C)
the bounding &-cycle group ЗУ®)· The group (?fc(©) is isomorphic to
the direct product (Art. А-З(Л)) © ? · · · ? © (?? factors). In this
book, we restrict ourselves to integral cycles, for which © = Z, the
additive group of the integers and, where explicitly stated, to cycles
mod 2, where © = Z2 (Art. А-ЦВ)).
EXERCISES
1. Verify Statements A) to C) in the proof of Lemma 1 with the aid of the
substitutions p{ = q{ (i ? j, j + 1), pj = qj+1, pj+1 = qj9 sk = p0 . . . pk.
2. Prove the last sentence of (D).
Fig. 5-3. The torus as an oriented complex.

96 INTRODUCTORY TOPOLOGY [Ch. 5
У 3. Show that a 1-cycle can be expressed as a sum of chains of simple closed
paths, which may intersect in sets of common vertices.
• 4. In the triangulation of the torus shown in Fig. 5-3, let all 2-simplexes be
oriented clockwise and show that ?&2 is a 2-cycle. Then reverse the orientation
of just one of the 2-simplexes and find <fLs\.
5. Write out ddt3 and ddfi, where t3 = q0q1q2(ls and ?4 = ЗД^г^зЗ^· Show that
they reduce to 0.
6. Let K2 consist of the oriented faces of a 2-simplex s2 = p0p1p2 together
with three oriented 1-simplexes ?2?$, РъР& РьРъ and their vertices. Discuss the
homomorphisms д of E.13) for & = 0, 1, 2.
7. Show that a homeomorphism mapping an oriented circle onto itself, with
orientation reversed, has exactly two fixed points.
8. Show that the unit disk x2 + y2 < 1 in E2 can be homeomorphically
mapped onto itself so that (a) the orientation of the boundary circle is reversed
and (b) there are exactly two fixed points.
5-2. Homology Groups of Finite Simplicial Complexes
The kth homology group §fc = $&k{K) of a finite oriented complex
is the factor group (Art. А-2(Б))
E.19) S* = 3*/& - ***.·
where 3^ an(l Ък are> respectively, the &-cycle group and the bounding
&-cycle group.
The fact that a &-chain Gk is the boundary of some chain will be
symbolized by
E.20) Gk ~ 0 о 3 Gk+1 such that Gk = dCk+1
<=>Cke%k.
(A) The symbol ~ (read is homologous to) will be used, more
generally, between two ?-chains, as follows:
E.21) G* ~ С? о Gk - Ck~ 0 => dCk = dCk.
While E.20) implies that Ck is a cycle, E.21) does not imply that G\
and G\ are cycles. As indicated, it does imply that G\ — G\ is a
cycle, or that дС\ = дС\.
(B) It is easy to see that ~ is an equivalence relation. The
corresponding equivalence classes are homology classes. The homology
classes of ^-cycles are the cosets of 3^ m°d ?/ь an(i are tne elements
of §л. The homology class of a cycle ? will be denoted by [Z]. By
definition of factor groups (Art. A-2),
E.22) [Zk] + [?5 = [Zk + Zk2].
(G) Since §л is a finitely generated (abelian) additive group (see

Art. 5-2] HOMOLOGY AND COHOMOLOGY GROUPS 97
Art. A-3), it is uniquely expressible as a direct product (or direct
sum)
E.23) S* = »* * 2, (or 23, + Xk)
where A) 33fc is a free abelian group with /?fc generators for some
/?fc > 0 and B) Xk is the direct product of a set, possibly vacuous, of
finite cyclic groups whose orders tJ, . . . , rkk have the property that
ткРк > 1 and тк+1 divides ?\ (i = 1, . . . , Pk - 1).
(D) The number /?fc of generators of 23fc is the kth Betti number of K,
and 25fc is the kth Betti group of if. The orders тк (г = 1, . . . , pk)
of the cyclic subgroups of Xk are the kth torsion coefficients of К and
Xk is the kth torsion group of K.
(E) A set (Ck, . . . , Ck) of ^-cycles of a complex if are A) linearly
independent if and only if
E.24) 2^О* = 0=>а. = О (г=1, ...,r),
» = i
that is, no non-trivial linear combination of them equals the null
&-chain; B) homology-independent (an abbreviation for linearly
independent with respect to homology) if and only if
E.25) i>,G*~ 0 =>a{ = 0 (i = 1, . . . , r),
i = l
that is, no non-trivial linear combination of them is a bounding
&-cycle.
Theorem 2. The kth Betti number /?fc of К equals the number of
?-cycles in a maximal set of homology-independent ^-cycles of К#.
Proof. In the first place, no &-cycle Ck can belong to a homology-
independent set if its homology class is an element of Xk. For Xk is
the subgroup of §fc made up of elements of finite order, which would
signify that some positive multiple of Gk is null-homologous,
contradicting E.25).
Secondly, a homology-independent set cannot contain two cycles
from the same homology class, for their difference would bound.
To complete the argument, one need merely note that a set of
cycles consisting of one from each of a set of /?fc homology classes
generating 33fc is a homology-independent set.
Corollary 1. The wth Betti number of an w-complex К equals
the number of w-cycles in a maximal set of linearly independent
m-cycles of K.
The only (w + l)-chain of К is the null chain. Therefore the only
bounding m-cycle is the null cycle. Hence homology independence
and linear independence mean the same thing for w-cycles.

98 ? INTRODUCTORY TOPOLOGY [Ch. 5
Goroliary^. An m-complex has no mth torsion coefficients.
Theorem 3. The kth homology group of a complex is the same as
that of its (k + l)-skeleton. r &??*??~?*?** % к*" , i/UW-~ * fc>)
Proof. The group 9)k is determined by the incidence relations
between the ^-simplexes and the (k + l)-simplexes, which are the
same on the (k + l)-skeleton as on the entire complex.
Theorem 4. The kth homology group of a complex К is the direct
product (or sum) of the kth homology groups of the components
(Kv ...,??) ?? ?. (See Art. 4-6(Я).)
Proof. If С is a chain on K, then С = ?? + · · · + ??> where Cj9
to be called a component of C, is the part of С into which simplexes
of ? ? enter (j = 1, . . . , ?). It is easy to verify that С is a cycle if and
only if each Gi is a cycle.
Consider a (k + l)-chain Ck+1 = ?#=1 Cj+1. Its boundary is
? ?
E.26) Ck = dCk+1 = j dCk+1 = ^Ck
3=1 i=l
where G1· = dCj + 1. Thus the components of a bounding cycle Ck on
К are bounding cycles on the components of K. Conversely, a linear
combination of bounding cycles on the components of К is a bounding
cycle on K. Thus the aggregate of a class of sets of generators, one
set for the kth homology class of each of the components of K, is a
set of generators of %k(K), and each homology relation among cycles
of К is a consequence of homology relations among cycles of Kl9 . . . ,
??. Theorem 4 follows with the aid of Art. A-3.
EXERCISES
9. Let K2 consist of the oriented boundary faces of an octahedron. Show that
P1P2 + P2P3 ~ P\Vi + Р\Ръ where the PiPj in the relation are 1-simplexes of K2.
fib) Find the Betti numbers ?1 and ?0 of the 1-skeleton of the complex
consisting of the faces of a 3-simplex.
5-3. Some Lower-dimensional Cases
Theorem 5. The 0th Betti number of a complex К equals the
number of its components.
Proof. The Kronecker index of a 0-chain C° = ?«°=1 a^i means
the sum ?^=? % of its coefficients. We employ two lemmas.
Lemma 3. The Kronecker index of a bounding 0-chain G° is zero.
Proof of Lemma. Let C1 = ?^? b{s} be a 1-chain such that C° =
dC1. IfSj is the initial vertex of s] and s^ the terminal vertex, then the

Art. 5-4] HOMOLOGY AND COHOMOLOGY GROUPS 99
term д(Ь^}) = Ъ^к — Ъ$ contributes b{ — b{ = 0 to the Kronecker
index. This being a typical contribution, the Kronecker index is zero.
Corollary. If (Cj, . . . , C°p) are the components of a bounding
0-chain (proof of Theorem 4), then the Kronecker index of C? is zero
(j = 1, · · · , ?).
Lemma 4. If К is connected, ?)?(?) is the free cyclic group.
Proof of Lemma. Since К is connected, there exists a path ??? from
s\ to s\ (i = 1, . . . , ?0). But the chain C\ of ?{ then has the boundary
dC] =^-5? (see Art. 5-l(F)). Hence s? ~ s°v It follows that
<*o ao
E.27) 2«?°~?«?°=?
г = 1 г = 1
where к is the Kronecker index of C°. As a consequence of Lemma 3,
ks® ~ u'sj if and only if к = к'. Thus the chain S® is a complete set
of homology-independent 0-chains, and the lemma follows from
Art. 5-2(C), (D) and Theorem 2.
Corollary. Two 0-chains on a connected complex are homologous
if and only if their Kronecker indices are equal.
Theorem 5 now follows from Lemma 4 and Theorem 4.
(A) Two 0-chains C° and D° on a complex are homologous if and
only if the Kronecker indices of their respective components C® and
D° (j = 1, . . . , ?) are equal (Exercise 11).
(B) There exist no 0th coefficients of torsion for any complex
(Exercise 12).
Theorem 6. The first Betti number ?? of a connected linear graph
K1 equals the cyclomatic number (Art. 1-3) ?? = ?? — ?0 + 1. For
a linear graph with ?0 components,
E.28) A = «i - «o + At
Proof. By Corollary 1 to Theorem 2, linear and homology
dependence are equivalent for 1-cycles on K1. Therefore, by Art. 1-3, ?1 =
? = ?? — ?0 + 1, in the connected case. Equation E.28) follows^
from Theorem 4 in the case where K1 has ?0 components.
EXERCISES
11. Prove (A).
12. Prove (B).
5-4. Homology Groups of a Surface
Let K2 be a triangulation of the closure of a convex plane region
В bounded by a polygon ?. Later, we will identify some of the edges

100 INTRODUCTORY TOPOLOGY [Ch. 5
of 77 in pairs, so as to obtain a polygonal representation of a surface
Ж, and the triangulation of ? so obtained will be called K*.
Lemma 5. The 2-simplexes of K2 can be so numbered {s2} =
(si, ... , si ) that the point set
E.29) Bj = U ?
г = 1
is, for each j e A, . . . , ?2), the closure of a region Bi bounded by а
simple closed polygon ?5.
Proof of Lemma. Geometrically, the lemma means that the
complex K2 can be built up, one closed 2-simplex at a time, so that
the growing subcomplex at each stage covers the interior and
boundary of a simple closed polygon. Let s\ be arbitrarily chosen from
{s2}. Suppose, for some positive integer к < ?2, that the 2-simplexes
sf, . . . , sf have been so defined that the lemma holds for j = 1, . . . , k.
To complete the proof, we need only select as sf+1 a 2-simplex of K2
whose intersection with Rk is either a single 1-simplex or the union
of two 1-simplexes, but not a 1-simplex and the opposite vertex
(Exercise 14).
Let Щ denote the subcomplex of K2 consisting of (s\, . . . , sj) and
their boundary simplexes. A l-simplex_pf Щ will be called a boundary
or an inner 1-simplex thereof, according as it is on iri or on R}\ that
is, according as it is incident with one or two 2-simplexes of Щ. The
subcomplex of Щ on iri will be called the boundary complex Bj of Щ.
Now let the 2-simplexes {s2} all be oriented clockwise, and let each
boundary 1-simplex of Kf have the orientation induced by the 2-
simplex of Щ incident with it. This orients K2, since each of its
1-simplexes is a boundary simplex of some Щ.
(A) As a direct consequence of Lemma 5 and the orientation just
defined, the boundary of the chain Gf = ?|=1 sf is the sum of the
boundary 1-simplexes of Щ.
Lemma 6. Let C1 be a 1-chain on K2 whose boundary dC1 is on
the boundary complex В of K2. Then C1 is homologous (Art. 5-2(A))
to some 1-chain D1 on B.
Proof of Lemma. We will pass from C1 to D1 by adding to C1 the
boundaries of certain 2-chains. The addition to C1 of such null-
homologous 1-chains keeps us, by definition, within the homology
class of C1. Our proof will be recurrent.
Hypothesis. For some positive integer к < ?2, there exist integers
(av . . . , ak) such that, for each j e A, . . . , k), no inner 1-simplex of
Щ appears in the chain
E.30) С} = С1+2,а<д$.
i = l

Art.5-4] HOMOLOGY AND COHOMOLOGY GROUPS 101
The hypothesis is automatically fulfilled for к = 1, since there
exist no inner 1-simplexes of K\, which consists of the faces of s\.
Figure 5-4 shows two possibilities for the relationship of sf+1 and
Bk. In either case, let ak+1 equal the multiplicity with which s1
appears in C\. Then, since s1 has multiplicity — 1 in dsk+1, it does not
(b)
Fig. 5-4. The 2-simplex s|+1 and the boundary complex Bk. (a) s?+1 having
just one edge s1 on Bk; (b) s?+1 having two edges on Bk.
appear in G\
= c\
This takes care of the case shown
k + 1 — ^k + ak+l Vsk+1·
in Fig. 5-4a. For the case in Fig. 5-4b, we show that t1 cannot appear
in C\+1. For if tl appeared without s1, then the initial vertex of s1
would appear in dC\+1, contradicting the fact that (Art. 5-2(Л))
дС\+1 = дС1, which is on B. The recurrency leads to a chain D1 = C*a
satisfying the lemma.

102
INTRODUCTORY TOPOLOGY
[Ch. 5
Corollary 1. If C1 is a cycle, then (Exercise 15)
E.31) D1 = mB where В = Э ? sf.
г = 1 .
Corollary 2. The first homology group of K2 consists of the null
element alone; that is, every 1-cycle bounds on K2 or, otherwise
expressed, the first Betti number of K2 is 0 and K2 has no first torsion
coefficients (Exercise 16).
Theorem 7. Let К * be an oriented triangulation of the standard
polygonal representation of a surface ? (Art. 2-4, Lemma 3). Then
К * has the Betti numbers and torsion coefficients shown in the
following table, corresponding to the possible structures of ? listed in the
left column:
A Sphere with ?0 ?1 ?2 Torsion Coefficients
h > 0 handles 1 2h 1 None
q > 0 crosscaps 1 q — 1 0 A first torsion
coefficient ?1 = 2
In > 0 handles,
r > 0 contours 1 2h -\- r — 1 0 None
q > 0 crosscaps,
r > 0 contours 1 g + r — 1 01 None
Proof. Corollary 2 to Theorem 2 implies that there are no second
torsion coefficients and Art. 5-3(Б) does the same for 0th torsion
coefficients. Since ? is connected, ?0 = 1 by Theorem 5.
The rest of the above table will be established with the aid of
Lemma 6, which applies to the present case, with suitable
modifications to take account of identifications of certain 1-cells of В in
pairs. Let Av . . . , Av denote these 1-cells, cyclically numbered and
oriented in the clockwise sense. An identification of two 1-cells can
be expressed in one of the forms
E.32) (a) A, = A„ (b) At = -Ajf
which will be taken to imply the identification of corresponding
points under the linear homeomorphism between At and A$ or —Aj.
Let К * be the complex obtained from K2 of Lemma 6 as a result of
these identifications. Let Af be the 1-cell of K* corresponding to
the 1-cell At of K2, taken with the orientation of A{. Then
E.33) (a) A, = Aj => Af = A*, (b) At = -A, => A* = -A*.

Art. 5-4] HOMOLOGY AND COHOMOLOGY GROUPS 103
Now (compare Lemma 6, Corollary 1)
E.34) Э2«? = л=24?.
г = 1 г = 1
where ?*=1 Af can be reduced by substitutions suggested by E.33).
In the case of the sphere with h handles, represented by the symbol
?&?^??1 · · · AhBhA^Bb\ E.34) reduces to
E.35)
д 2 sf = At + B* - A\ - Bt + · · · + At +Bt-A*h- B*h = 0.
г = 1
Similarly, for the other standardized forms (Art. 2-4),
(a) d 2 *? = ? 2С'г* = 2 ? °г* (? > 0 crosscaps),
г = 1 г: = 1 г = 1
?2 Г
E.36) (b) d 2 5* = ? A (A > ° handles, r > 0 contours),
г = 1 г = 1
(с) Э J «sf = 2 2 С* + 2 D* (д > 0 crosscaps, г > 0 contours).
г = 1 г: = 1 г = 1
Let Б* be the complex of those cells of K* which were obtained
from cells of B.
Lemma 7. Each 1-cycle on K* is homologous to a 1-cycle on B*9
hence to a 1-cycle of the form
h r
(a) 2 (aiA* + biB*) + 2 diDt (h >0 handles, r > 0 contours), or
г = 1 г = 1
F.37) у
(b) 2 cfi* + ? &??>* (q > 0 crosscaps, r > 0 contours).
г = 1 г = 1
Proof of Lemma. We first note that, since if* was obtained from
K2 by identifications of certain 1-cells of В by pairs, the given 1-cycle
C1 might, without the said identifications, become a 1-chain with
boundary on B. Nevertheless, it is easily seen from Lemma 6 that
C1 is homologous on К * to some cycle С on B*. Since the only 1-cells
of ifin Case (a) are the Af, Bf, and Df, and the only ones in Case (b)
are the Of and Df, it follows that the lemma is true.
Lemma 8. Each bounding 1-cycle on K* is homologous to a
multiple of
r
(a) BQ = 2 D* (h >0 handles, r contours), or
E.38) ' = 1
? ' q г
(b) Bx = 2 2 Cf + 2 A* (^ > ° crosscaps, r contours),
г=1 г=1
and each multiple of B0 or Бх bounds.

104 INTRODUCTORY TOPOLOGY [Ch. 5
Proof of Lemma. By Lemma 7, each bounding 1-cycle on K* is
homologous to a bounding 1-cycle all of whose 1-cells are on B*.
(B) If the boundary of C2 = ??1? et-s? is on B*, then all the
coefficients et are equal (Exercise 18).
The lemma now follows with the aid of E.36b, c).
Corollary 1. Let B* be the subcomplex of if* on the boundary
of the polygonal representation. In the case of a sphere with h > 0
handles and no contours, the only bounding 1-cycle on B* is the null
cycle, which is the boundary of m ?^? s? for each m e Z.
Corollary 2. For a sphere with q > 0 crosscaps and no contours,
the only bounding 1-cycles on B* are 2m ??=1 С* = dm ?^? sf (m e Z).
Corollary 3. For a sphere with h > 0 handles and no contours,
?^? sf and all its multiples are the only 2-cycles. For the other
surfaces, the only 2-cycle is the null cycle.
From Corollary 3, the values of ?2 in Theorem 7 are deduced. By
Corollary 1 and Lemma 7 the 1-cycles (Af, Bf, . . . , Af, Bf) are a
maximal set of homology-independent 1-cycles in the case h > 0,
r = 0, so that ?? = 2h. In the case h > 0, r > 0, we see from E.36b)
and Lemmas 5 and 7 that there is a single homology relation, ??=1 Df
~ 0, among (Af, Bf, . . . , A%, Bf, 7)?, . . . , Df) on which all other
homology relations depend, so that ?? = 2h + r — 1. In the case
q > 0, r = 0, all homology relations among cycles of the form
??=1 cfif depend on 2 ??=1 С? ~ 0. Hence ? ?|=1 Of + ???} сД*
yields one cycle from each homology class as ? = 0, 1 and the c's
range independently over the integers. Accordingly, the first homology
group is the direct sum of a cyclic group of order 2 and q — 1 free
cyclic groups, so that there is a torsion coefficient ?1 = 2, and the first
Betti number is ?? = q — 1. In the case g > 0, r > 0, there are
only the relations mB ??=1 Gf + ?$=1 Df) ~ 0 (me Z) among the
cycles (C?, . . . , C*, Z>*, . . . , Z)*), so that ?? = q + r - 1.
EXERCISES
13. Let jFC2 be a triangulatien of the annulus 1 < ?2 + ?/2 < 4 into six
2-simplexes and their boundary faces. Show that each 1 -cycle on K2 is homologous
to some 1-cycle on the circle x2 + y2 = 1.
14. Complete the proof of Lemma 5.
У 15. Prove Corollary 1 to Lemma 6. Suggestion: Show that if D1 = ??? = 1?$
where (tl, . . . , t\) are the boundary 1-simplexes of K2, and if dD1 = 0, then all
the coefficients at are equal.
У 16. Prove Corollary 2 to Lemma 6.
17. In the notation of E.30) and E.31), give a 2-chain of which C1 is the
boundary, for the case where C1 is a cycle.

Art. 5-4] HOMOLOGY AND COHOMOLOGY GROUPS 105
18. Prove (B) by showing that, under a contrary assumption, some inner
1-simplex of the polygonal representation appears on dC2.
19. Figure 5-5 shows a disk with two holes in it, along with a particular
triangulation. Show that the 1-cycles which are the sums of the 1-simplexes on
Fig. 5-5. A surface and a triangulation of it.
CQf Cl9 С2 are in different homology classes, but that these 1-cycles are not
homology-independent.
20. For the same triangulation, show that the homology classes represented by
Cx and С2 generate §x; that is, that any other homology class equals a linear
combination of them.

106
INTRODUCTORY TOPOLOGY
[Ch. 5
5-5. Surface Topology
Theorem 8. Two closed surfaces are homeomorphic if and only if
they have the same Betti numbers.
Outline of Proof. In the first place, by Theorem 7, orientable
closed surfaces (r = 0) are distinguished from non-orientable closed
surfaces by the value of the second Betti number, ?2 = 1 and 0,
respectively. The orientability class being thus determined, one can
deduce the number of handles h or of crosscaps q in a standard
representation from ?? = 2h or q — 1, respectively.
Theorem 8 is still not proved, however, for the Betti numbers
were deduced from a standard representation, and we have not shown
that the same numbers would result from an arbitrary polygonal
representation.
(A) Theorem 8 will follow in all its generality from the work in
Chapter 6, as will the following statement. Two surfaces with contours
are homeomorphic if and only if they have the same number r of
contours, the same orientability class, and the same first Betti number ??
In order to bring out some of the topological properties of surfaces, we
assume these results.
A convex polygonal region R bounded by a polygon ? can be
triangulated by introducing as 1-simplexes all the chords joining a
vertex ? of ? to other vertices. This leads to a complex К = {s}
covering R where A) the 0-simplexes {?0} of ? are the vertices of P;
B) its 1-simplexes {s1} are the edges of ? and the chords from p; C)
its 2-simplexes are the triangular regions into which the chords from^p
separate R. We will take К to be oriented.
Now let К be converted into a different surface K0 by the
identification in pairs of some of its 1-simplexes on В and, of course, the
induced identification of some of the vertices (Arts. 2-1, 2-2). But
then {t}, denoting {s} with the identifications, is not a complex,
because some pairs of closed 2-simplexes intersect in more than the
closure of a single common face. For example, in the case where ? is a
quadrilateral (Fig. 5-6a), К = {s} consists of two 2-simplexes, five
1-simplexes, and four 0-simplexes, to be numbered and oriented as
indicated. We now convert the surface into a torus by the
identifications 4 = sl> sl = sh Then (Fig. 5-6b) K0 = {t} consists of two
2-simplexes, t\ = s\, t\ = s\\ three 1-simplexes, t\ = s\, t\ = s\ = s\,
t\ = s\ = s\, and one 0-simplex, t° = s° (i = 1, 2, 3, 4). Since t\ and t\
have all their faces in common, {t} does not satisfy the definition of a
complex. If, however, each simplex of {t} is subjected to its second
barycentric subdivision (Fig. 5-6c), the resulting set of simplexes is a
complex. So is the simpler subdivision shown in Fig. 5-3 above.

Art. 5-5] HOMOLOGY AND COHOMOLOGY GROUPS 107
? Fig. 5-6. Subdivisions of a rectangle and a torus.
(B) In the general case, the set of simplexes obtained from {t} by
subjecting each simplex of {t} to its second barycentric subdivision
is a simplicial complex (Exercise 23).
Theorem 9. Let К be a triangulation of a surface M, and let a, be
the number of ^-simplexes of K. Then, in the notation of Theorem 7,
E.39) a0 - ai + a2 = ?0 - ?? + ?2 = ?
[2 -2h- r (h > 0, г > 0)
2 — q — г (q > 0, г > 0).
Proof. Consider the standard polygonal representation of ? in the
orientable case, as given in B.6a). It is easy to verify that there are
o$ = 1 + r vertices, af = 2h + 2r edges, and ocf = 1 region, the
interior of the polygonal representation. One can pass to an arbitrary
triangulation of К by a sequence of steps each consisting of either A)

108 INTRODUCTORY TOPOLOGY [Ch. 5
introducing a vertex which divides an edge into two edges or B)
introducing a new edge, perhaps in the form of a broken line, which
divides a region into two regions. Each step adds one unit each either
to the number of edges and the number of vertices or to the number of
edges and the number of regions, and hence does not affect the value of
the quantity a0 — ax + a2. Therefore, at each stage,
oc0 — 04 + a2 = a* — af + a*
E.40) = l+r — 2h — 2r+l = 2~2h — r
The non-orientable case can be similarly proved (Exercise 24).
e
a a
Fig. 5-7. The polygon FAECGBHDC-HDB^A-1.
(C) Given a polygonal representation, not necessarily in standard
form, for a surface M, it is possible to deduce its standard form from
its orientability class, its number of edges ocf, its number of vertices a*,
and its number of contours (see (D) below), which can easily be
counted with the aid of a diagram of the representation.
We illustrate (C) by carrying out the described process for the
symbol
E.41) 77 = FAECGBHDG-HDB^A-1 (Fig. 5-7).
It is non-orientable because D appears twice with the same
superscript. In the diagram, the same letter is used for all copies of a set of
identified vertices, since they count as a single vertex. A count yields
a* = 5, af = 9, and there is just one region, so that a* = 1. The
contours are made up of free edges. Since the end points of F are
identified, F represents a contour. The edges EIH^G represent a

Art. 5-5] HOMOLOGY AND COHOMOLOGY GROUPS 109
four-edged contour, from 6 to с along E, thence along / to e, along ?~?
to d, and back to b along G. Since this exhausts the free edges, r = 2.
Substituting in E.39) for the non-orientable case (compare E.40)), we
find
E.42) a* - af + a* = 5 - 9 + 1 = -3 = 2 - q - 2,
so that q = 3. Thus E.41) represents a sphere with three crosscaps and
two contours, so it is reducible to C1C1C2C2C3C<iK1D1K^1K2D2K2i.
(D) The contour number r is the number of closed curves which can
be traced along the free edges of a polygonal representation.
(E) The number ? = ?0 — ax + a2 is the (Euler-Poincare)
characteristic of an arbitrary 2-complex K2. If \K2\ is a closed manifold, then
\K2\ is determined topologically by its orientability class and its
characteristic N. For, since r = 0 for a closed manifold, K2 is a sphere with
h = |B — N) handles if orientable and with q = 2 — N crosscaps if
non-orientable.
(F) Each even integer not exceeding 2 is the characteristic of some
orientable closed surface. Each integer not exceeding 1 is the
characteristic of some non-orientable closed surface. There is just one closed
surface of characteristic 2, the sphere, and just one of each odd
characteristic <1. For each even characteristic <0 there are just two
closed surfaces: one orientable and one non-orientable.
(G) The genus of a closed surface is g = h (the handle number) or
g = q (the crosscap number). A closed surface is topologically
characterized by its orientability class and its genus.
EXERCISES
~* 21. Show that two surfaces with contours can have the same Betti numbers
without being homeomorphic.
22. Show that the first barycentric subdivision of KQ in Fig. 5-6b is not a
complex.
23. Prove (B).
24. Complete the proof of Theorem 9 by dealing with the non-orientable case.
25. In E.41), replace the second D by ?~? and use the illustrated method for
deducing the standard form.
26. Characterize eaoh of the following by finding its orientability class, its
contour number, and its handle number or crosscap number, as appropriate: (a)
ABCDA-^EFC-^GF-m, (b) AEAFDCGBD^HC^B-1.
27. Find the orientability class and characteristic of (а) А ВАС ВС, (b)
ACA-'LC-'LDD, (c) ABACB-W-1.
28. Find the orientability class and genus of (a) ABCDDEBACLE'1, (b)
ADCACD-^B-1, (c) ABA-WB-W.
29. Find the Euler-Poincare characteristic, contour number, orientability
class, and standard form for the following surfaces: (a) ABCADBEFC^G,
(b) ABCCA^DEB^ED, (с) АВСАСВ.

110 INTRODUCTORY TOPOLOGY [Ch. 5
30. (a) Two square regions have their edges matched in pairs as suggested by
ACB~1D~1 and ADB~1C~1. Characterize the resulting surface, (b) Do the same
for two pentagonal regions ABCDE and ??-???~??.
31. A sphere with three twisted handles (see Figs. 2-22 and 2-23) is equivalent
to a sphere with how many orosscaps?
32. To what form H(h, r) or C(q, r) is each of the following equivalent:
(а) АгА2 . . . AkA-*A-i . . . Л? (b) AXA2 . . . A^A^ . . . A'1! (c)
АгА2. . .АкАгА2. . .Ak1
33. Figure 5-8 shows a disk with
a simple bridge, a twisted bridge, and
a double bridge, the terminology being
clarified by the figure. Prove the
following statements:
(a) A sphere with r > 0 contours
and h handles is equivalent to
a disk with r — \ simple
bridges and h double bridges.
(b) A sphere with r > 0 contours
and q crosscaps is equivalent
to a disk with r — 1 simple
bridges and q twisted bridges.
This problem uses terminology and
states results to be found in Kerekjarto
[K2]. Note that it gives a method
whereby one can make a model, free
from self-penetrations, for any surface
having a boundary. Fig. 5-8. Simple, twisted, and double
34. The same as Exercise 29 for the bridges,
surface represented by a disk with
three double bridges and three twisted bridges.
35. The same for the surface represented by a disk with a double bridge
modified (a) by putting a twist in one of the two bands composing it, (b) by putting
a twist in each of these bands.
36. The same for a Moebius band with a handle.
37. Show the equivalence of the following: (a) a sphere with one crosscap
and t twisted handles (Fig. 2-23); (b) a sphere with one crosscap and t ordinary
handles; (c) a sphere with 2t + 1 crosscaps.
38. Show that a sphere with ? > 1 twisted handles and у ordinary handles is
equivalent to a sphere with ? + у twisted handles.
39. Describe, as spheres with handles or crosscaps, all closed surfaces (a)
with characteristic —6; (b) with characteristic 1 — 2k.
5-6. Pseudomanifolds
In Em, with coordinate system (xv . . . , xm), consider an oriented
simplex sm = +V0V1 · · · Pm- Let (an> · · · > aim) be tne coordinates of
p. (j = 0, . . . , ??), and let A(sm) be the determinant
E.43) A(sw) = |1 an ajZ . .. aim\ (j = 0, . . . , m).

Art.5-6] HOMOLOGY AND COHOMOI.OGY GROUPS III
The statement that p0, pv . . . , pm are linearly independent is
equivalent to the statement that A(sm) ? 0 (Chapter 4, Lemma 2, and
Art. 4-3). Reversing the orientation of sm reverses the sign of A(sm).
That is, A(sm) = — ?(— sm), a consequence of the fact that an odd
permutation of vertices of sm corresponds to an odd permutation of
the rows of A(sm).
(A) Two oriented m-simplexes sm and tm in Em have like ( = similar)
or unlike orientations according as A(sm) and A(tm) agree or disagree in
algebraic sign. The whole space Em (or any region of Em) is oriented
when one m-simplex sm in it is oriented. All other oriented m-simplexes
are then described as positively or negatively oriented, according as
they are oriented like or unlike sm.
(B) The selection of a coordinate system also serves to orient Em
or any region of Em, each oriented sm in such region being
described as positively or negatively oriented according as A(sm) > 0 or
A(sm) < 0.
(C) Let (pv . . . , pj, but notp0, be given, and suppose (pv . . . , pj
determine an (m — l)-simplex. Let ? : (?) = (?? . . . , xm) be an
arbitrary point, let s™ = +ppx . . . pm, and let ?(#) denote A(s™).
Then ? e A(pv . . . , pm), which is the (m — l)-plane of pv . . . , pm
(Art. 4-1), if and only if ?(?) = 0. Hence ? : ?(#13 . . . , xm) = 0 is an
equation for* ? = A(pv . . . , pm). Therefore ? separates Em into two
parts (Exercise 44), called the half-spaces ?(#) > 0 and ?(?) < 0.
These are the loci of a point ? : (?) such that the orientation of
PPi · · · Pm is positive and negative, respectively.
Lemma 9. If (a) sm = psm~1^na tm = g«5w_1 are oriented alike and
(b) A(sw_1) separates ? from q, then sm and tm induce opposite
orientations on sw_1 (Exercise 45).
Thus far, we have discussed orientations of m-simplexes in an Em.
We turn now to oriented m-simplexes in an En (n > m).
(D) Consider, in En (n > m), two oriented m-simplexes sm =
ps™-1, tm = qs™-1 such that sm ? tm = ё™-1. Guided by Lemma 9, we
say that sm and tm have like (= similar) or unlike orientations
according as they induce opposite or equal orientations on sw-1.
(E) A rectilineal simplicial m-complex ? in En is called a (closed)
m-pseudomanifold if A) each (m — l)-simplex of ? is incident with
exactly two m-simplexes and B) any ty/o m-simplexes of ? can be
used as first and last members of a sequence of such m-simplexes,
consecutive members having a common (m — l)-dimensional face.
If ? satisfies A) and B) with the exception that each member of a
non-vacuous subset {?™-1} of its (m — l)-simplexes is incident with a
corresponding single m-simplex, then ? is an m-pseudomanifold with
boundary B, where В is the complex consisting of all the faces of the

112 INTRODUCTORY TOPOLOGY [Ch. 5
simplexes {?171-1}. We describe two m-simplexes of ? as adjacent to
each other if they have a common (m — l)-dimensional face. We say
that ? is orientable or non-orientable according as it is or is not
possible to orient all its m-simplexes so that if two m-simplexes are
adjacent, they are oriented alike.
A pseudomanifold ? is coherently oriented if it is oriented so that
A) every two adjacent ra-simplexes are oriented alike and B) each
(m — l)-simplex of the boundary В of M, considered vacuous if ? is
closed, has the orientation induced by the incident ra-simplex.
Lemma 10. An m-pseudomanifold ? can be coherently oriented if
and only if it is orientable. In such case, exactly two coherent
orientations are possible. If sm e M, each of the two possible orientations of
? is determined by one of the two possible orientations of sm. If Ж has
a boundary В and sw_1 e B, an orientation of sw_1 similarly determines
an orientation of M, and conversely.
This lemma follows directly from the definitions.
(F) By the chain of an oriented m-complex Km, with oriented m-
simplexes sf (i = 1, . . . , aw), we mean the chain Gm{Km) = ?^? sf.
Lemma 11. If Ж is a coherently oriented ra-pseudomanifold, then
дСт{М) = Ст-г{В)9 where В is the possibly vacuous boundary of M.
For, if an (m — l)-simplex sw_1 of ? is incident with two m-
simplexes sm and tm of M, then s™-1 appears on d(sm + tm) with
coefficient 0, and sm~1 appears with coefficient 0 on drm for any rm not
incident with sm~1. If an (m — l)-simplex is on В it appears with
coefficient 1 in dCm(M), by definition of coherent orientation.
Lemma 12. Let ? be an m-pseudomanifold whose simplexes of all
dimensions к Ф т — 1 are arbitrarily oriented. If sw-1 e ? is incident
with two similarly oriented m-simplexes, let it be arbitrarily oriented.
If 5W_1 e ? is incident with two m-simplexes with unlike orientations,
let s™-1 have the common induced orientation and let Km~x be the
oriented complex consisting of the faces of the latter (m — l)-simplexes.
Let each (m — l)-simplex of В have the orientation induced by the
incident m-simplex of M. Then
E.44) дСт(М) = 2Ст-1(Кт~1) + Cm-\B).
The proof of Lemma 11 with obvious modifications establishes
Lemma 12 (Exercise 47).
Theorem 10. If Ж is a closed orientable m-pseudomanifold, then
E.45) Ьо(Щ & §т(М) ъ the free cyclic group,
where ^ is the isomorphism relation.

Art.5-7] HOMOLOGY AND COHOMOLOGY GROUPS 113
Proof. In the case of §0, this follows from the connectedness of ?.
For §w, it follows from the facts that A) the chain Gm(M) is a cycle if
? is coherently oriented (Lemma 11), B) each m-cycle on Ж is a
multiple of Gm(M), and C) no m-cycle ? 0 can bound, since the only
(m + l)-chain on an m-complex is 0.
Theorem 11. If Ж is a closed non-orientable m-pseudomanifold,
then its mih Betti number is ??? = 0 and it has an (m — l)th torsion
coefficient equal to 2 (Exercise 49).
EXERCISES
40. In Definition (B) let m = 2. Suppose, in E2, that a clockwise rotation
through ?/2 takes the positive a^-axis into the positive x2-a,xis. Show that
Definition (B) then means that clockwise oriented 2-simplexes are positively
oriented.
/ 41. Triangulate a Moebius strip into the faces of three 2-simplexes, orient
them, showing them in a diagram, and apply Lemma 12, indicating on the
diagram the 1-chains which are involved.
42. (a) Show that Definition (A) implies that two 2-simplexes in E2 are
oriented alike if and only if both are oriented clockwise or both counterclockwise.
(See Exercise 40 above.) (b) Interpret the definition for m = 1.
43. Expand ?(?), in Statement (C), by minors so as to put A(x) = 0 into the
form of a linear equation in (x).
44. Show that ? in Statement (C) separates Em into at least two parts, by
showing that a curve from a point where A(x) < 0 to a point where A(x) > 0 must
pass through a point where A(x) = 0. Show that if ? (?) and ? F) are of like sign,
then A(x) is not zero on the segment ab^
45. Deduce Lemma 9 from (C).
u 46. (a) Show that a 2-sphere with two of its points identified is a pseudomani-
fold. (d) Do the same for an arbitrary surface S with arbitrary identifications
among the vertices of a triangulation of 8. *****?? Д^Г^ ??· ^-»
47. Prove Lemma 12. г V
48. Let ck be a convex polyhedral k-cell (Art. 4-12). Show that each linear
simplicial subdivision of ck (Chapter 4, Theorem 8) is an orientable &-pseudo-
manifold, with a boundary consisting of the subcomplex on ck — ck.
% 49. Prove Theorem 11.
5-7. Homology Bases and Incidence Matrices
Let К = {s} be a finite oriented m-complex. Using the notation of
E.5), let the boundary chain of s^+1 be written in the form
E.46) &?+1=|ф? (j=l,..-,aw).
(A) If ef is not incident with «s*+\ then ek} = 0. If sf < dj+1,
(Art. 4-5(D)), then ekj = +1 or — 1. In the former case, 8* has the

114 INTRODUCTORY TOPOLOGY [Ch. 5
orientation induced by Sj+1 (Art. 5-l(D)). The number ?|· is the
incidence number of s% and Sj+1.
The kth incidence matrix of К As the matrix
E.47) Jk = (?%) =
(*?)
«)
D+1) ·
к
Fk
fcl«A+l
(& = 0, 1, . . ., ? — 1)
C?») be
where (sf) and (s|+1) are used to label the rows and columns.
Now let {C^1} = (C*+1, . . . , Gj+J) and {Ck} = (Ckv ...,^
arbitrary bases for &k+1 and (?fc, respectively. Since {Ck} is a base for
the ^-chains, the boundary дС\+1 of C\+1 can be uniquely expressed in
the form
E.48)
dGk^= 24f/7f (j=l,...,afc+1).
i = l
The &th incidence matrix of К for {Ok+1}, {Ck} is
E.49) Jk({Ck+l}, {Ck}) = ? = (Gk)
*+l\
4tf
(I?) Л characteristic property of incidence matrices is that the boundary
of the chain С|+1 labeling a column is the chain ?^? ?% G] in which the
chains labeling the rows appear with the respective elements in the column
as coefficients.
Our next objective is to show how bases can be selected for the
chain groups so as to simplify the corresponding incidence matrices
as much as possible and to put them into a canonical form. Exercises
50-54 illustrate the general procedure for 1-complexes.
j. Theorem 12. Let (J0, . . . , Jm-i) be the incidence matrices
corresponding to arbitrary bases {Ck} for Gfc (k = 0, . . . , m). Then, in terms
of matrix products (see Art. A-3),
E.50)
Jjc-l^k — Ofc
where Ok is the null matrix (each element is zero) of oc^ rows and
afc+1 columns.

Art. 5-7] HOMOLOGY AND COHOMOLOGY GROUPS 115
Proof. Applying the boundary operator twice, and using the
formulations E.48) and E.49) for к = @, . . . , m), we find
<Xk <*k a/c-i ?/c-i /?* \
E.5i) еде** = 2 п\т = 11 nt^cf1 = 1A пыхп% YV
U= 1» · · · у ад+?).
The coefficient of Cj-1 (in parentheses in E.51)) is the element in row h,
column j oiJk_1Jk. Since ddCk+1 is the null cycle (Theorem 1), the
present theorem follows.
(C) Indeed, E.50) is a way of formulating the general property
aacfc+i = 0.
я Theorem 13. It is possible to choose bases for the chain groups
(?0, (?1? . . . , (?w in terms of which the incidence matrices have the form
afc+i - 7k columns
0
0
< 0
.
o· ,
Ty*
0
where the t's are positive integers, with ?\ divisible by ?*+1.
Proof. The matrices Jf will be called the normal or normalized
incidence matrices of K. In the Appendix, the matrix {eu) of Eqs.
(A.29) having m rows and ? columns is normalized into a form exactly
like E.52) save that the submatrix of the t's appears in the upper left
corner instead of the upper right (Appendix, Theorem 5).( Property
E.50) is inconsistent with Jk_1 and Jk being of the latter form.) Rows
can, however, be permuted and columns can be permuted (Art.
A-3(E)(F)) to obtain such forms as E.52).
The simultaneous normalizing of the matrices (J0, . .,. , Jm_i) into
(«/*) J*, · · · yJm-i) differs from the normalizing of the single matrix
in the Appendix because A) the labels for the columns of Jk_1 are also
the same as the labels for the rows of Jk for an arbitrary set of bases
{Gh} (h = 0, . . . , ??) and hence B) an alteration of the columns of
Jjc-i entails a corresponding alteration of the rows of Jk.
(D) Starting with a base (CJ, . . . , G* ) for (?fc, we define certain
elementary changes of base, and note the effects on the incidence
matrices:
A) Replacing G\ by D\ = —G\ corresponds to changing all signs in the
гш column of Jk_x and the гш row of Jk.
= 0, 1, . . . ,m — 1)
— 7fc rows

116 INTRODUCTORY TOPOLOGY [Ch. 5
B) Replacing G\, G1· by Dkh = Gj, D1· = G\ corresponds to interchanging
columns h and j in Jk_v rows h and j in Jk.
C) Replacing U\ by V\ = U\ + rUj for some integer r and some
j ? h corresponds to making the replacements
(new col h) = (old col h) + r(col j) in Jk_v
(new row j) = (old row j) — r(row h) in Jk.
Parts A) and B) of (D) are trivial to verify. Part C) results from
dV\ = dU\ + rdV) and the fact that if dU*+1 = ?^ rfigU*, then
Э175+1 = ?^.^?/f + 1&F$ -Htf, " rrfiJU).
(E) The matrix J0 is the incidence matrix of a linear graph, the
1-skeleton K1 of K. As such, it can be normalized into a form Jjf, like
E.52), in which each ?? = 1 (Exercise 54).
In accordance with (D), operations on the columns of J0 during its
normalization (see Art. A-3) involve corresponding operations on the
rows of Jv Let J[ be the incidence matrix into which Jx is thus
carried.
Hypothesis. For some h > 0, it is possible to select bases for the
chain groups (?0,. . . , (?w for which the incidence matrices are (JJ , . . .,
</*_!, J'h, Jh+V · · · , Jm-i)- Here the starred matrices are of the form
E.52), while Jh+1, . . . , Jm-i are the matrices for the initial bases
{sh}, . . . , {sm}.
For h = 1, the conditions of the hypothesis are fulfilled by the
definitions of J*Q and J[ with the corresponding new bases for (?0
and Q,v
Lemma 13. The last yh_x rows of J'h consist entirely of zeros.
Proof of Lemma. If a non-zero element ? appeared in the jth of the
last yh_x rows of J'h, then the product matrix J^-iJh would contain the
element ?]~??\ ? 0, by E.52) for к = h — 1. But this would contradict
Theorem 12. Hence Lemma 13 is correct.
Let the normalizing procedure of the Appendix (Art. A-3) be
applied to carry J'h into the form J%, with corresponding changes of
bases and of the matrices J%_19 J'h in accordance with (D). Normalizing
operations on the last yh_x rows of J'h are unnecessary, by Lemma 13.
Operations on its first ?? — ??_? rows affect only the first ?? — ул_1
columns of </*_!· Since they contain only zeros, by E.52) for к =
h — 1, their elements remain unchanged, only their ?-chain labels
being affected. The normalizing operations on the columns of Jh
involve a change of base for (?л+1 and a transformation of Jh+1 into
J'h+1. This completes the general step of the proof. The step with
h = m — 1, in which, of course, there is no Jh+1 to consider, yields
Theorem 13.
(F) Among the numbers ?*, those which exceed 1 are the kth torsion
coefficients (?*, . . . , ?? ) of К.

Art.5-7] HOMOLOGY AND COHOMOLOGY GROUPS 117
(G) Consider a normalized matrix J?, к > 0. Let (A\, . . . , A*)
denote the &-chain labels of its first yk rows. Let the last ук_г rows be
labeled (C\, . . . , C* ). They consist entirely of zeros, by Lemma 13,
hence do not overlap with the first yk rows. There may be rows of
zeros between row (??) and row {G\). If so, let them be labeled
(B\, . . . , BkPk). As for «/*, let its rows be labeled (A°lf . . . , A°/q, B°v . . . ,
I?f) ). Since the columns of JJ_X are labeled like the rows of Jf, it
remains only to assign symbols to the column labels of J*t_v Let
them be {B\\ . . . , Щт, G\\ . . . , G^J.
This labeling system is displayed as follows for a typical Jf. The
sets (AJ[+1, . . . , ^l^1) are vacuous if к = ?? — 1; and so are (C\, . . . ,
C* )iffc = 0.
The labels at the right and bottom are inserted for reference in Art.
5-9. They are to be ignored for the present.
E.53)
A*
k,c
Ay\
Bk
Ak+1 . A'c+1
1 " ' ' Vi+i
0
0
0
0
xk+l... xk+i
1 Vk+l
0
0^
0
0
Vk+l Vkt 1
? · At+i
(Jk+l Ck+1
0
0
<
0
0
0
Z?+1 . . . Zk+1
1 Pk
j · · · S*
0
1
. 0
0
1
0
0
•••^+1
4k
Yk
Z\-i
(k = 0, . . . , m - 1)
(H) Since Jf has afc rows and afc+1 columns,
, E·54) ? = a* - y* - y,_! (* = 0, 1, . . . , m\ y_x = ym = 0),
the values y_x = ym = 0 being conventional, since only y0, . . . , y,

118 INTRODUCTORY TOPOLOGY [Ch. 5
are defined as the ranks of the Jk. From E.54), it follows that
(Exercise 59)
m m
E.55) N = 2 (-1)%= J (-1L·
fc=0 fc=0
This number N is the Euler-Poincare characteristic of K. We adopt
the notation
?*} = (A\,...,A*k), {вк} = (в\,...,щк), {ck} = (с* ...,cknj,
E.56) {Ak, Bk) = {Ak} U {Bk}, {Ak, Bk, Ck) = {Ak, Bk} и {Ск},
{rAk} = DAk1>...>rkPkAkPk,Akk+1,...,Akk).
For any &-chain Dk, Dk will denote its homology class. Let
{JP, Б*} = {Ak} и {5*}.
Note that the last subscript in the definition of {Ak} is pk, not yk.
Theorem 14. The following sets of chains and homology classes are
bases** for the indicated groups:
Group
(a) d*
(b) 3*
(c) D*
(d) 3f*
(e) S* = 3*/S*
(f) X,
(g) 33,
Base
{Ak, Bk,Ck}
{A", B"}
{A"}
{tA*}
{Ak, B"}
{A"}
{B*}
Description of Group
&-chain group
к -cycle group
kth boundary divisor* group
bounding &-cycle group
kth homology^group
kth torsion group
kth Betti group
* A boundary divisor cycle is a cycle, some non-zero multiple of which bounds.
X Sometimes, but not here, called the kih Betti group.
Proof. Part (a) follows from the definition of {А, В, С}. A &-chain
Dk e (?fc can therefore be uniquely expressed thus:
* 7k Pk 7k-l ,
E.58) Dk = 2 <h4t + ? Ъ,% + 2 cfi).
A=l i = l 3 = 1
From the matrix Jf, by E.53) with к replaced by к — 1,
Ук-? Ук-?
E.59) dDk = 2 с,д<% = ? c^Aj-1
where ?)-1 = 1 for j > рк_г; so, the A*'1 being linearly independent,
• In the sense of minimal sets of generators.

Art.5-7] HOMOLOGY AND COHOMOLOGY GROUPS 119
Dk is a cycle if and only if all the coefficients c} are zero; hence Part (b).
Replacing к by к + 1 yields
7k
E.60) dDk+1 = 2 <уг*4* =>Parts (c) and (d).
Comparing the right side of E.60) with E.58), we deduce that
= 0modr? (A=1, ...,yfc)
E.61) Dfc
'\ = с, = 0 (<=l,...,j8fc;j=l,...,yM).
Parts (e), (f), and (g) follow with the aid of Art. A-3. See especially
Lemma 4 and Theorem 7 of the Appendix (Exercise 60).
Corollary 1. The groups involved in Theorem 14 are related as
follows:
(a) gfc с Dfc с 3, с (?fc
(b) & = 3*/&
(D·62) (c) 2, = war*
(d) 23* = «Л = 3*/»*
Corollary 2. Two ^-cycles ??*=1??^$ + ?^? Ъ{В\ and ??*=1 а;Л* +
???? 6г'Б^ are in the same homology class if and only if a'h = ah mod ?\
and &· = b{. That is, any &-cycle Zk satisfies a unique homology of
the form
pk Д* [ah = an integer mod ??,
E.63) ?»~2???+ ? MS * " .
? = ? i = i [bj = an integer.
The set (A\, . . . , Ak , B\, . . . , Б^ ) is called a homology base of
dimension к or a &th homology base. Also, (A\, . . . , AkPk) is a &th
torsion base and (?{,..., 5^ ) a &th Betti base. It is, however, the
homology classes of these respective sets of cycles which are bases
of the groups ?fc, %k, and 93fc.
EXERCISES
50. Show that a 1-complex ? which is a tree (Art. 1-3) can have its vertices
(Pv · · · > J°a ) an(i its 1-simplexes (sj, . . . , s^ ) so numbered that (a) 5^ = PhPj+1
for some /г < j (j = 1, . . . , ?? = a0 — 1) and (b) the subcomplex (pv . . . , Pj+i,
s\, . . . , sj) is a tree for each j ? A, . . . , ??). With such a numbering, what is the
form of the incidence matrix J0?
51. Using a numbering and orientation as described in Exercise 50, let C\ be
the sum of the 1-simplexes on the simple broken line from p1 to pi+1 (i = 1,
. . . , ??). Show that the 1-chains C\ and the 0-chains ? J = ?>1? A\ = pi+1 — p1

120 INTRODUCTORY TOPOLOGY [Ch. 5
(i = 1, . . . , ??) are bases for the chain groups (?х and d0 respectively. Suggestion:
Show that the s* and the pri can be uniquely expressed in terms of the C\ and the
Cj respectively.
52. Write the incidence matrix Jn for the bases (Л?, . . . , A0 , 5?) and
? ? ?7 7 ??7 l'
(Gp . . . , C\ ), labeling the rows and columns, respectively, with these chains in
the order named.
53. An arbitrary connected 1-complex K1 can be regarded as a tree ?
(Art. 1-3, Theorem 2) and a set (cj, . . . , c1) of 1-simplexes, each joining two
vertices of T. Let B1. be a 1-cycle into which only cj and simplexes on Tenter,
(a) Show that (B\, . . . , B^9 C\, . . . , C\ _x) is a base for (?x, where the notation of
Exercise 51 defines the C\ relative to ? c= К. (b) Write the incidence matrix for
the bases (A\, . . . , A^, Щ) and (B\, . . . , B\, C{, . . . , 0\^_?)9 labeling the rows
and columns with the elements of these bases in the order named.
54. Prove (E) either (a) directly from the fact that each column contains just
two non-zero elements, equal to +1 and —1 respectively, or (b) with the aid of
the result of Exercise 53, noting however that K1 is no longer assumed to be
connected.
^r 55. Let jFC3 be the 3-complex consisting of all the faces of s3 = р0р1р2Рз each
positively oriented when its vertices are written in order of increasing subscripts.
Write out the incidence matrices of K3.
56. Reduce to normal form the incidence matrices of K3 in Exercise 55.
V- 57. Triangulate a*polygonal representation CC for a projective plane, orient it,
set up its incidence matrices, and give their normal forms.
58. In Exercise 57, specify a first torsion basis for the complex.
59. Deduce E.55) from E.54).
60. Complete the establishment of Parts (e), (f), and (g) of Theorem 14.
5-8. Connectivity Groups and Numbers
(A) In this article we outline homology theory mod 2, in which
the coefficient group ? of Arts. 5-1 to 5-7 is replaced by Z2 (see Art.
5-1 (G)). We represent Z2 by the numbers 0 and 1 with 0 + 0 =
1 + 1=0, 0 + 1=1+0 = 1.
The incidence matrices mod 2 are
E.64) Jk = (e*·) (i = 1, . . . , ak;j = 1, . . . , afc+1; h = 0, . . . , m - 1)
with e^ = 1 or 0 according as s* and Sj+1 are incident or not.
(B) The integral theory, meaning the theory based on integral
coefficients, reduces to the mod 2 theory if two chains Ck = ?^? а^\
and Dk = ?**=? ?? are identified if and only if at = 6г mod 2. Thus
Ck is equivalent to the &-chain obtained by changing each even
coefficient to 0 and each odd coefficient to 1.
(C) A yfc-chain Ck mod 2 can be interpreted as the set of ^-simplexes
appearing in.it. Then Ck + Dk consists of the u-simplexes each in
Ck or Dk but not in both. Orientations of simplexes have no
significance in the mod 2 theory, since — 1 = +1 mod 2.

Art. 5-8] HOMOLOGY AND COHOMOLOGY GROUPS 121
(D) The &-chain group (?fc mod 2 contains exactly 2a* elements.
It is the direct sum of the ak groups of order 2 generated by (s\, . . . ,
The mod 2 boundary of a simplex sk is the sum dsk of its (k — 1)-
dimensional faces. In terms of E.64),
E.65) 94+1 = J e>*
г = 1
and the boundary of a &-chain Gfc = ??^? a>$ is 9G7c = ??^? afis1·.
In the interpretation of (C), a (A; — l)-simplex belongs to 9G7c if and
only if it is incident with an odd number of &-simplexes of Ck.
As in the case of the operator 9, 9 can be interpreted as a homo-
morphism (see Art. 5-l(E)) 9 : (?fc -> ?fc_i. Its kernel is the mod 2
&-cycle group 3fc> since a mod 2 &-cycle is by definition a cycle C/c for
which dCk = 0.
(E) The operator 9 has the property ddCk+1 = 0, and the mod 2
incidence matrices have the property Jlc_vJk = 0k (see Theorem 12)
(Exercise 61).
As in the integral case, the bounding mod 2 cycles form a group §fc.
They are said to be null-homologous mod 2, symbolized by Zk ~ 0
mod 2. The relation G| ~ G\ mod 2 is defined precisely as in Art.
5-2{A) with 9 replacing 9, and mod 2 homology classes are defined by
analogy with the integral homology classes of Art. 5-2(Б). This
leads to the kth connectivity group
E.66) ik = 3j§;
analogous to the kth homology group. For some integer Д., called
the kth connectivity number of K, Hk is the direct product (or sum)
of /?fc groups of order 2. This can be seen by replacing ? with Z2
throughout Art. A-3.
(F) For any complex, ?0 = ?0 (Exercise 62).
By the procedure of Theorem 11, J0, . . . , Jm_1 can be normalized
into matrices J^, . . . , J^_1 of the form E.52), with the top right
square in E.52) replaced by a unit matrix of yk rows and columns
where yk = the rank of Jk (also of Jf).
(G) In the notation of Art. 5-4, especially that used in the proof
of Theorem 7, the 2-chain C2 = ?^? sf is a 2-cycle mod 2 for the
sphere with q crosscaps, since its boundary in the integral theory is
2Gf (see Lemma 8) which becomes the null cycle in the mod 2 theory.
This remark has a bearing on Exercise 64.
The rows and columns of Jf can be labeled exactly as in E.53)
except that A) the non-zero elements in the top right square are all

122 INTRODUCTORY TOPOLOGY [Ch. 5
l's, so that there is no pk and B) each А, В, С, y, ? should be modified
by putting " above it (see Exercise 64 for ?). As in Art. 5-7(#),
we find
(a) At = 4 ~ Уте ~ Утс-? (* = 0, . . . , w; у_! = ym = 0),
m m
(b) N= J (-1)%= I (-1L·
fc=0 fc=0
(#) By analogy with the integral homology theory, {Ak, Bk) =
(A\, . . . , A\k, Ё\, . . . , Ёк» ) is a base for 3fc> and {^fc} is a base for
§fc. The cycles (i?i, . . . , ?? ) = {Bk} are representative of a set of
mod 2 homology classes generating Hk. They are a &th connectivity
base.
EXERCISES
61. Prove (E).
62. Prove (F).
63. (a) Find the connectivity numbers ?0, ?? ?2 for a sphere with q crosscaps.
(b) Show that, for an orientable surface, ?? = ?? (г = 0, 1, 2). (с) How are ??
and pi related for a non-orientable surface?
64. (a) Show that yk = yk — ek, where ek is the number of even kth torsion
coefficients of Km. (b) Show that fik = ?? -l· ek + ek_v See (G).
65. Show that 9)k is determined by §)k_1 and §>k.
66. Let the three edges of a triangle, oriented coherently, be identified,
creating a space (not a 2-manifold) with the polygonal representation AAA. It
can be subdivided into a complex K2 with the aid of a second barycentric
subdivision of the given triangle. What are (a) the Betti numbers and torsion
coefficients of ?2? (b) its connectivity numbers?
67. Give the Betti numbers, torsion coefficients, and connectivity numbers of
a square with all four edges identified as suggested by the symbol AAA A.
5-9. Cohomology Groups
We continue to study a finite oriented simplicial complex Km =
m
W = U {sk} with 'oriented &-simplexes {sk} = (s{, . . . , s* ). We
k=0
again turn to the theory involving integral chains, although the
discussion requires little modification to make it applicable to chains
with coefficients in an arbitrary abelian group.
We also continue to use e|, defined in Art. 5-?(?), for the incidence
number of sf and Sj+1. In cohomology theory, each &-chain Ck has
a so-called coboundary chain 6Gk which is a (k + l)-chain, while the
boundary chain dCk is a (k — l)-chain. The coboundary operator ?
plays a role in cohomology theory analogous to that of д in homology

Art. 5-9] HOMOLOGY AND COHOMOLOGY GROUPS 123
theory. For purposes of comparison, we formulate the boundary
and coboundary of the simplest &-chain Gk = sk, as follows:
(a) Э«* = 2 e^sf-1 (see E.46)),
E.68) i=1 (¦ = 1, . . ., afc).
(b) &* = 2 **X+1
Л = 1
Thus dsk is the sum of the (k — l)-faces of sf, each with the induced
orientation, and dsf is the sum of the (k + l)-simplexes with к for a
face, each with the induced orientation. The coboundary of a A-chain
Ck = ??=1 arf is the (k + l)-chain 6Ck = ?^=1 ?№·
(A) As in homology theory (see Art. 5-7(A)), but with the roles of
rows and columns interchanged, for any bases {Cfc}, {Cfc+1} of (?fc, dk+lf
the incidence matrix Jk({Ck+1}, {Gk}) has the property that the
coboundary of the chain G\ labeling a row is the chain ??^1 ?^?^+1 in
which the chains labeling the columns appear with the respective
elements in the row as coefficients (see Art. 5-7(B)).
A &-cocycle is a yfc-chain Ck for which 6Gk = 0.
Theorem 15. Each coboundary chain is а сосуcle; that is, ???*-1 =
0 (the null (k + l)-chain) for any (k — l)-chain Ск~г (Exercise 72).
(A proof alternative to that of Exercise 72 follows the proof of
Theorem 16 below.)
(B) Because ? is a linear operator, it can be interpreted as a
homomorphism (Art. A-2) ? : Gk -> Ck+V The &-cocycles are, as a
direct consequence of their- definition, the kernel of this
homomorphism.
Theorem 16. The mappings д : Gk -+Ck_x and ? : Ск_г ->Gk are
dual homomorphisms (Art. A-5).
Proof. Let the chain groups be interpreted as integral modules
(?h = [sh] = [?*, . . . , e*j with {sh} for initial basis (h = 0, . . . , m).
Consider two elements
E.69) C* = 2 atfeG» 0*-1= ? ^ГЧ^.
г=1 * j=l
Evaluating the two sides of Eq. (AV7t) with f = dy S = Ck, g = ?,
W = Ck~\ we find
'<Xk-l / <*k \
j=l \г=1 /
(Gk^Gk-1)= ?*? $l%-
i=l j=l
Hence
E.71) (dCk · C*) = (Gk · ОС*),
which formulates Theorem 16.

124 INTRODUCTORY TOPOLOGY [Ch. 5
We also obtain the following alternative proof of Theorem 15.
For any Si+1 g {sfc+1}, since the dot product is commutative,
E.72) {???*-1 · s^1) = (dC*-1 · ds^1) = (C*-1 · dds*+1) = 0,
which implies that the coefficient of sj+1 in the expression ddC*'1 =
? а^+1 is zero.
(C) The &-cocycles and the cobounding &-cocycles are subgroups,
to be denoted by 3fc and 5fc> respectively, of (?fc, and
E.73) g* с 3* с (?fc (go = 0).
Here, and throughout, the groups of cohomology theory will be
denoted, as is customary, by the symbols for their analogs in homology
theory (see the notation in Theorem 14) with superscripts in place of
subscripts. The &-chain group (?fc = (?fc plays the same role in both
theories. Thus the kth cohomology group is denoted by
E.74) $* = 3*/g*.
We say that two ^-chains G\ and G\ are cohomologous, a relation
symbolized thus:
E.75) C\»Cl
if there exists a (k - l)-chain C*-1 such that ОС*-1 = G\ - C\.
As a consequence of Theorem 15, the relation E.75) implies
6G\ = dC\, a condition automatically fulfilled if G\ and C\ are k-
cocycles.
The classes into which <* separates the &-cocycles are, by E.74),
the elements of §fc and are called cohomology classes.
(D) The kth cohomology group of a complex К depends only on
its (#-&keleton. (Compare Theorem 3.)
EXERCISES
J 68. Write out ??0 and ?(?0??) for the complex consisting of all the faces of
s4 = VoViPbPbPii orienting each simplex according to the order of increasing
subscripts on the symbols for its vertices.
v 69. List all the &-cocycles (k = 0, 1, 2) for the complex consisting of all the
faces of s2 = p0p1p2-
? 70. By a second direct application of ?, show that ???0 and ??(?0??) are nuU
in Exercise 68.
4 \> 71. In the complex of Exercise 68, discuss the group §3.
72. Prove Theorem 15, by an argument analogous to the proof of Theorem 1.
73. In the notation of this article and Art. 5-7, find, in all dimensions, dual
bases {A, B, C} and {X, Y, Z} for the complex K2 consisting of the faces of
s2 = ?0???2 an(l of two 1-simplexes ?0?^ and p^pv

Art. 5-10] HOMOLOGY AND COHOMOLOGY GROUPS 125
5-10. Dual Bases
(^4) A set (C\, . . . , Gkr) of &-cocycles is called cohomology-independ-
ent, an abbreviation for linearly independent with respect to cohom-
ology, if
E.76) J а?*" 0 => «< = 0 (i = 1, . . . , r).
» = 1
Consider now the basis {A, B, C} = (A\9 . . . , Akk\ B\, . . . , BkPk\
C{, . . . , G* x) of Gj mentioned in Theorem 14. Let {?, ?, Z) =
(X{, . . . , i* ; 7*, . . . , Yfc Z*,..., ZkkJ be the basis of the
module &k = [sk] dual to {A, B, G) (see Art. A-5, Theorem 9) with
A\ dual to X\9 Bk to Yl and G) to Z) (h = 1, . . . , 7k; i = 1, . . . , /?fc;
j = 1, . . . , у;^). The homomorphism ? : (?fc -> (?fc+1 is determined by
??*=?*?« (i = 1, . . . , />fc),
E.77) aZ*=Z^ (j=P*+l,---,yfc),
07,fc = bZ\ = 0 (A = 1, . . . , ?,; г = 1, . . . , ум).
This statement, justified by Lemma 14 below, corresponds to the
labeling, in E.53), of the rows and columns of J% at the right and
bottom, in accordance with Art. 5-9(A).
To shed further light on the basis {?, ?, ?}, consider again the
stepwise procedure whereby the incidence matrices (J0, Jv . . . , Jw_!)
were normalized into JJ, J\, . . . , J^-i- Consider, in particular, the
elemjentary changes of base as described in Art. 5-7(D). The
correspondence there described between modifications in the incidence
matrices and changes in the labels of columns and rows was
determined by Condition (B) of Art. 5-7. The analogous principle from
the viewpoint of cohomology is stated in Art. 5-9(A).
Lemma 14. Let the incidence matrices be normalized precisely as
in Art. 5-7, save that the labels shall be determined in accordance with
Art. 5-9(Л). Assume at a given stage that the bases for each (?fc
corresponding to д and ? are dual. Then an elementary Operation on a
matrix results in new bases for each (?fc, for д and ?, which are again
dual.
Proof of Lemma. When changing labels for the sake of д we write
them at the top and the left, as in Art, 5-7. When changing labels
for the sake of ?, we write them at the bottom and the right. We use
Jk in place of Jk for the kth matrix with the rows and columns labeled
for the sake of ?. At the start of the process, we have
D+1)
E.78) Jb({a}) = (ej) |i 4 || (?*) = J*({*}).

126 INTRODUCTORY TOPOLOGY [Ch. 5
At a typical later stage, suppose
E.79) J^uUvb II 7* = J*.
ylc+l
}
As basis for an induction assume that, at the stage in normalizing
represented by E.79), {17*} = (U\, . . . , U*k) and {Vk} = (Vklf . . . ,
Vak) are dual bases of &k, and {Uk+1}, {Ук+1} are dual bases of &k+1.
This hypothesis is verified at the initial stage E.78), since the initial
basis {sk} for each module &k = [sk] (k = 0, . . . , m) is self-dual (Arts.
A-4, 5). For elementary operations of types A) and B) in Art. 5-7(D),
the conclusion of the lemma is easily verified, and we omit the details.
We now reformulate Part C) of Art. 5-1 (D) and combine it with its
analog corresponding to Art. 5-9(A) to obtain the following.
(B) Corresponding to the elementary matrix transformation
(new col j) = (old col j) + a(col h) in E.79) are the following changes
of basis:
(a) Э: Replace Uk+1 by (Uk+1 + aUl+1),
(b) ?: Replace Vkh+1 by (Vkh+1 - aVk+1),
where д and ? indicate the respective changes made, as in Art. 5-7,
for the sake of д and, as here, for the sake of ?. Corresponding to the
elementary matrix transformation (new row i) = (old row i) + 6(row h)
are the following changes of basis:
(а) Э: Replace Uk by (Uk - bU%
E.81) (b) ?: Replace Vk by (Vk + bVkh).
The proof is now completed by Exercises 74 and 75.
EXERCISES
74. Establish (B). Note that E.80a) and E.81a) are a reformulation of Part 3
of Art. 5-7(Z>), and that E.80b) and E.81b) are analogous statements from the
cohomology viewpoint.
75. Complete the proof of Lemma 14, showing that the changes of basis in
E.80) are a dual pair (Art. A-5) and, similarly, so are the changes in E.81).
5-1 I. Comments on Cohomology Groups
By E.77), the chains Yk and Z1· are &-cocycles, but the chains X\
are not. Let a chain Gk be expressed in terms of the basis {?, ?, ?}
thus:
У к ? к Ук-1
E.82) G* = 2 *А* + ? У ? + ? *&
Л=1 » = 1 3 = 1

Art. 5-11] HOMOLOGY AND COHOMOLOGY GROUPS 127
Its coboundary is
Pk Ук
E.83) 6Gk = ? V%+1 + ? *Л+1·
h = l h = pk+l
Thus Gk is a cocycle if and only if xh = 0 (h = 1, . . . , yk).
Furthermore, from E.83) with к replaced by к — 1, Ck is a coboundary
cocycle if and only if it is a linear combination of (r\~1Z\,... , ?^_1?^ ,
?* +1, . . . , ZkVki). Hence a cocycle Sfi-^Ff + ?^?,?* cobounds
if and only if ?/г· = 0 and ^ = 0 (mod ?*-1) (j = 1, . . . , /o^).
(A) Hence, A) {Xk, Yk, Zk) is a basis for (?/c = (?fc, B) {Yk, Zk} is
a basis for 3*, C) (?*^, . . . , ?^?* , ?* , . . . , ?* ) is a basis
for g7c, and D) the cohomology classes of (Y\, ... , Y^, Z*, . . . , Z* )
are a basis for §fc. The following theorem is a consequence.
Theorem 17. The kth cohomology group $&k{K) is isomorphic to
the direct sum of the (k — l)th torsion group Хк_г and the kth Betti
group 93*.
While the cohomology groups are thus deducible from the
homology groups, for finite complexes, they are valuable A) because their
use facilitates the statement and proof of various properties of
manifolds, some of which will be established in Chapter 7, and B)
because they have generalizations to spaces other than complexes,
where cohomology has led to the discovery of invariants not revealed
by homology considerations.
Cohomology theory entered into topology relatively recently,
within the past two or three decades, as compared with the homology
theory of classical topology. In recent years, cohomology has played
an increasingly important role, and an acquaintance with it is of
fundamental importance in algebraic topology.
EXERCISES
^ 76. Interpret Theorem 17 for §7c(P2), where P2 is the projective plane
(k = 0, 1, 2).
У 77. Show that a 0-chain C° on ? is a cocycle if and only if all the vertices in
each component of К -enter into C° with the same multiplicity.
y 78. Discuss §^(К) for an arbitrary linear graph.
r 79. Discuss the 1-cocycles on the 2-complexes shown in Figs. 5-3 and 5-5(b).
Find some which are coboundaries and some which are not.

6
Topological Invariance of
Homology Properties
A compact topological polyhedron was defined as a topological
space ? homeomorphic to a finite rectilinear simplicial m-complex
in an En for some natural numbers (m, n). Given a particular homeo-
morphism
F.1) h: \Km\ -> ? Km с Еп
and the corresponding triangulation Я™ of ?, we can apply the
considerations of Chapter 5 to Km and thus arrive at a set (/?/c, ?\, . . . , ?* )
(к = О, . . . , т) of Betti numbers and torsion coefficients for Rm.
While these numbers are derived with the aid of a particular
triangulation, the following basic result holds.
Theorem 1. The Betti numbers and torsion coefficients of a
topological polyhedron ? are independent of the triangulation Km
and are topological invariants of ?.
A proof of Theorem 1 will occupy most of the present chapter.
The method consists in giving topologically invariant definitions of
so-called singular simplexes, chains, homology groups, and so on. It
is then shown that the topologically invariant homology, Betti, and
torsion groups of the singular theory are, for any homeomorphism
F.1), isomorphic to the corresponding simplicial groups of Km. The
first proof of this fundamental theorem was by J. W. Alexander.*
The present improved proof was made possible by Theorem 4
below.
6-1. Singular Simplexes
Let sk denote a rectilinear ^-simplex, and let ? be an arbitrary
topological space.
* "A Proof of the Invariance of Certain Constants in Analysis Situs," Trans. Am.
Math. Soc. 16 A915), pp. 148-154.
128

Art.6-1] INVARIANCE OF HOMOLOGY PROPERTIES 129
(A) In our applications, ? will always be a topological polyhedron.
In work beyond the scope of this book, however, singular homology
theory like that about to be defined has led to the discovery of
important properties of more general spaces.
By a singular ^-simplex on ? we mean a pair (yk, /) where/ : sk —> ?
is a continuous mapping and where
F.2) /=/(«*)·
Although we will use yk as a symbol for this singular ^-simplex, it is
to be understood that the simplex is not merely a subset of ? but
consists of the subset yk together with its defining mapping. Closed
singular simplexes {yk,f) are similarly defined by / : sk -> ?.
Examples: A) A mapping of sk onto a single point defines a
singular ^-simplex. B) If ?1 is mapped onto a Peano curve (Art. 8-1)
covering a closed 2-simplex s2, then a singular 1-simplex is defined
which coincides, as a point set, with s2. C) Let the boundary of a
2-simplex s2 be mapped onto a point ? of a 2-sphere S2, and let the
inner points of s2 be homeomorphically mapped onto S2 — p. Thus
a closed singular 2-simplex covering S2 is defined.
(B) We say that two singular ^-simplexes, yk = f(sk) and dk = g(tk)
are equal, yk = dk, if there exists a linear homeomorphism ? of sk
onto tk such that q = ?(?) =>f(p) = g(q) (? e sk, q etk). Otherwise
expressed, yk = dk if and only if the inverse images, sk and tk,
correspond under some lhlear homeomorphism such that corresponding
points have identical images on yk and dk.
If a subset of sk is a linear simplex r3' of arbitrary dimension j < k,
then f(rj) will be referred to as an/-linear (or yfc-linear) j-simplex on
the closure of the singular simplex yk = f(sk).
(G) Thus the defining mapping / carries the concept of linearity
from sk onto yk. If yk — dk, then yfc-linearity agrees with ofc-linearity,
as a direct consequence of (B).
If sk is an oriented simplex, then yk =f(sk) will be called an
oriented singular ^-simplex and a distinction must be made between
F.3) +/=/(+**) and _/=/(_**).
This is done by changing the definition E) of equality, so as to
require that the linear homeomorphism ? map sk onto tk so that
orientations agree. This implies that, if the condition in Definition
(B) holds but orientations are opposite, then — yk = dk.
(D) A singular ^-simplex yk (k > 0) will be described as degenerate
if yk = —yk (Exercise 5); that is, if a linear, orientation-reversing
homeomorphism ? of sk with itself exists, such that if ? = X(q),
where p, q e sk, then f(p) =f(q).

130
INTRODUCTORY TOPOLOGY
[Ch.6
EXERCISES
1. Show that (a) a closed singular 1-simplex can be defined which covers an
arbitrary compact 2-manifoid, M2, assuming the existence of Peano curves,
defined in Art. 8-1 below, and (b) for an arbitrary к > 0, a closed singular
^-simplex can be defined which covers ikf2.
2. Give an example of two continuous mappings of the unit interval @ < ? < 1)
onto the unit interval @ < у < 1) which define unequal 1-simplexes.
3. Given a singular oriented ^-simplex yk =/(+sfc) (k > 0) show that there
generally exist \{k + 1)! different mappings of sk which define* yk.
4. In the previous example, give ayk = /(-\-sk) (k > 0), such that all mappings
of s* onto the set/( -\-sk) define (a simplex equal to) yk.
у 5. Which of the following singular oriented simplexes are degenerate, and
why? (a) The 1-simplex ?1 = /(s1) where "^ is the segment — 2 < t < 2 on the
tf-axis and wheref(t) = —t. (b) The same, where f(t) = \t\. (c) The same where
f(t) = |1 + t\. (d) The singular simplex defined by a linear mapping of sk
onto sj (see Art. 4-10B?)), where j < k.
6. Let s2 = abc be a rectilinear 2-simplex in the euclidean plane, let m be the
midpoint of ab and h the foot of the altitude from с onto ab. Suppose In ? m.
Which of the following mappings of s2 define degenerate simplexes? Justify
your answers, (a) The projection of s2 onto s1 = ab by lines parallel to ch.
(b) The same by lines parallel to cm. (c) The continuous mapping which is the
identity on acm and which maps bcm linearly onto acm. (d) The same as (c),
but with In in place of m.
7. Show that an arbitrary non-degenerate ^-simplex yk = f(sk) (k > 0) can
coincide, as a point set, with a degenerate ^-simplex dk = g(tk). Give necessary
and sufficient conditions on g for dk = — dk in the case yk = sk = tk.
6-2. Singular k-Chains and Groups
A singular &-chain on ? means a finite set (?\, . . . , у*) of oriented
singular ^-simplexes where y\ is associated with an integer at or,
more generally, with an element of an arbitrary given additive
(abelian) group. Such a chain is expressed as a linear form
F.4) Gk = atft + · · · + ???%
which is an appropriate expression in view of the following definitions
of equality and addition.
(A) Two chains are equal if they can be reduced to identical forms
by a sequence of elementary operations of the following types: A)
replacement of a singular simplex by an equal simplex; B) omission
of a term а{у\ if at = 0; C) omission of a term a^y) if y) is degenerate;
D) replacement of агук by (— af){— y\)\ E) permutation of terms;
F) replacement of a{y\ + Ъ{у\ by (аг· + Ьг)у\.
* Or, more completely, which define simplexes equal to yk. However, as here, we
will generally not distinguish between equal simplexes.

Art.6-2] INVARIANCE OF HOMOLOGY PROPERTIES 131
Consider next a second &-chain
F.5) Dk = Ъгд1 + · · · + bpdk.
The sum Gk + Dk is then defined as
F.6) Gk + Dk= J ai7k + J Ь,й*
г=1 г=1
or any chain equal thereto.
Lemma 1. The singular ^-chains form an additive group (??
(Exercise 8).
The group is called the &-chain group of ? over the coefficient group
Z. The &-chain group of ? over any additive abelian coefficient group
© is similarly defined.
Now consider a continuous mapping / : sk -> ? of a closed oriented
simplex sk into a topological space. Then yk =f(sk) is a singular
^-simplex. Its j-faces (j = 0, 1, . . . , k) are the singular j-simplexes
?? =f(sj), where sj = a j-dimensional face of sk. If j < k, yj is a
proper or boundary face and yj, yk are said to be incident. Let the
(k — 1)-dimensional faces of sk, with the induced orientations, be
denoted by sf_1 (i = 0, 1, . . . , k) so that dsk = ?^=04_1> and let
yk~x = /(sf_1). Then the boundary chain of yk is the singular (k — 1)-
chain dyk = ?*=0 y\_1.
(Б) Since the boundary operator is essential to homology theory,
we confine ourselves to lingular simplexes whose defining mappings
have continuous extensions over the closures of the inverse images.
Given a space ?, let (?? = (?*(?) denote the &-chain group of ? over
a coefficient group ©. We are primarily concerned with integral or
mod 2 chains, but the invariance proofs are independent of the
coefficient group. The set of all singular ^-simplexes which can be defined
on ? generate (?jf. The null element 0 of (?| is the chain in which
each coefficient is the zero element of ©. The relation
F.7) yk = 0 if yk is degenerate
is a group relation.
A singular &-chain Gk = ?"=1 а{у\ is a cycle if dCk = 0.
As a corollary to Theorem 1 of Art. 5-1, the boundary of each
singular (k + l)-chain is a singular &-cycle. The relation Ck ~ 0 will
mean that Ck = dDk+1, where Dk+1 is a singular (k + l)-chain on ?.
Also, G\ ~ G\ will mean that G\ - G\ ~ 0.
(C) Just as in the simplicial case, A) the singular ^-cycles constitute
a subgroup 3* of &*> B) the bounding ^-chains form a subgroup gf of
3*, and C) the ^-chains having non-zero integral multiples which
bound form a subgroup Djf з д* of 3*· Homology and cohomology
are defined and symbolized exactly as in the simplicial case.

132 INTRODUCTORY TOPOLOGY [Ch. 6
The kth singular homology, torsion, and Betti groups are (see
E.59))
F.8) Ь*к=ЗЖ> *? = ?>**/8?? »? = &/3? = 3?/»?·
One also writes, for example, §*(?) and §*(?, ©), where ? is the
space and © is the coefficient group.
Theorem 2. A continuous mapping ? : ?, —> ?2 of a topological
space ?? into a topological space ?2 induces a homomorphism фк :
C^) -> ??(?2) (A = 0, 1, . . .), with the property дфк = фк д.
Proof. If yk = f(sk) is a singular ^-simplex on ?? so is r\k = фк/(як)
on ?2 as a result of Theorem 5 in Chapter 3. If yk is degenerate, so is
r\k. The mapping ф defined by ?^? а{у\) = ? a$k(y\) clearly has the
properties фк(Ск + Dk) = фк(Ск) + фк{Бк) and is therefore a
homomorphism. It is a direct consequence of the definitions that фк
commutes with the boundary operator; that is, for each Ck e (??
F.9) дфкСк = фк дСк.
Corollary. The mapping фк carries singular ^-cycles of ?? into
singular ^-cycles of ?2, bounding cycles into bounding cycles, and
homologous cycles into homologous cycles.
(D) Hence фк induces a homomorphism
F.10) 9?,*:^*(?1)-§*(?2), »*(?1)-95?(?8), 3^) - ?*(?8).
Theorem 3. The groups §f, 33*, and 2f are topologically invariant;
that is, $?*(??) and §*(?2), for example, are isomorphic if ?? and ?2
are homeomorphic.
Proof. If ? is a homeomorphism, then it and its inverse, ?-1,
induce a homomorphism ?*, as in F.10), and an inverse
homomorphism ?*-1 between each of the pairs of groups in question. But a
homomorphism with an inverse which is also a homomorphism is
easily seen to be an isomorphism.
(E) Also, (?*, 3*> ?*> 3* are topologically invariant, which implies
the theorem, by precisely the same argument; but we must pass to
the factor groups (§*, 23*, ?*)> to obtain topological invariants.
EXERCISES
8. Establish Lemma 1 by verifying the group axioms for Hk.
? 9. Show that equal singular &-simplexes have equal boundary chains. Deduce
that dyk = 0 if yk is degenerate.
10. Show that a non-degenerate singular ^-simplex can have degenerate
boundary faces.

Art. 6-3] INVARIANCE OF HOMOLOGY PROPERTIES
133
6-3. Sperner's Lemma; Invariance of Dimension
(A) If К is a simplicial subdivision of a topological polyhedron
? = \K\, then each point ? of ? is on a unique simplex of \K\, called
the carrier of ? in K. Suppose that К is linear simplicial in E, and
let L be a linear simplicial subdivision of K. Let ?0 be a mapping
of the vertices {q} of L into the vertices of К such that, for each
q e {g}, A0(g) is a vertex of the carrier of q in K. Then A0 is easily seen
to be a simplicial mapping, in the sense of Art. 4-10, of the abstract
complex of L into that of К. ^ ^
Theorem* 4. Given an oriented тя-simplex sm9 suppose that K(sm)9
the complex consisting of the faces of sm9 is subdivided into a linear
simplicial complex Km whose m-simplexes are oriented like sm (Art.
5-6). Let/ be a mapping of the 0-cells of Km into those of sm under
which the image of each 0-cell of Кт is a vertex of its carrier in K(sm).
Then the number of ra-simplexes of Km whose vertices are mapped onto
all m + 1 vertices of sm with orientation preserved exceeds by one the
number whose vertices are mapped onto all m + \ vertices with
orientation reversed.
Proof.f We take sm polyhedral in a euclidean га-space^ Using
mathematical induction, take m = 1 so that sm = s1 is an oriented
1-cell and let ds1 = sj — sg. Let {t}} be the 1-cells of K1, oriented like
s\ Then /(?,. dt}) =/(*? - s°0) = ej - s°0. Now suppose that ? of
the t} have end points mapped onto s? and s[j so that orientation of the
1-cell is preserved, that у are mapped with orientation reversed, and ?
have both end points mapped onto s\ or both onto s{J. ????/(?, dt}) =
*(*? - *o) + У(*о - s°i) +z0 = (x - y){4 - s°0). Hence s* - s°0 =
(x — y)(Si — Sq), from which we infer that ? — у = 1, completing
the proof for m = 1.
Now assume that the theorem has been proved for ?? = 1, 2, . . . , r.
Let sr0, s[, . . . , srr+1 be the boundary r-simplexes of sr+1, with the
orientations induced by that of sr+1. Let {^+1} be the (r + l)-simplexes
of Kr+1 oriented like sr+1 and let [u]} be the r-simplexes of Kr+1
which are on sr0, . . . , 5j+1, each oriented like the one of the latter
on which it lies.
With the mapping by / of the vertices of each oriented simplex of
Kr+1 we associate in an obvious way a mapping of the oriented
simplex itself.
????/(9?^+1) = /(?^), which, by the theorem for m = r, equals
(so + si + ' ' ' + K+i) + degenerate r-simplexes. On the other
hand, /(9?$+1) = ?,/C^+1). Now if the number of distinct 0-cells
* Theorem 4, which represents a strengthening of Sperner's lemma (see (B) below),
and Lemma 7 below, which is a corollary, are due, together with their present proofs, to
Professor Arthur B. Brown (see Preface).

134 INTRODUCTORY TOPOLOGY [Ch. 6
among the images under / of the vertices of t\+l is exactly r + 1, it
is easily verified that /(9^+1) contains exactly two non-degenerate
r-simplexes of opposite orientations, so that/(9^+1) = 0 + degenerate
r-simplexes. If the number is less than r + 1, there are no non-
degenerate r-simplexes in /(9^+1). Hence the only cases in which
/(9^+1) ? 0 when degenerate r-simplexes are dropped are those in
which the r + 2 vertices of ^+1 are mapped onto the r + 2 vertices
of sr+1. Suppose that this happens for ? of the tr{ + 1 with orientation
preserved and for у with orientation reversed. Then ?{/(9??+1) =
{? — y)(sr0 + s{ + · · · + srr+1), + degenerate r-simplexes and, as in
the proof for the case that w = 1, we infer that ? — у = 1. This
completes the proof.
(B) It follows from Theorem 4 that an odd number, hence at least
one, of the m-simplexes of Km have their vertices mapped onto, not
merely into, the vertices of sm. This result is Sperner's lemma.
Theorem 5. If a compact topological polyhedron ? can be
triangulated into an m-complex, then ? has Brouwer dimension m.
Proof. Since Brouwer dimension is topologically invariant, it is
sufficient to give the proof for a linear simplicial polyhedron ? =
\Km\ in a euclidean space En.
Lemma 2. A closed linear m-simplex sm = ?0?? . . . pm in En is
of Brouwer dimension at least m.
Proof of Lemma. It is to be shown (Art. 3-8) that, for ? > 0
sufficiently small, a closed finite covering {A} of sm is of order >m if
diam ? < ? for each A e {A}.
(G) Let ? > 0 be so small that A) no A e {A} intersects all the
(m — l)-dimensional faces of sm and B) no A e {A} intersects both a
vertex Pj of sm and the closure sj1'1 of the opposite face (Exercise 11).
Since sm с и {A} and since (C2) implies that no A e {A} contains
two vertices of sm, it follows that m + 1 of the ^'s can be so numbered
that p5 e A j (j = 0, 1, . . . , m). Let 3l0 be the subset of {A} consisting
of those elements which do not intersect sff~l, and let ? ? be the union
of the sets 3l0. For some j e A, . . . , m) assume (A$, . . . , Af_^
defined, where Af is the union of a subset 3^ of {A} (i = 0, . . . , j — 1).
Let 31,· be the subset of {A} — \J 31г· consisting of all elements thereof
which do not intersect sf~l, and let Af be the union of the sets %.
(D) The recurrent definition just given leads to a partition of {A}
into subsets 3l0, . . . , 3lw with the property that a closed face sk =
к
Pi · · · Pik °f ?™ is covered by U 31г-л and is not intersected by % for
j ? (»o> · ·*· > **) (exercise 13). A=0
Now let Km = {s} be the set of all faces of sm. Let ? be an integer

Art. 6-3] INVARIANCE OF HOMOLOGY PROPERTIES 135
so large that the mesh of the vth barycentric subdivision K™ of K™ is
less than the Lebesgue number ? of the covering A$, . . . , А*ъ (Arts.
4-11(C) and 3-8(Я)). Let each vertex q of K™ be mapped onto pj
where j is the smallest number such that q e Af. By (G) and (Z)),
Pj is a vertex of the carrier of q on K™. By Sperner's lemma, there
exists an га-simplex tm e K™ whose vertices are mapped onto (p0,
Pv · · · > Pm)- But tm is a se^ °f diameter less than ? intersecting all
the sets (A$, . . . , Л*). Therefore (Chapter 3, Theorem 22) these
sets have a non-vacuous intersection; hence, the covering {A} is of
order at least m + 1, proving Lemma 2.
Corollary. If ifw is a finite linear m-complex in En, then dim |iiw|
> m (Exercise 14).
To complete our proof of Theorem 5, it remains to show that
dim \Km\ < m.
The star St{q) of a vertex q of Km means the union of q and all
the simplexes of Km incident with q.
Lemma 3. Let ?? be the first barycentric subdivision of К = Km,
and let {q} be the vertices of K. Let St(q) be the star of q on Kv Then
F.11) {Si} = {Si(q)\qe{q}}
is a closed covering of ? = \K\ of order m + 1.
Proof of Lemma. By definition, St{q) is the union of the simplexes
of Kx incident with q. Let sk = q0qx . . . qk be a ^-simplex of if, and
let (w0, %, . . . , uk) be barycentric coordinates on sk relative to
(?o> Vv · · · > ?*)· _
(E) The part of sfc covered by St(q0) is defined (Exercise 16) by the
inequalities u0 > иг (г = 1, . . . , k). Hence, the barycenter of sk is
к
the only point on sk which belongs to the intersection f\ St(qi).
г=0
Applying this argument for к = т, we find that each barycenter of
an m-simplex of Km belongs to exactly m + 1 of the sets {St}, and
that these are the only points where (m + 1) of these sets intersect.
Corollary. For an arbitrary ? > 0, there exists a finite covering
of ? = \K\ which is of mesh ? and order m + 1.
For, by Art. 4-1 f(C), we can choose ? so large that Kv+1 is of mesh
E/2. This implies that each star on Kv+1 is of diameter less than ?.
Since Kv+1 is the first barycentric subdivision of Kv, the lemma
implies the corollary. The latter, in turn, implies dim \K\ < m,
completing the proof.
(F) Theorem of the invariance of dimension: The dimension number
m of an m-complex Km is a topological invariant of the space ? =
|UTW|. This follows from Theorem 5 and the topological invariance of
Brouwer dimension (Chapter 3, Theorem 20).

136 INTRODUCTORY TOPOLOGY [Ch. 6
This result applies to the case where Я is infinite, hence ? not
compact, if the Brouwer dimension of ? is defined as m provided ?
contains as a subset some compact topological polyhedron of
dimension m but none of dimension m + 1.
(G) The dimension number ? of euclidean тг-space En equals its
Brouwer dimension.
To establish (G), first subdivide En into the infinite complex К
consisting of all the faces of all the тг-cells of the form rni < ?. <
т{ + 1 (г = 1, . . . , ?), where the ra's are integers. Even though the
polyhedral complex К is infinite, it can be subdivided into a simplicial
тг-complex by the process used in the proof of Chapter 4, Theorem 8.
Statement (G) now follows from Theorem 5.
EXERCISES
11. Show that there exists a number ? which satisfies (C).
12. Draw figures to illustrate the sets A. and A}- in the proof of Lemma 2 for
the case where m = 2 and {A} is a set of overlapping disks.
13. Prove (D).
14. Prove the corollary to Lemma 2 by proving the more general result that
A c: S => dim S > dim A, where A and S are compacta.
15. Draw a diagram to illustrate the covering {St} in Lemma 3.
16. Prove (E), with the aid of the corollary to Chapter 4, Theorem 6.
6-4. The Brouwer Fixed-Point Theorem
We include this article here because it involves an application of
Sperner's lemma.
A large body of topological theory with important applications in
other branches of mathematics is concerned with fixed points of
mappings. Let ? be a topological space, and let A c= T. Consider a
mapping / : A —> ? of A onto a subset /(^4) of the same topological
space T.
(A) A fixed point of/is then defined as a point ? e A which coincides
with its image; that is
F.12) ? =f(p)op is a fixed point of/.
Obviously, ? must then be in the intersection A nf(A).
(B) The graph of/ is the following subset of ? ? ?:
F.13) F = F(f) = {(p,q)eT ? T\q=f(p)} (see Art. 3-2 (F)).
The graph of the identity mapping i, which maps each ? e ? onto
itself, ? = i(p), is called the diagonal D of ? ? ?:
F.14) D = F(i) = {(?, ?) ? ? G ?} с ? ? ?.

Art. 6-4] INVARIANCE OF HOMOLOGY PROPERTIES 137
The fixed points of/then correspond to the intersection F ? D, under
the projection ? ? ? -> Т.
To clarify the foregoing, consider the case where ? is the real
number system, and let ? ? ? be interpreted as the (x, ?/)-plane, so
that ? is represented by the x-axis X and also by the ?/-axis Y. A
mapping/ of А с X = ? onto В =f(A) с 7 = ?7 has a fixed point
on X corresponding to each point where the graph F :y =f(x)
intersects the diagonal, which is the line D : у = ? (Fig. 6-1).
У
h
*1
ai
Fig. 6-1. Fixed points for/: A —* Б с= А
Theorem 6. Any continuous mapping/of a segment ?: ?? < ? <
?2 onto В ^ A possesses at least one fixed point.
As an aid in proving this theorem, consider Fig. 6-1. Geometrically,
the theorem means that the graph J7 of a continuous mapping cannot
cut across the square A x A from the left side to the right side
without intersecting the diagonal (Exercise 17).
Theorem 7 (Brouwer's Fixed-Point Theorem). Let ВЪе a space
homeomorphic to the solid m-sphere ?™=1 x\ < 1 in Em, and / : D —* D
a continuous mapping of D into itself. Then / has at least one fixed
point.
Proof. Consider first the case D = s, where s is the closure of a
simplex s = р0рг . . . pm.
Let (u) = (u0, uv . . . , um) be barycentric coordinates relative to
(Po> Pi> · · · ' Pm)> and 1е^ (K> · · · > um) denote the image under / of a
point (u). Let Ap j e @, 1, . . . , m), be the set of all points (u) for each
^ ?

138 INTRODUCTORY TOPOLOGY [Ch. 6
of which u'j < иj. Suppose (u) ? pio . . . pi]c = sk. Then, by definition
of barycentric coordinates,
F.15) uio ? + uik = u'0-\ + u'm= 1.
But each и and each u' is non-negative. Hence, u{ > и- for some
& ' lh lh
h e @, 1, . . . , k), for otherwise ul + · · · + u{ would be less than
u'io + · · · + u'ik < 1. It follows that each point of sk = pio . . . p{
belongs to at least one of the sets A{ . That is,
F-16) S* = pit...pik<zUAik.
/1 = 0
Now let К = {s} be the set of faces of sm; and let Kv be the vth
barycentric subdivision of К, where ? is so large that the mesh of Kv is
less than the Lebesgue number ? of the covering A0, . . . , Am of sm
(Arts. 3-8 and 4-11). Let ? : Kv -> К be a simplicial mapping such
that, for each vertex q of Kv, A) 99(g) is a vertex of the carrier of q in К
and B) q Cmi 99(g) are common to some set A}. The possibility of
imposing this second condition is assured by F.16). By Sperner's
lemma (Art. 6-3E)) there exists at least one m-simplex tm e Kv, whose
vertices are mapped onto (p0, ръ . . . , pm). But, by the definition of 99,
this implies that tm intersects (A0, . . . , Am). Since diam tm < Я, the
sets (A0, . . . , Am) have a common point p. At p, 0 < u\ < u{ (i =
0, 1, ... , ??), by definition of A{, which, with ??'{ = ??{ = 1, implies
u[ = ub so that ? is a fixed point. By a homeomorphic mapping, this
proof can be applied to the general homeomorph D of s.
EXERCISES
17. (a) Prove Theorem 6, using properties of real continuous functions,
(b) Show that Theorem 6 becomes false if ? is replaced by an open interval ?? <
? < a2 or a half-open interval ?? < ? < a2.
18. Show that a continuous mapping of the circle x2 + y2 = 1 onto a proper
subset of itself has a fixed point.
19. Let / be a homeomorphism of the clockwise oriented circle x2 + y2 = 1
onto itself, oriented counterclockwise. Show that/ has a fixed point.
20. Show that a continuous mapping / of a segment A : ?? < ? < a2 onto a
segment В : \ < ? < b2 containing A has a fixed point, but that this is not
generally so if / : A -> В is merely into.
21. Show that a continuous mapping of the x-axis onto a bounded subset of
itself has a fixed point, but that a continuous mapping of the x-axis onto the
subset ? < 1 need not have a fixed point.
22. Show that the 2-sphere x2 + y2 + z2 = 1 can be homeomorphically
mapped onto itself by a mapping with exactly one fixed point.
23. Show that the closure s2 of a 2-simplex can be homeomorphically mapped
onto itself so that the only fixed point is a vertex.

Art. 6-5] INVARIANCE OF HOMOLOGY PROPERTIES 139
24. Go through the proof of Theorem 7, with figures to show the sets A^ for
the case m = 2, where/ (u0, uv u2) = (uv u2, u0).
25. Show that the solid sphere x2 + y2 + z2 < 1 in E3 can be homeo-
morphically mapped onto itself (a) so that @, 0, 1) is the only fixed point, (b) so
that @, 0, 0) is the only fixed point.
6-5. Invariance of Regionality
Theorem 8. If h is a homeomorphism which maps the closure s of a
simplex s = ?0?? . . . pm into Em, then h maps each point of s onto an
inner point of h(s).
Proof. We commence with a lemma.
Lemma 4. Let С be a closed bounded subset of Em, and let {A} =
(Аг, . . . , Aa) be a closed covering of С such that there is just one point
? eC which belongs to m + 1 of the sets {A}. Then, if ? is not an inner
point of C, there exists, for each ? > 0, a closed covering [A'} =
(A[, . . . , A'a) of order m, where A{ and A\ coincide outside the
spherical ^-neighborhood of p.
Proof of Lemma. Let 8 be the (m — l)-sphere of radius ? about p.
Suppose the A's so numbered that (Av . . . , Ac) is the subset of \A}
consisting of all elements each of which intersects S, and let A* =
ifnS(i = l c). Then {A*} = (Л*, . . . , A*) is a finite closed
covering of С П S of order at most m. By Lebesgue's lemma (Chapter
3, Theorem 22), there exists a number ? > 0 so small that the solid
sphere p(p, q) < ? about a point q e S intersects at most m elements of
{A*}.
Since S is of dimension m — 1, it admits a closed finite covering
(Bv . . . , Bv) = {B} of order m, where diam B{ < ? (i = 1, . . . , v).
We define a partition ?? . . . , /?c of {B} as follows: A) ?? is the set of
all the sets B{ each of which either intersects A\ or else intersects
none of the sets (AJ, . . . , A*). B) Once (??, . . . , ??_?) are defined,
for some h > 1, ?? is defined as the set of all the sets B{ each of which
intersects Af and does not belong to one of the sets (?? . . . , /??_?).
This recurrency applies for h = 2, . . . , с
Now let 49 = A* U ?? (г = 1, . . . , с). Then (A°v . . . , A°c) covers
С V
S, since (J ?? = (J Бг- covers #. If g e S, then the union of the sets
г = 1 г" = 1
Вг containing q is on the spherical ?-neighborhood of q, which
intersects at most m of the sets A%. It follows that q belongs to at most m
of the sets A\, since in order to belong to a set A\ it must belong to
Af or to a set Bt which intersects A\.
(A) We have just shown that (AJ, . . . , A%) is a covering of S of
order <m.

140 INTRODUCTORY TOPOLOGY [Ch. 6
Since ? is not an inner point of C, there exists a point p0 of ? —С
inside S. Let A" be the union of all the line segments from p0 to points
of A® (i = 1, . . . , c). Then (A'[, . . . , A"c) is a closed covering of the
solid sphere bounded by S, and p0 is the only point where more than
m of the sets (A'[, . . . , A") intersect. Let A\ be the union of A"{ with
the part of A{ outside S (i = 1, . . . , c), and let A\ = A{ (г = с + 1,
. . . , a). Then (A[, . . . , Л^) satisfies Lemma 4.
Now suppose that, contrary to Theorem 8, there exists a point
q e sm = р0рг . . . pm such that ? = h(q) is not an inner point of
h(sm). Let ? be the complex consisting of the boundary faces of sm
and let ?? be its first bary centric subdivision. Let ?,- be the star of p}-
on ?? (j = 0, . . . , m). Then (?0, . . . , ???) is a closed covering of
^m _ sm 0? or(jer m (Lemma 4). Let Dj be the union of all the
segments joining q to ?,·.
(B) Then (D0, . . . , Dm) is a closed covering of ?w, such that ? is
the only point common to all m + 1 of the sets Dj.
(C) There exists a number ? > 0 so small that if A) (D'0, . . . , D^)
is a closed covering of sm and B) the parts of ?'· and Dj outside the
?-sphere about q are identical, then D'0 ? · · · ? D^ is not vacuous
(Exercise 26).
Let A, = h(Ds) (j = 0, . . . , m). Then {A} = (A0, Al9 . . . , Лт)
fulfills the hypotheses of Lemma 4 for С = h(sm). Hence, ? = h(q)
being assumed not to be an inner point of h(sm), there exists, for each
? > 0, a closed covering A'0, . . . , A'm such that A) A'· and A5 coincide
outside the ^'-sphere about ? and B) A'0 ? · · · ? Л^ = 0.
Now let ? be so small that A-1 maps the intersection with h(sm)
of the spherical ^-neighborhood of ? into the spherical ?-neighborhood
of q, and let Щ = h-ЦА}) (j = 0, . . . , m). Then A'0 П - · · П A^ = 0
implies D'Q ? · · · ? D^ = 0, contradicting (C) and proving Theorem 8.
Corollary 1. Under the hypotheses of Theorem 8, each point of
sm — sm maps onto a non-interior point of h(sm) (Exercise 27).
Corollary 2. If h is a homeomorphism mapping a set С с Ет
onto A(C) cz E™9 where Em and ^ are euclidean m-spaces, then h
maps each inner point of С onto an inner point of h(C).
Tn particular, if С is a region* of Em, then h(G) consists entirely of
inner points. Thus, since h(C) is easily shown to be connected, Theorem
8 implies the topological invariance of the property of being a region.
EXERCISES
26. Prove (C). The material following Art. 6-3 (D) can be used.
27. Establish Corollaries 1 and 2 to Theorem 4.
* That is, a connected subset made up entirely of inner points.

Art. 6-6] INVARIANCE OF HOMOLOGY PROPERTIES
141
6-6. Singular and Simplicial Groups on a Topological Polyhedron
Consider an arbitrary compact topological m-dimensional
polyhedron ?, and let h be a homeomorphism mapping ? onto a simplicial
polyhedron ? = \K\ с En, where К = {s} is an oriented simplicial
complex. By Chapter 4, Theorems 8 and 9, К is finite and
m-dimensional.
(^4) Theorem 1 will, by Theorem 3, follow in all its generality if it
be shown that the singular homology group §*(P) is isomorphic to
the simplicial homology group $k(K) as defined in Chapter 5. Symbols
for simplicial homology groups and chains are modified by asterisks
to obtain symbols for the corresponding singular groups and chains.
The first barycentric subdivision (Art. 4-11) of a complex К will
be denoted by Kx = ??, where ? is the (barycentric) subdivision
operator. Thus the rth barycentric subdivision of К is
F.17) Kr=CK (r=l,2,...), ?0=?°? = ?.
(B) We need to define ? for an oriented complex K. We do so by
requiring that all the ^-simplexes of Kx = ?? on a ^-simplex sk e К
shall be oriented like sk (Art. 5-ЦА)). The requirement has meaning,
since all these ^-simplexes are on A(sk), which is a euclidean k-
dimensional subspace of En.
(C) The operator ? induces a subdivision operator ? for chains,
where Uk is the chain of ?s* (Art. 5-6(i)), and where, if Gk = ??*=1 а$,
(ck1=lCk=Z«l1aM
F.Г8) ...
[ Ck = ?Ck = ??^?* (r = 2, 3, . . . ), Ck= ?°0*.
Lemma 5. The operators д and ? commute, that is:
F.19) dtrCk = trdCk (r = 1, 2, . . . ).
Proof of Lemma. Let the (k — l)-faces of an oriented ^-simplex
sk be oriented like sk, and let the j-faces (j < к — 1) be arbitrarily
oriented. Then dsk is the (k — l)-chain of its boundary complex.
The faces of the simplexes in ?^ constitute, by the definitions, a
coherently oriented &-pseudomanifold, with the subdivided boundary
simplexes of sk for boundary. Lemma 5 now follows readily from
Art. 5-6, Lemma 11 (Exercise 29).
(D) The concept of barycentric subdivisions of simplicial complexes
and simplicial chains is carried over to singular complexes and chains
in a direct way with the aid of continuous mappings of subcomplexes
of К into an arbitrary topological space. Lemma 5 holds for the
singular case.

142 INTRODUCTORY TOPOLOGY [Ch. 6
The ideas underlying the present proof of Theorem 1 are as
follows. It is to be shown (see (A) above) that §*(P) ъ ЬАК) where
я& denotes the relationship of being isomorphic. Simplicial simplexes
and chains can be considered as special cases of singular simplexes
and chains, where the function f (Art. 6-1) maps each point onto
itself, and similarly for cycles and boundary cycles. That is,
F.20) G? => cb 3* => 3,, d* 3 Db g* 3 gb
But the singular groups in F.20), for к > 0, contain vastly more
elements than the simplicial.
EXERCISES
28. Verify Lemma 5 directly in the case where r = 1, Ck = -\-s2 and s2 =
P0P1P2· Illustrate with a diagram.
29. Explain how Lemma 5 follows from Art. 5-6, Lemma 11.
6-7. Simplicial Subsets of Singular Homology Classes
We continue to use the barycentric subdivision operator ?, subject
to Art. 6—6B?). As a consequence of Art. 6-1B)), ? can operate on
singular complexes as well as on linear simplicial complexes, where a
singular complex is the set of all the faces of a set of singular simplexes.
Different simplexes of a singular complex may intersect. They may
also have common vertices without having in common the respective
boundary faces they determine.
Tf yk is a linear or singular ^-simplex, then ???]? will denote the
chain of ????€ (Art. 5-6B")). The operator lv is applied to a singular
or simplicial chain as follows:
F.21) ???^ = ??^??
By the complex K{G\) of a singular or simplicial chain G\ = ??__г агу\
we mean the set of all the faces of all the simplexes y\ (i = 1, . . . , ?).
Now let 2Г* be a singular ^-complex on the linear simplicial complex
K. We will say that K\ is star-related to Kk if the closure St*(p) of
the star on Kk^ of each vertex of 2Г* is on the open star of some vertex
???. That is,
F.22) St(q) =5 St*(p) for some q ? ?.
We proceed to obtain a consequence of this relationship.
(A) Extending the definition of star, we define the star St(sh) of a
simplex sh of К as the union of all the simplexes of К having sh for a
face. If p0, . . . , ph are the vertices of sh, then
F.23) St(sh) = П St(p{)
i = 0
since the simplexes of К with sh for a face are precisely the simplexes
thereof which have all the points (p0, . . . , ph) among their vertices.

Art. 6-7] INVARIANCE OF HOMOLOGY PROPERTIES 143
h
Conversely, if (p0, . . . , ph) is a set of distinct vertices with f| St(pj) ?
j=o
0, then (p0, . . . , ph) are the vertices of an ?-simplex of К (Exercise 30).
Lemma 6. Let K\ be star-related to K, and let yk = qQqx . . . qk be
a simplex of K\. Let pj be a vertex of К such that Stipj) => St*(qj)
(j = 0, 1, . . . , k). Then the distinct vertices in the set (p0, . . . , pk)
determine a simplex sh e K, and St(sh) => yfc.
Proof of Lemma. It follows easily from the hypothesis that
к к
Г\Щр3) ? 0; hence, by (A), that П St(Pi) = #*(**) f°r some sh e K.
j=o _ j=o
Since yk cz St*(qj) (j = 0, . . . , &), we have
к к
F.24) ??(?») = П St(Pi) з Гй*(<7,·) = f
3 = 0 3 = 0
as was to be shown.
Now let {q} = (q0, qv . . . , qa) denote all the vertices of K\, and
let g0 be a mapping of {q} into the vertices of К such that ?И(д0(д^) =>
St*^^. As a consequence of Lemma 6 and Art. 4-10, Lemma 7,
there is a unique extension g : \K\\ -> |if | of g0 such that <7(yfc) is a
closed simplex of К for each y/c cz j^* and # is linear in terms of
singular barycentric coordinates on yk and barycentric coordinates on
9(yk)'
(B) For each yk e if*, we will call $л = g(yk) the approximating
simplex to yk (with respect to gr). The set of all such approximating
linear simplexes to the singular simplexes of K\ will be called the
approximating simplicial complex д*(Кк^) to K^, where g* is the
simplex mapping induced by g.
Theorem 9. A singular homology class of ^-cycles on \K\ contains
at most one simplicial homology class of ^-cycles on K.
Proof. We will deduce the theorem from the following lemma.
Lemma 7. If a simplicial &-cycle Zk is the boundary of a singular
(k + l)-chain C* + 1, it is also the boundary of some simplicial (k + 1)-
chain Dk+1.
Proof of Lemma. Let ? be so large that the complex K^+1 of ?PC*+1
is star-related to K. Let Kh = д*(Кк^+1) be an approximating
simplicial complex to K^+1, in the notation of (B). We will denote
with g the chain mapping induced by g*. As a consequence of Theorem
4, gl»Zk = Zk. But glpZk = dglpCl+1, by Theorem 2 and Lemma 5.
Hence Zk is the boundary of the simplicial chain Dk+1 = ^PG*+1.
Now let Z\ and Z\ be two simplicial cycles in the same singular
homology class, so that Z\ — Z\ bounds a singular chain. By Lemma
7, Z\ — Z\ also bounds a simplicial chain. Hence Z\ and Z\ are in
the same simplicial homology class, and Theorem 9 follows.

144
INTRODUCTORY TOPOLOGY
[Ch. 6
EXERCISE
30. Prove the last sentence in (A), perhaps by an induction.
6-8. Chains on Prism Complexes
The work of this article is preliminary to a proof that each singular
&-cycle on |If | is homologous to some simplicial &-cycle on К (Art. 6-9,
Theorem 10).
Let (xv . . . , xn, y) be a rectangular cartesian coordinate system
in an En+1 (n > 0) and let En be the subspace у = 0. If ? e En has
coordinates (xv . . . , xn, 0) then (p, y) will mean the point (xv . . . , xn, y).
Let ak = (ak, 0) be a linear ^-simplex in En, and let ? be the
interval 0 < у < 1 on the y-axis. The (k + l)-prism (or prism cell)
with floor ak will mean the convex polyhedral (k + l)-cell (see Fig.
6-2a and Art. 4-12)
F.25) 77fc+1 = ak ? ? = {(?, у) I p e ak, 0 < у < 1}.
The roof of 77fc+1 is the ^-simplex rk = (ak, 1) = {(p, 1) | ? e ak}.
Now let Fk = {a} be a finite linear simplicial complex in En with
?,-simplexes {ah} = <rj, . . . , a\n (h = 0, . . . , k). Let ?^+1 = a\ ? 7
be the prism cell with of for floor, and let r\ be the roof of ?^+1.
(a)
Fig. 6-2. A simplicial prism ?3 and its triangulation ???*. Lines in interior
partially shown. Lines on rear rectangle not shown, (a) ?3; (b) ?0??; (с) ?^;
(d) ?2?*.

Art. 6-8] INVARIANCE OF HOMOLOGY PROPERTIES 145
(A) The set of all the faces of all the prism cells ??-+1 is a special
sort of polyhedral (k + l)-complex, ^/c+1 (Art. 4-12), to be called a
prism complex. Its cells fall into three categories:
A) the floor simplexes, constituting the floor complex Fk = {a};
B) the roof simplexes, constituting the roof complex Bk = {r} =
{r?|A = 0,...,*; < = 1,...,??};
C) the prism cells or wall cells {77} = {??^+1 | h = 0, . . . , к; г =
1, . . . , ??}.
(d)
Fig. 6-2. (cont.)

146 INTRODUCTORY TOPOLOGY [Ch. 6
(B) Since our homology theory has been developed for oriented
simplicial complexes, we next define some standard simphcial
subdivisions and orientations of ф/с+1. We will denote the minimal
central subdivision (Art. 4-l2(F)) of ?+1 by ?0?+1 (Fig· 6-2b).
More generally, for ? =0, 1, 2, ... , we will denote by ? tyk+1 the
triangulation obtained by first replacing Ek by its pth barycentric
subdivision, ????€ (?°?}? = Ek), then applying the minimal central
subdivision process (Fig. 6-2c, d, where the subdivisions of the
interior, and of the rear rectangle, are not shown). Thus ??^?}€+1 has
???}? and Fk for subcomplexes. If h > 0, it has just one vertex on
each 77-f+1, namely the barycenter of 7rf+1, which is the point (m, |)
if m is the barycenter of the floor of ??\+1 (Art. 4-12(Z))).
(C) The standard projection of |^3fc+1| is the mapping ? : |^fc+1| ->
\Fk\ defined by ?(?, у) = (?, 0) (? e \Fk\). It maps each element
? ? ? = {(?, y)\pe \F% 0 < у < 1} of |?*+4 = \Fk\ ? ? onto
its initial point (p, 0). For each ? e @, 1, 2, . . .), ? induces an
isomorphism 99* between the г-simplexes of ???}? and those of ?^?€ (i = 0,
1,...,*).
(D) Let Fk be arbitrarily oriented, and let {a} now denote its
positive simplexes. Consider a typical ?-simplex ah = +р0рг . . . ph.
We orient the roof rh of ??+1 = ah ? ? by defining rh = +#0#i · · · 4n
where qt = (p{, 1) (г = 0, . . . , h). The A-simplexes of ???? are, of
course, oriented like rh. The isomorphism 99* : ???* —>- ?pFk is obviously
orientation-preserving.
(E) To orient the (h + l)-simplexes of ???+1 (h > 0) on the
typical 7??+1 = ah X Y, we first let m denote the barycenter of ??+1
and define ah+1 = +mah = +mp0p1 . . . ph, thus assigning the
orientation induced by ah to the incident (h + l)-simplex of ????+1
on 7??+1. The other (h + l)-simplexes of ?????+1 on ??+1 are to be
oriented like ah+1, which has meaning (Art. 5-6(A)) because they are
(h + l)-simplexes on the Eh+1 containing ??+1 (Fig. 6-2). If h = 0,
with pt on Fk and qi the corresponding 0-cell on Ek, the incident
wall cell is oriented +(^рг.
(F) Applying (D) and (E) to all the ah e {?} (h = 0, . . . , к), а
standard orientation is induced by the orientation of Fk on the roof
complex ????€ and on each (h + l)-simplex of ^p^37c+1 on a prism
(h + l)-cell. The other simplexes of ^p^3fc+1 can be arbitrarily
oriented.
Lemma 8. Each roof simplex r\ has the opposite of the orientation
induced by the incident (h + l)-simplex th+1 of ?0^ on ?^+1 under the
standard orientation.
Proof of Lemma. For h = 0 the result is obvious from (E). Now
suppose h > 0. We drop the subscript as in (D) and (F). If Oq+1 is

Art. 6-8] INVARIANCE OF HOMOLOGY PROPERTIES 147
obtained by translating ah+1 = mah (see (E)) one unit in the ^/-direction,
then ?E+1 has rh for a face with the induced orientation. Furthermore,
Oq+1 is oriented like ah+1, hence like th+1, and is adjacent to th+1 with
rh as common face. Hence (Art. 5-6, Lemma 9), th+1 and Oq+1 induce
opposite orientations on rh, and therefore rh has orientation opposite
to that induced by th+1, completing the proof.
(G) We adopt the following notation and terms:
(a) lpr\ = the chain of ???\ (Art. 5-6 (F));
F.26)
(b) *7РИ7+1 = the chain of the oriented (h + 1)-
cellsof7yp^*+1on7rJ+1.
If Ch = ^lh=i a^i is an arbitrary ?-chain whose oriented cells are
cells of Fk, then, letting Dh = ?^? а У}, we have
????? _ ^ ?·??^ is the corresponding roof chain,
F.27)
rjpWh+1= 2 ????^??+1 is the connecting chain,
г = 1
and we will refer to (Ch, ?*jpWh+1, tpDh) as a triple of related chains.
Lemma 9. Let ef·-1 be the incidence number of of and of-1, and let
(C\ ????+1, tpDh) be a triple of related chains, with h > 0. Then, in
the notation of (G)
F.28) dr%Wn+1=Ch- ????- j 2 ^?*?4^7·
i: = l j? = ?
Proof of Lemma. The lemma will follow (Exercise 32) in all its
generality if we show that
F.29) dijpWJt+1 = of - ??? - 2 ejr1'^*
Now rjpWf- + 1 is the chain of the similarly oriented (A + l)-simplexes
of a triangulation of 7rf+1. It is to be shown that the right side of
F.29) represents the chain B\ of the coherently oriented boundary
A-simplexes on ??·+1. By (E), of appears in B\ with coefficient +1.
By Lemma 8, the simplexes of ???}- appear in B\ with coefficient —1.
In the chain ddijpWf-+l, da] contributes ef;_1of ~\ which, in F.29), is
canceled, as it should be, by the term in of_1 from the last sum (see
(E)). Since any incorrect coefficient in the last sum in F.29) would
clearly give the wrong coefficient of the corresponding of_1 in the
boundary of the right-hand side of F.29), we conclude that all the
coefficients in F.29) are correct, and it is clear that each ?-simplex of
B\ appears on the right side of F.29).
(Я) If Gh is a cycle in Lemma 9, then
F.30) dfjpWh+1 = Ch - t"Dh (p = 0, 1, 2, . . . ),

148 INTRODUCTORY TOPOLOGY [Ch. 6
so that Gh and lpDh are homologous cycles on Pk+1. For, since Gh is
a cycle,
<*ft «ft-1
dch = j ? «4r4_1 = 0.
г=1J=l
which implies ?^? a^f1 =0 (j = 1, . . . , ?^?). Relation F.30)
now follows from F.28).
Lemma 10. In terms of singular homology theory on |^fc+1|,
Ch~lpC](P =0, 1,2,...).
Proof of Lemma. Let ? be the chain mapping induced by the
projection ?. Tt is the identity on the floor chain Gh. By the last
statement in (D), ?^?? = ????. Hence, by (#) and Theorem 2,
ф dijpWh+1 = d<pijpWh+1 =Gh - lpCh, and Lemma 10 follows.
EXERCISES
31. Characterize the "other simplexes" mentioned in (F).
32. Explain how F.29) implies F.28).
6-9. Invariance of Homology Properties
Lemma 11. Let Z\ be a singular &-cycle on ? = \K\9 in the notation
of Art. 6-2(Л). Then Z\ ~ ???\.
°i Proof of Lemma. It is not difficult to show that the complex of Z\
can be defined as the image K\ =f(Fk) of the floor complex of a
prism complex. If/is the chain mapping induced by/, then Z^. = fCk
for some cycle Ck of Fk and lpZk^ =flpCk. Lemma 11 now follows
from Lemma 10, read with h = k.
Theorem 10. Let Z\ be a singular &-cycle on ? = \K\. Then
there exists a simplicial &-cycle Zk homologous to Z*.
Proof. Case 1. The complex K\ of Z\ is star-related to K. Let
K0 = д*(Кк^) be an approximating simplicial complex to K^, in the
notation of Art. 6-l(B). We recall that g* is the simplex mapping
induced by a mapping g which is linear between each closed simplex
yk of K^, in terms of yMinearity, and the corresponding linear simplex
*"A = g(yk\
(A) Let p' be a point on a closed simplex yk of K\, and let q' =
g(p'). Then, by F.24) , p' is on a simplex s of К with sh for a face. The
line-segment p'q' is therefore on s, hence on the polyhedron ? = \K\.
Now let / : ak -> yk be a defining mapping for fk, where ak is a
linear ^-simplex in En с En+1, and let €k+1 = ak X Y, with roof rk
and standard projection ? : тгк+1 -> ak (Art. 6-8(C)). Then the mapping
? == g/^ :rk ^sh (Art. 3-l(i^)) defines a simplex dk = y(rk) which

Art. 6-9] INVARIANCE OF HOMOLOGY PROPERTIES 149
covers sh. The linearity of g implies that ? is a linear mapping of rk
onto sh.
(B) A) If h < k, then дк = гр{гк) is a degenerate (oriented) singular
^-simplex*; B) if In = k, then dk = +sk or —sk.
By / and ?, the floor and roof of ттк+1 are mapped onto yk and its
simplicial approximation g(yk) respectively in such ajWay that if /
maps ? e ak onto p' =f(p) eyfc6ahd if q = (p, l)L(equivalent to
V = <р{я)> Я. e rk) then ? maps qerk onto q' = ip{q) = g(p) e g(y). u
Hence, in the terminology of Art. 6-8@),/and ? map the respective
end points ? = (?, 0) and q = (p, 1) of an element of ттк+1 onto a pair
of points p' and q' = g(p). By (A), the line segment p'q' is on St(sh).
@) Let ? be defined on ттк+1 by the requirements A) ? | ak = f so
that Ф(дк) = fk, B)Ф|г*=у so that $(rk) = g{yk) = sh (though
g(fk) = cp(rk) will sometimes be interpreted as a closed singulars-simplex
dk covering^71), and C) ? maps each element до of tt7c+1 linearly onto the
segment p'q' on St{sh). Then ? can be shown to be continuous. Hence
Ф(ттк+1) can be regarded as a singular prism on St(sh) with linear
elements p'q\ with yk for floor and its approximating simplex sh =
g(yk) for roof (Fig. 6-3).
Let K\ be regarded as the image K\ = f(Fk) of the floor complex
of a prism complex Рк+г as in the proof of Lemma 11. The mapping
? defined in @) can be extended in an obvious way over all the prisms
of Pk+1. We are thus led to a singular prism complex Ф(Рк+1) with
K\ = Ф(Рк) for floor, and with the approximating simplicial complex
Kk = g*(Kl) for roof. Each closed singular prism cell ?(^ + 1) of Ф(Рк^) ^
is a union of line-segments joining its floor ?((?) to its roof (yk).^\o X
Let ? be the chain mapping induced on ?0?}€+1 (Art. 6-8(B)) by ?.
Then, for some cycle Ck on Fk, we have Z\ = Ф(Ск). The roof cycle
Dk corresponding to Gk maps onto a chain Ф(Вк) which, as a
consequence of (B), reduces to a simplicial S-chain Zk ?? ? when
degenerate simplexes are dropped (see (B)). Theorem 10 now follows from
Lemma 10, since the homology Gk ~ Dk = ?°Z)fc (see Art. 6-8(#)) is
carried over by ? into Ф(Ск) = Z\ ~ Zk = Ф(Вк).
Case 2. The complex K\ of Z\ is not star-related to K. Let ? be
so large that ???\ is star-related to К (Exercise 33). By the proof
of Case 1, there exists a simplicial chain Zk ~ 1??^. Theorem 10
follows from Lemma 11 and the transitivity of the homology relation.
(D) Proof of Theorem 1. Since simplicial homologies are a special
case of singular homologies, the singular homology class [Z*] of a
cycle Z\ on ? = \K\ contains the simplicial homology class of a cycle
Zk ~ Z\ (see Theorem 10). By Theorem 9, [Z*] therefore contains
* Compare Exercise 5d above. This is a simple consequence of the definitions and
Art. 4-10G?).

Fig. 6-3. Simplicial approximations to singular complexes, (a) ??+1;
(b) Ф(^-Ы)(Ь = 0, / = 1); (с) Ф(^-Ы)(Ь = 1, / = 1); (d) ?(?*+*).
150

Art. 6-10] INVARIANCE OF HOMOLOGY PROPERTIES 151
exactly one simplicial homology class. Otherwise expressed, each
element of §fc is obtained from an element of §? merely by dropping
from the latter all the non-simplicial ^-cycles belonging to it. The
mapping which maps each singular homology class onto the simplicial
homology class it contains is then clearly an isomorphism. Theorem 1
now follows, by Art. 6-6(^4). By Chapter 5, Theorem 17, cohomology
groups, as well as homology groups, are invariant.
(E) A polyhedron ? is connected if and only if it has no proper
non-vacuous subset which is both open and closed.
To verify (E), let К be a triangulation of П. Then (Art. 5-3,
Theorem 5), the 0th Betti number, ?0, of К equals the number of its
components. The latter are characterized as those connected subsets
of the complex K, each of which is both open and closed (see Art.
4-6(E) and (F)). The present statement полу follows from Art. 4-6,
Lemma 4, and the invariance of ?0.
EXERCISES
33. Show that ? can be chosen as specified in the proof of Theorem 10, Case 2.
34. Show that if Kj is a subcomplex of К of dimension j > 0, then 1РЮ is
star-related to К if and only if" ? > 1.
6-10. Classes of Mappings
Let ? be a topological space, and consider the product space
? ? ? where ? is the unit segment 0 < t < 1. Let ? also be a
topological space, and let
F.31) /:? ? ?->?
be a continuous mapping of ? ? ? into ?.
The points of ? ? ? are representable in the form (p, t), where A)
? = (p, 0) ?? and B) t e T. We will denote with
F.32) ?:? —? (teT)
the mapping of ? into ? denned by
F.33) ft(p)=f(P,t) (???,???).
Then ft is a continuous one-parameter family of continuous
mappings @ < t < 1). It will be called a (homotopic) deformation
of /0 into fv It is helpful intuitively, to regard t as the time and to
think of the image of a point ? e ? as "moving" from f0(p) to fx(p)
through the positions ft(p) as t increases from 0 to 1. Its deformation
path is a continuous curve, defined by the restriction of / to the
segment ? ? ?.

152 INTRODUCTORY TOPOLOGY [Ch. 6
A mapping gx : ? —> ? is homotopic to a mapping g0 : ? —> ? if it
is possible to define a deformation of g0 into gv This relation will be
symbolized by gx ^ g0.
Lemma 12. The relationship ^ is an equivalence relation.
Proof of Lemma. It is reflexive, since ft=f0, 0 <t < 1, is a
deformation of/0 into^ = f0. It is symmetric, since iift is a
deformation of/0 into/l3 then/1_i is a deformation ?? fx into/0. To prove its
transitivity, let/f be a deformation of/0 into/! and gt a deformation
of 9o = ? into 9?·
(Л) Then i^, defined for 0 < t < \ as/2i and for J < t < 1 as <72*-i>
is easily seen to be a deformation of /0 into gx (Exercise 35).
(B) As a consequence of Lemma 12, the continuous mappings
? -> ? fall into homotopy classes, generally called merely classes of
mappings.
(C) If ? = \K\ where К is a linear simplicial polyhedron, then
ft induces a mapping ff : К —> ? defining a family Kf @ < ? < 1)
of singular complexes. This leads to the concept of homotopic
singular complexes and homotopy classes of singular complexes on ?.
Note that ?+1 = ? ? ? is a prism complex, and P7;+1 =/ *$*+*
is a singular prism complex with K$, Kf for floor and roof, where /*
is the simplex mapping induced by /. This leads to the following
result.
Lemma 13. Two singular complexes K% and Kf are in the same
homotopy class if and only if there exists a singular prism complex
?^+1 with K$ and Kf for floor and roof complexes.
(D) The mapping ff induces a chain mapping/^ of simplicial chains
on К into singular chains on ?, defining, for each such chain Gk a
family C\t of singular chains. This leads to the concept of homotopic
singular chains and homotopy classes of singular chains on ?, where
two cycles are homotopic if one can be deformed into the other.
Theorem 11. If two cycles are in the same homotopy class, they
are in the same homology class.
This follows from F.30) applied to singular cycles with ? = 0.
(?) The work of this article is applicable if ? = ?, in which case
continuous mappings of a space into itself are involved. A mapping
/ : ? -> ? is called a deformation of ? if it is in the deformation class
of the identity mapping. As a corollary to Theorem 11 above and
Theorem 17 of Chapter 5, a deformation of ? induces the identity
isomorphism of each homology group, and of each cohomology group,
onto itself.
EXERCISE
35. Prove {A).

7
Manifolds
An га-manifold, defined below, is a generalization of a 2-manifold,
or surface. Manifolds are of frequent occurrence in analysis and in
applications of mathematics. Their topological properties are of
fundamental importance. The m-dimensional pseudomanifolds (Art.
5-6) are a broader generalization of surfaces than desired, for they
include such geometric objects as the 2-sphere with two points
identified.
7-1. Some Homology Properties of Pseudomanifolds
Theorem 1. If Ж is a closed orientable ra-pseudomanifold, then
G.1) S0(Jf) ^ Ьт(М) ъ ?>°(?) ъ Ьт{Щ ъ the free cyclic group,
where ъ denotes the isomorphism relation.
Proof. Part of this result was proved as Theorem 10 of Chapter 5.
The rest will be deduced from the following lemma.
Lemma 1. Let ? be coherently oriented. Then, there exists a linear
graph L whose incidence matrix J0(L) is the transpose of Jm_1(M).
Proof of Lemma. Let s% (k = m — 1, m; i = 1, . . . , ock) be the
A-simplexes of M. The vertices of L will be (p1} . . . , pa ). If sf~l is
incident with (sf, sf) and is oriented like sf, then sj = +PjPk will be a
1-simplex of L and all 1-simplexes of L are thus defined. Then
column i ofJ0(L) has —1 in row j, +1 in row k, and 0's elsewhere;
likewise, row i of Jm_1(M) has —1 in column j, +1 in column k, and
0's elsewhere.
Since a linear graph has no torsion coefficient, it follows from
Lemma 1 that ? has no (m — l)th torsion coefficient. Hence (Art.
5-11, Theorem 17), ?>™{?) ** 93Ш(Ж) = §w(if). The relation $°(Х) -
Ь0(К) is valid for all complexes.
(^4) As a corollary to Theorem 1, the 0th and mth Betti numbers of
an orientable m-pseudomanifold are ?0 = ??? = 1 and there are no
(m — l)th torsion coefficients.
Theorem 2. If Ж is a closed non-orientable m-pseudomanifold,
then its mth Betti number is ??? = 0 and it has an (m — l)th torsion
153

154 INTRODUCTORY TOPOLOGY [Ch. 7
coefficient equal to 2. Its mod 2 homology and cohomology groups
satisfy
G.2) %Q(M) ъ 'Ьт{М) ъ ?>°(?) ъ §)т{М) ъ the group of order 2
and its 0th and first connectivity numbers are ?0 = ?? = 1 (Exercise 2).
EXERCISES
1. Draw a diagram showing (a) a triangulation ? of a tor-us; (b) a
corresponding linear graph L satisfying Lemma 1.
2. Prove Theorem 2, with the aid of Art. 5-6 and the proof of Theorem 1.
7-2. The m-Sphere
Let Em+1 be a euclidean (m + 1 )-space, with a rectangular cartesian
coordinate system (x) = (x1} . . . , xm+1). The unit m-sphere #™ in
?'w+1 is denned by
m+l
G-3) ?0W: ? *?=1.
г = 1
A (topological) m-sphere means a space Sm homeomorphic to S™.
Lemma 2. The boundary Bm = 7fw+1 — ?™+1 of a convex polyhedral
(?? + l)-cell ?™+1 in Em+1 is an m-sphere.
Proof of Lemma. Choose the system (x) so that its origin О is on
7TW+1. Each ray from О meets Bm in a single point ? and #™ in a single
point q. The mapping defined by ? -> g is easily seen to be a
homeomorphism.
Corollary. A triangulation of Bm and a homeomorphism h : Bm ->
8m define a triangulation of an m-sphere Sm.
(A) In particular, the boundary Bm = ^w+1 — sw+1 of an (m + 1)-
simplex sw+1 is an m-sphere, and the set of boundary faces of sw+1 is a
triangulation of Bm.
We will describe a simplicial complex Kn as a cone complex or a
closed star with center p0 if A) p0 is a vertex of Kn and B) each simplex
of Kn is a face of a simplex incident with p0. In other words,* ifw =
&t(Po)> where #?(?0) is the star of p0 on Kn (Art. 6-3) and ??(;?0) is the
complex consisting of all the faces of simplexes of St(p0). The sub-
complex B4-1 = St(p0) — St(p0) will be called the outer boundary of
St(p0) or of Kn (relative to p0). A given complex Kn may be a cone
complex with several possible choices of its center. In particular, the
set of all faces of a simplex is a cone complex with respect to an
arbitrarily selected vertex as center.
* We are here using St(p0) to denote a set of simplexes, rather than their point-set
union.

Art. 7-3] MANIFOLDS 155
Theorem 3. If Kn is a cone complex, then ?>0{??) is free cyclic, and
?>?(??) reduces to the null element (i = 1, 2, . . .).
Proof. This follows from Chapter 6, Theorem 11, and the fact that,
in a polyhedral representation of Kn, an arbitrary cycle on Kn can be
deformed into the center p0 along segments from p0.
Theorem 4. The 0th and mth homology groups of an m-sphere are
free cyclic. Its other homology groups reduce to the null element.
Proof. By Chapter 6, Theorem 1, it is sufficient to give the proof
for the complex Bm consisting of the boundary faces of an (m + 1)-
simplex sm+1 = p0px . . . pm+i- The complex Km+1 consisting of all the
faces of sm+1 is a cone complex, with Bm for its m-skeleton. Therefore
Bm has the same kth homology groups (k = 0, . . . , m — 1) as Km+1.
Since the boundary faces of sm+1 are an orientable m-pseudomanifold,
Theorem 4 now follows from Theorems 1 and 3. As a corollary,
G.4) fi0(Sm) = fin(Sm) = 1 &0П = 0 (гф 0, т).
Theorem 5. No orientation-reversing self-homeomorphism of an
m-sphere is a deformation (Art. 6-lO(E)).
Proof. We continue to represent Sm by Bm. Consider first the
linear self-homeomorphism ? of \Bm\ which interchanges ?0 andj^ and
leaves Pi fixed (i > 1). Lfet s™ = {-1)<р0 . . . Vi-iVi+i · · -Vm+? {i =
0, 1, . . . , m + 1) and let Zm = ?™^1 sf. The homology class of Zm
generates $)m(Sm), and §>m(Sm) is the free cyclic group. On the other
hand, ? induces a mapping which carries Zm into — Zm. However,
Zm ~ — Zw => 2Zm ~ 0, which is impossible, since no cycle kZm
(к ф 0) can bound on Sm (Exercise 5).
EXERCISES
3. Show that if Km in a linear simplicial complex in En с En+1 and p0 ?
En+1 — En, then р0 and Km determine a unique cone complex with pQ as center
and Km as outer boundary.
4. Show that (a) the only simplicial ?-cycle on a cone ?-complex Kn is the
null ?-cycle and (b) at least one simplex of the outer boundary of Kn appears on
each simplicial &-cycle on Kn, if к < п.
5. Justify the last sentence in the proof of Theorem 5.
7-3. Projective m-Space
Projective m-space Pm is the space of proportionality sets of m + 1
real numbers. That is, an ordered set (x) = (xv . . . , xm+1) of real
numbers, not all zero, represents a point, and the same point is
represented by (yv . . . , ym+i) = (txv tx2, . . . , txmV1) for each t ? 0. Given
(x), (y) can be any point, except the origin O, on the line determined
by О and (x) in Em+K

156 INTRODUCTORY TOPOLOGY [Ch. 7
(A) Hence Pm can be interpreted as a space whose elements are the lines
{?} through 0 in Em+1. A neighborhood of Я = Op is the set of all lines
Oq with q in a neighborhood of ? in Em+1.
(B) Each ? e {A} intersects the unit m-sphere S™ <= Em+1 in a pair
(p, p') of antipodal points; that is, points symmetric to each other in
O. Hence Pm can be interpreted as S™ with ? and p' identified for each
antipodal pair (p, p').
In a euclidean space Ej with coordinate system (xv . . . , Xj), the
unit j-cell will mean the spherical region
G.5) cri: 2>?<1.
г = 1
Its boundary is the unit (j — l)-sphere
G.6) -SS··! *?=1, -Si = ?{, - ??·
г = 1
Theorem 6. Projective m-space Pm can be represented topologically
as Oq* with antipodal points of S™~x identified in pairs.
Proof. Consider the representation in (B) above. Let S™ be
analyzed into A) its lower hemisphere ?™, on which xm+1 < 0, B) its
upper hemisphere ?™, on which
^m+i -^«0, and C) its equatorial
(m — l)-sphere #™_1, on which xm+1 = 0. Of a pair of identified
antipodal points (p,pf) on S™ either both are on #™_1 or one is on ?™ and
one on ?™. Hence, Pm can be represented as ?™ with antipodal points
on S™'1 = ?2 - ?^ identified. But ?™ can be defined by
G.7) ?2™ : хт4_г = Vl - {x\ + · · · + x2J, (xv . . . , xj e ?™.
Hence, ?™ is homeomorphically mapped onto off by the projection
(xv . . . , xm, xm+1) -> (xl5 . . . , #w, 0), which is the identity on S™-1.
The theorem now follows easily. (See Exercise 7.)
Theorem 7. Projective m-space Pm is orientable if m is odd, non-
orientable if m is even.
Proof. Let 77™-1 and Tw be defined, in Em, as follows:
m
2""-1 : ?>;| = 1,
G.8) where \x\ means absolute value of x,
m
i = l
(G) Then tw is a convex polyhedral m-cell inscribed in S™, with
the unit points pt (i = 1, . . . , m) on the #raxes and the antipodal
points p- as vertices (Exercise 9).
(D) The (m — l)-dimensional faces of tw consist of the (m — 1)-
simplex в = pL . . . pm> the antipodal simplex s' = p[ . . . jo^, and all

Art. 7-3] MANIFOLDS 157
simplexes of the form rx . . . rw? where rt = p{ or p\ (i = 1, . . . , m)
(Exercise 10).
(E) Let two points on Tm~x be called antipodal if they are on a line
through the origin. Then fw is converted into a projective m-space Pm
by the identification of antipodal pairs of points on Tw_1 (Exercise 11).
The identifications just described identify s = +px . . . pm with
s' = +p[ . . . p'm. Now s has the orientation induced by t =
Opi · · · Pm and s' nas the orientation induced by t/ = Op[ . . . p'nv
where О is the origin. The linear mapping of Em onto itself which
interchangespi and p\ has jacobian (— l)w. Hence t and V are oriented
alike or oppositely in Em according as m is even or odd.
(F) In order that Pm be orientable, it is necessary that when t and
/' are oriented alike, they induce opposite orientations on s and s'
respectively (Exercise 12), in other words that m be odd.
(G) The condition given in (F) is also sufficient (Exercise 13), which
completes the proof.
Theorem 8. The homology groups of Pm are
(a) §0(PW) = the free cyclic group,
(\\\ & pm\ _ /tne grouP °f order 2 (k odd and <m)
G.9) ( > ®k( } ~ (the null group (k even and >0),
(С) Ьт(Рт) =
the free cyclic group (m odd)
the null element alone (m even).
Proof. Part (a) of G.9) is valid since Pm is connected. Whether Pm
is represented as in (B) or as in Theorem 6, the identification of points
converts S™-1 into a projective (m — l)-space P^'1 с ?™. But aS^
can, like S™, be analyzed into its lower hemisphere ?™-1, where
xm < 0, its upper hemisphere ?™-1, where xm > 0, and its equatorial
(m — 2)-sphere S™~2, where xm = 0; and the identifications of
antipodal points convert S%~2 into a Pn0l~2 <=. P^'1 <=. Pm.
Continuing thus, we are led to
?? : \xi+2 = · · · = xm+i = 0, 2 »f = 1, xj+1 < 0 j,
G.10) ?{ : (xj+2 = - - - = xm+1 = 0, Д>? = 1, *,+1 > ?) ,
$0 : \X3+1 = Xj+2 = * ' * = Xm+i = 0, 2,Жг = 1 I
for each j = 1, . . . , m. In the case j = 1, #$ is the pair of points
?[ : ?? = —1 and ?% : хг = +1 on the a^-axis, separating the unit
circle of the (xl9 #2)~plane into the arcs ?\ and ??.

158 INTRODUCTORY TOPOLOGY [Ch. 7
The identification of antipodal points converts each of the spheres
Sq (k = 1, . . . , m — 1) into a projective &-space Pj, and
G.11) pi cz p2 cz · · · cz P^-1 cz Pm.
Lemma 3. If Z\ is a singular &-cycle on Pw and if к < m, then
Z\ ~ Yk^ for some singular &-cycle Y1^ on P[f_1.
Proof of Lemma. Let Pw be represented as fw with antipodal points
of Tm~x identified, converting the latter into P™-1 (see the proof of
Theorem 7). Let Pw be triangulated (compare Exercise 12 for
example) into simplexes {t} which are linear in the representation fw.
Let Zq be a simplicial approximating cycle on {t} to Z^ and let p0 be a
point on an m-simplex of {?}. Since Z\ is a chain on the ^-skeleton of
{?}, its complex K(Zq) does not contain jp0. Each point g on |7il(Zq)|
but not on Tm~x is therefore on a unique ray jp0g which, since fw is
convex, intersects P™-1 in a unique point p, represented by a point on
jTm-i jf ^ js on jim-i^ there are two such rays, one to each of the two
antipodal positions of q, and the two corresponding points ? are
identified with each other, and with q. The set of all segments qp can
be used in defining a deformation of Z^ into Y* on P™~\ and the
lemma follows from Theorem 11 of Chapter 6.
Corollary. The cycle Z\ is homologous to some cycle X\ on P\.
This corollary results from successive applications of Lemma 3,
with ?I and P™-1 replacing Z\ and Pw, and so on.
Let ?1? be a triangulation of ??, and let X\ be a simplicial
approximating cycle on ?* to X1^. If & is even and positive, then, by (D) and
Theorem 2, Xq must be the null cycle, since that is the only &-cycle
on a non-orientable &-pseudomanifold.
If к is odd, Xq is homologous to an integral multiple of the chain VkQ
of ?^ taken with either possible coherent orientation, and Vq is a
non-bounding cycle whose double bounds a chain on Pq+1- Theorem 8
now follows readily.
(H) As a corollary to Theorem 8, the connectivity numbers of Pw
are ?0 = ??? = 1 and ?? = 0 (i = 1, . . . , m — 1) (Exercise 14).
EXERCISES
6. Draw diagrams to illustrate, in the cases m = 2 and m = 3, the proofs of
Theorems 7 and 8.
7. (a) Show that Theorem 6 leads to the earlier definition of P2 as the sphere
with crosscap. (b) Show that P1 is a circle.
8. Show that ?™ = ?™ U ?™_1 is converted into an m-sphere when pairs of
points on S™'1 are identified if and only if they have the same (xv . . . , Хщ^)-
coordinates. Their xw-coordinates will then be xm = ± Vl — (x\ + · · · + ^-i)·
Suggestions: (a) Consider first the cases m = 1, 2. (b) Let ?™ be the
part of ?0 where xm < 0 and ?% the part where xm > 0. Consider the mapping

rt. 7-4] MANIFOLDS
: (xv . . . , xm) -* (xv . . . , хт_19 ут, ym+1), defined as follows,
'1 -H + ··· +*4-?):
[Ут = ~*xm ~~ *
I (хл. . . . , xm) ? ?™,
\Ут+1 = ~ Vk2 - Ут
\Ут = *xm ~ tc
159
where к =
[Ут+l = ^k2 ~ У\
\XV · ¦ · » Xm) ^ ?2
9. Prove (C).
10. Prove (?>).
11. Prove (#), with the aid of Theorem 6.
12. Prove (F). Note that the faces of the 2W simplexes Orx . . . rm (notation as
in (D)) are a triangulation of rw into an m-pseudomanifold with boundary
Tw_1. After the identifications in (E), the second barycentric subdivision affords
a triangulation of Pm into an m-pseudomanifold, without boundary.
13. Prove (G). Suggestion: Consider the effect of orienting positively, relative
to Em, the 7tt-simplexes of the suggested triangulation of Pm.
14. Deduce (H) from Theorem 8.
7-4. Local Homology Groups
Given a topological polyhedron ? and a point ? e ?, let a
neighborhood of ^9 be triangulated so that ? is a vertex, and let #?B9) be the star
of p. The outer boundary В of the closed star St(p) (see Art. 7-2 for
definitions) is a surrounding complex of 2?, and its kth homology group
Ьк(В) will be called the kth local homology group of ? at p, denoted by
G.12) Ьк(П,р) = Ьк(В).
Theorem 9. The kth local homology group §fc(Il, p) is a topological
invariant.
Proof. (A) This theorem means that $>к(И, р) ъ §?(?, g) if ? and
? are homeomorphic with ? and q corresponding. This is equivalent
to the statement that §>к(В) ъ §>к(В*) if В and Б* are two surrounding
complexes of ? (Exercise 17).
Let St(p) and St*(p) be closed stars with В and Б* for respective
outer boundaries. Linearity on the simplexes of St(p) and St*{p) (see
Art. 6-1 (C)) will be called #?-linearity and #?*-linearity, respectively.
Then \St(p)\, for example, is the union of the ^-segments pq0 (q0 e B).
By the y-contraction yr of \St(p)\ in the ratio r, we mean the
deformation under which each point q on a segment pq0 moves toward
q along that segment to the point qr such that pqr\pq = r @ <
r < 1) in terms of ^-linearity. If St(p) were in a euclidean space En,
with origin ? and coordinates (xv . . . , xn), yr would move qt : (txv
. . . , txn) from a position q : (xv . . . , xn) to qr(rxv . . . , rxn) as t

160 INTRODUCTORY TOPOLOGY [Ch. 7
decreases from 1 to r. The contraction yr of \St(p)\ deforms St(p) into a
star yr{St(p)) a y-contraction of St(p).
Similar considerations^ terms of St* -linearity lead to the concepts
of the ? * -contraction of \8t*(p)\ in the ratio r and of the ? *-contraction
Yr(St*(p)).
Given an arbitrary neighborhood N of ? on ?, there exists an r so
small that N contains the ^-contraction of \St(p)\ and the y*-
contraction of \St*(p)\. Hence it is possible to define y-contractions
Sti(p)_of St(p) (i = 0, 1) and. a y*-contraction St*(p) of St*(p) so that
W\St0(p)\ cz \8t*(p)\A2)\Sti(p)\ c l«o(p)l, and CI^A1I cz \St*(p)\
(see Fig. 7-1, where ^-linearity is shown as straight).
Then Б* cz B01 = \St0(p) - Щр)\, where B01 shows in Fig. 7-1
as the closed polygonal band between B0 and Bv
(B) Let ?^(pq0) be the #/-segment from ? to a point q0 e B0. It
intersects Вг in a single point, which will be called the ^-projection
7r1(g) of each point q on X{pq^). Thus the $?-projection ттх is defined
as a continuous mapping {\St0(p)\ — p) ^ Bv Similarly, by using
^^-projections, the St*-projection ?* is defined as a mapping
(\St*(p)\ -p)^B*.
Let у be the mapping В* -> 2?x defined by ip(q) = 7r1(g), q e Bf,
and let ? be the mapping 2?? —> 2?f defined by 9?(g') = ?*(qf), q' e Bv
Lemma 4. The mapping gx = ?? : В* -> В* is a deformation of B*
(Art. 6-10).
Proof of Lemma. Let ?7 = [0, 1], the unit segment on a i-axis.
First note that the mapping ? : В* —> Вг is a deformation of 2?f on
\St${p)\- Indeed, if /((Б*) =/(Б*, t) where / : (Б* x_T) - |M0Cp)|
maps_each segment q ? ?, q e Bf, linearly in terms of ^-linearity on
the ^-segment from q to q' = y>(q), then f1\B*=xp and f0\B* =
(the identity). The deformation path of a point q e B* shows as a
line-segment X(q) in Fig. 7-1. Now letj/ : Ef X T) -> 2?f be defined
by </(#, 0 = ?*?(<1, 0> ?* being the #?*-projection into B*. Then
g0(B*) = g{B*, 0) is the identity and f/B?f, 1) = дг(В*) = ??, so that
gt(B*) = g(B*, ?), 0 < ? < 1, is a deformation of the identity into gv
establishing the lemma.
Hence (Art. 6-10B?)), gx induces the identity isomorphism
$k(B*) -+ $k(B*). But the isomorphism g* induced by gx is the
product g* = ?*?* of the homomorphism ?* : $k(B*) -> bk(^i)
induced by ? and the homomorphism ?* : $k(B*) —> bk(Bi) induced
by ?.
(C) Since g* is the identity isomorphism, ?* and ?* are both one-to-
one. Hence ?* and 99* are isomorphisms since a one-to-one
homomorphism is an isomorphism (Exercise 18).
Theorem 9 now follows, because A) &(!!, p) = %>k(B) is isomorphic

Art. 7-4]
MANIFOLDS
161
Fig. 7-1. Two nested pairs of star neighborhoods of p on ?.
to ??^??), since a y-contraction of Б is a homeomorphism and B)
^(?, ?) = %k(B%) is for similar reasons isomorphic to §)k(Bf).
EXERCISES
15. Draw a 2-dimensional St(p) whose outer boundary is a closed polygon with
? interior to it. Show y±(St(p)) and yi(St(p)) on your diagram.
16. Let ?2 be the surface of a double cone, vertex at p. What are the local
homology groups at pt
17. Prove the italicized part of (A), with the aid of a homeomorphism
18. Establish (С).

162
INTRODUCTORY TOPOLOGY
[Ch. 7
7-5. Topological Manifolds and Homology Manifolds
A (closed) topological m-manifold ?™ is a connected compact
topological space on which each point has a neighborhood homeo-
morphic to an m-simplex. For m < 4, it is known that ?™ is a
topological polyhedron; in other words, ?™ can be triangulated.
For the general га-manifold, the triangulabihty question is one of
the major unsolved problems of topology.
A (closed) homology* m-manifold Ш is a connected compact
topological m-dimensional polyhedron (Art. 4-6(C?)) on which the
local homology groups $к{УЛ, p) at each point are isomorphic to the
respective homology groups §fc(#m_1) o'f an (m — l)-sphere (Theorem
4). This is clearly a topologically invariant definition.
(^4) Every triangulable topological га-manifold is a homology
m-manifold (Exercise 21).
If m > 3, there exist homology га-manifolds which are not
topological manifolds; thus homology manifolds are more general than
triangulable topological manifolds.
Theorem 10. Each triangulation of a homology m-manifold Ш
is a closed m-pseudomanifold.
Proof. It is to be shown that if Ж is a triangulation of Ш and
sm ? G jf then sm_1 is incident with exactly two m-simplexes of M.
We start with a lemma.
Lemma 5. If ? is the number of ra-simplexes of ? incident with
sm~19 and ? is the barycenter of sw_1, then §W_1(SR, p) is the free
abelian group with ? — 1 generators.
Proof of Lemma. Let ? be regarded as linear simplicial, which,
by Chapter 4, Theorem 2, involves no loss of generality in proving
Lemma 5. Let sw_1 = p0 . . . pm_x and let sf = g^w_1 (i = 1, . . . , v)
be the m-simplexes of M, possibly reoriented, incident with sm~1.
The reader should follow this proof, drawing diagrams for m = 2.
Let sf -2 = (-l)^o ...&... Vm-\ (j = 0, . . . , m - 1) where -
denotes omission of p,. Then ds™'1 = ?]1^1 sf~2. Let sf^1 = q^f'2
(i = 1, . . . , v; j = 0, . . . , m - 2).
(B) The faces of the simplexes s|·-1 constitute a surrounding
complex B™'1 of ? on Ш (Exercise 22).
Let Bf -1 = Xf=-02 q^f'2 (i = 1, . . . , v). Then the boundary of
st - sf is (Exercise 23)
G.13) да™ - ds™ = Bf'1 - В™'1 (A = 1, . . . , v; i = 1, . . . Л, · · · , v).
Hence Bf'1 - B^'1 is an (m - l)-cycle on В™'1.
* The term combinatorial manifold is thus used in much of the literature. In recent
research, however, a combinatorial manifold has been denned with the stronger condition
that the outer boundary of each star be combinatorially equivalent (Art. 4-12(Cr)) to
the boundary complex of a simplex.

Art. 7-5] MANIFOLDS 163
(C) The complex K{Bf~x) of B™~x consists of the boundary complex
of sf minus the face sm~1r and K(Bf~1) is similarly the boundary
complex of s™ with sm~1 deleted. Hence the complex of Bf ~1 — B™~x
is a closed orientable (m — l)-pseudomanifold (Exercise 24).
If Zm~x is a cycle on Bm~x and sf^1 appears in Zm~x with coefficient
??, then each of the (m — l)-simplexes s^f1 (k = 0, . . . , m — 1)
appears in Zw_1 with coefficient at. Otherwise, some pair of adjacent
(m — l)-simplexes on K{Bf~1) would appear in Zm~x with unequal
coefficients and their common (m — 2)-face would appear in dZm~1,
contradicting its definition as a cycle. Hence each cycle on B171'1 is of
the form ?™'1 = ???=1 a.Bf'1.
(D) Furthermore, ???=1 a{ = 0, since bZm~x = -(???=1 at) ds™'1
(Exercise 25).
(E) It follows readily that Bf'1 - B™'1 (h = 2, . . . , v) is a
maximal set of linearly independent (m — l)-cycles on Вт~г (Exercise
26).
Lemma 5 now follows with the aid of Corollary 1 to Theorem 2
of Chapter 5 (Exercise 27).
By definition of homology m-manifold, §w_i(SR, p) is the free
cyclic group, which is the free abelian group with one generator.
Therefore ? = 2, and Lemma 5 implies Theorem 10.
Theorem 11. If a topological polyhedron ? can be triangulated
into an m-pseudomanifold M, then every triangulation of ? is an
m-pseudomanifold.
Proof. Suppose ? has a triangulation in which some (m — l)-sim-
plex s™-1 is incident with ? ? 2 m-simplexes. Then §W_1(II, p) = 0
if ? = 1, and it has ? — 1 > 1 generators if ? > 2. But, from the
triangulation ? of П, §Ш_1(П, р) is free cyclic, which yields a
contradiction.
Corollary. If ? can be triangulated into an m-pseudomanifold
with boundary B, then every triangulation of ? is an
m-pseudomanifold with a boundary which is a triangulation of \B\ (Exercise 28).
(F) Theorem 11 justifies us in referring to ? as a (topological)
m-pseudomanifold and, in the case of the corollary, as a (topological)
m-pseudomanifold with boundary В == |5|.
Lemma 6. If ? e Ш and Ш is a topological m-pseudomanifold,
then each surrounding complex of ? is an orientable (m — l)-pseudo-
manifold.
Proof of Lemma. In the first place, ? is on a face of an m-simplex
in each triangulation of a neighborhood of ? on Ш. Otherwise, the
local (m — l)th homology group would be null. From the fact that
each (m — l)-simplex sw_1 incident with ? belongs to exactly two
m-simplexes s™ and s% of St(p) it follows that the face sm~2 of s™-1

164 INTRODUCTORY TOPOLOGY [Ch. 7
opposite ? belongs to just two (m — l)-simplexes of the outer
boundary B171'1 ofSt(p), namely the faces s™_1 and s™_1 ?? s™ and s™ opposite
p. Hence Bm~x is an (m — l)-pseudomanifold. Its orientability
follows from the fact that §w_i(i?w_1) is a free cyclic group.
EXERCISES
19. Let M2 be the union of two euclidean 2-spheres which intersect in a
circle S1. Find the structure of ^(M2, p) (a) for ^G^1 and (b) for ? ? ?2 - S1.
20. Let ?3 be the union of two euclidean 3-spheres which intersect in a
^-sphere^2. Find the structure of ?2(M3, p) (a) for ? ? ?2and (b) for ? ? ?3 - S2.
21. Prove {A).
22. Prove (B). Suggestion: First show that the faces of the simplexes ps™~*
are a subdivision of the closed star of «w_1, which consists of all faces of all
ra-simplexes of ? incident with «w_1.
23. Establish Eq. G.13).
24. Give details of the argument outlined in (C).
25. Verify (?>).
26. Prove (E).
21. Show, in connection with the proof of Lemma 5, that each pair of
ra-simplexes of ? are first and last members of a sequence of m-simplexes,
consecutive pairs of which are adjacent. Suggestion: Given sm, let K(sm) be the
subcomplex of ? consisting of all faces of all m-simplexes of M, each of which can
be joined to sm by a sequence as described. Assume ? — K(sm) non-vacuous
and deduce a contradiction. Note that \M\ is connected, which has certain
implications for \K\ and \M — K(sm)\. The reasoning in the proof of Lemma 6 is
relevant.
28. Prove the corollary to Theorem 11.
7-6. Cell Complexes
During the rest of this chapter, we will work with homology
manifolds, partly because there is no known set of properties
depending only on the incidence numbers which enables us to determine
whether a given m-complex (m > 3) is a triangulation of a topological
m-manifold. Another reason is that there exist important results
depending only on the defining properties of homology manifolds
and hence applicable to all triangulable topological manifolds, by
Art. 1-5{A).
(A) Let Bm be a finite orientable m-pseudomanifold (Art. 5-6(E)).
Then 23w = \Bm\ is a homology* m-sphere if it shares with an m-sphere
Sm the properties
%JBm) ъ ?>0(Вт) is free cyclic,
GЛ4) Ьк(Вт) = ® (кф0,т).
* Called a combinatorial m-sphere in much of the literature, but the latter term has
lately been used for a complex combinatorially equivalent (Art. 4-12(Cr)) to the boundary
complex of an (m + 1)-simplex.

Art. 7-6] MANIFOLDS 165
In case m < 3, 33w is a topological га-sphere, but examples exist
showing that this is not necessarily so if m > 3. A homology 0-sphere
is a pair of points.
Let Bm, with properties G.14), be used as outer boundary of a
closed star Stm+\ let am+1 = \Stm+1\ and let am+1 = dm+1 - 93w. Then
am+1 is a homology (m + l)-cell, m e @, 1,2,.. .). A homology
0-cell is a point. If 33W is a topological sphere, then am+1 is a
topological га-simplex (Art. 4-6), but not otherwise.
Lemma 7. If ak is a homology ?-cell, then <;* is an orientable
A-pseudomanifold (Art. 7-5(.F)) with boundary ©fcl = 5* — ak.
Proof of Lemma. This follows easily with the aid of a triangulation
of ak into a closed star with outer boundary Б*-1, where Б*-1 is a
(A — l)-manifold and, since 9)k_1(Bk~1) is free cyclic, is orientable.
Compare the proof of Lemma 6.
E) As a consequence of Lemma 7, the homology ?-cell ak can
be oriented, by assigning an arbitrary orientation to its boundary
33?_1 and stipulating that a ?-simplex of a triangulation of dk shall be
positive or negative according as it is oriented like or unlike 23?_1.
We associate +ak with the positive orientation and — ak with the
negative. If +ак~г с: ($>к~1 is oriented like +ak, it will be said to
have the orientation induced by ak, otherwise the opposite orientation.
m
(C) A set {a} = \J {ak} of disjoint homology cells will be called
k = 0
a cell m-complex Гт if A) the set {?™} of m-cells* is non-vacuous,
B) each boundary dk — ak of a ?-cell ak e {a} is a union of elements
of {a}, called boundary faces of ak, and C) each intersection ?{ ? of
of the closure of two elements of {?} is the closure of an element of
{?}. An oriented cell complex is such a complex Гт with an arbitrary
positive orientation assigned to each of its cells.
Let Ym be a finite oriented cell m-complex, and let {ak} = (?^, . . . ,
a^k) be its positive ?-cells. The incidence number ?% is 0 if of is not
a face of of+1 and is otherwise +1 or —1 according as it is a face with
the orientation induced by of+1 or with the opposite orientation.
(D) Cell chains ?^? a,of and all the associated concepts of Chapter
5 will be regarded as defined for ?™. The boundary and coboundary
operators д and ? are determined for Tm by
*k
(a) 3?*+1 = ?>* ?*,
G.15) i=1
(b) ?<? = ? ?№+1,
3 = 1
respectively. As in Chapter 5, we are led to the concepts of cycle
* We generally write cell for homology cell during the rest of this chapter.

166 INTRODUCTORY TOPOLOGY [Ch. 7
groups 3/с(Гт), cocycle groups Зл(Гт)> bounding cycle groups %к(Гт),
cobounding cocycle groups <$к(Гт), and finally to homology and
cohomology groups
G.16) ыгт) = 3*(ГШ)ШР») §fc(rw) = 3*(rw)/5*(rw),
of a cell complex.
Lemma 8. A cell m-complex Гт = {a} can be subdivided into a
simplicial m-complex Km = {t} such that each of e {?} is covered by
the star Tk of a vertex gf on Km, where qk e of.
Proof of Lemma. (E) The ^-skeleton of Гт is the cell ^-complex
к
Гк = (J {aj} consisting of the cells of Гт of dimensions <k.
j=o
The proof is inductive, commencing with Г0, for which Л° = Г0
trivially satisfies the lemma. Assume ??_1 subdivided into ??_1 for
some positive h < m, so that the lemma is satisfied with h — 1
replacing m. Then the boundary of each of e Гл is covered by a
subcomplex B^1 of ??_1. It is a simple matter to introduce a vertex
q% on of and to subdivide of into a star T\ so that the hypothesis of
the recurrency is preserved.
Theorem 12. In the notation of Lemma 8,
G.17) ЫП^ЫЛ Ьк(Гт) ъ bk(Rm) (к = 0,1, 2,...).
Proof. Let the &-simplexes of if be oriented like of, and let of
be the chain of Tk. Then
<XJc
G.18) 3??+1=2>?·<7* (see G.15a)).
г = 1
Let 99 be the chain mapping from Гт to ftw determined by ?(<?) =
of, and let {Ck} be the image ^-chains. Then {Ck} contains a &-chain
Gk = ? ЪД of ftw if and only if the terms of Gk can be grouped so as
to express Gk in the form ?^=1 а$.
It follows from G.18) that ? : ? ??\ -> ? а$ (к = 0, 1, 2, . . .)
induces a homomorphism ?* : $к(Гт) -> %k(Km). We proceed with
two lemmas.
Lemma 9. Let Ck = ? Ъ?к be a &-chain on the subcomplex ftfc of
Я™ covering the ^-skeleton Г* of Tm. If ЭС^1 e {С1*-1}, then C* g {Ck}
(k = 1, 2, ... , m).
Proof of Lemma. If if e Тд appears in Ck with coefficient b{, then
all the &-simplexes on T\ appear in Gk with this same coefficient b{.
This follows readily from the facts that A) T\ is, by definition, a
coherently oriented &-pseudomanifold with boundary and B) dCk~x
contains no (k — l)-simplexes incident with the center gf of T\, since
dCk~l g {С*}.

Art. 7-7] MANIFOLDS 167
Lemma 10. Let Ck = ? Ъ$ be a &-chain on Я* (к > k), in the
notation of Lemma 9. If dCfc_1 belongs to {Gfc_1}, then there exists a
?-chain Dk on Я* such that Dk ~ C*.
Proof of Lemma. Suppose a ^-simplex incident with the center q\
of a star Гд appears in Ck, and let G\ be the subchain of Gk consisting
of all the terms involving simplexes incident with q^. Then dC\ is a
(k — l)-cycle on the outer boundary BKh~l of TKh. Since bk-ii^l^) = $>
there is a chain Z)| on Bl~x with boundary ЭСд. But the cycle G\ — D^
bounds, because TKh is a cone complex and therefore, by Theorem 3,
§k(TKh) = 0. Hence Ck ~Ck - (C\ - Dkh), and the latter chain
involves no ^-simplex incident with qKh. By repetitions, we obtain a
&-chain homologous to Gk involving no simplex incident with any q\
(Ji = 0, . . . , am), hence lying on Я*-1.
(E) As a corollary to Lemma 10, obtained by repeated applications,
there exists a &-chain t)k on Rk where A) Ok ~ Dk and B) Dk e {Ck},
as a consequence of Lemma 9.
In the case dOk = 0 of Lemma 10, (E) implies that each kth
homology class of Km contains cycles belonging to the set {Zk}
consisting of the cycles in {Ck}. In other words, each kth homology class
[Zk] of Я™ contains the image under ?* of at least one kth homology
class of Tw. Now if Z\ and Z\ are cycles of the set {Zk} both in [Z]
then, by (E) read with k + 1 in place of k, Z\ — Z\ bounds a chain
DkJrl e {Gk+l}. Hence [Zk] contains the image under ?* of at most one
homology class of ?™. Therefore ?* is a one-to-one homomorphism,
which is an isomorphism, and Theorem 12 is proved for homology
groups. No separate proof is needed for cohomology, because the
cohomology groups can be expressed in terms of the homology groups.
EXERCISES
29. Give a subdivision Rm satisfying Lemma 8, where Tm consists of all the
faces, including the interior, of a cube in euclidean 3-space.
30. Use the subdivision of Exercise 29 to illustrate the proof of Theorem 12.
7-7. Cellular Subdivisions of a Homology Manifold
Let ? = {s} be a coherently oriented triangulation of a compact
orientable homology m-manifold 9И, and let Mx = {t} be the first
barycentric subdivision of M.
(A) Let sm = +pioPi . . -Pi be a typical m-simplex of M, let
0*o> · · · ,3m) be a permutation of (г0, г15 . . . , im), and let pJo...h
denote the barycenter of the face pi . . . pj of sw. Then, by Art.
<7Л9) th...Jm=±PhVhh'"PHh..^m

168 INTRODUCTORY TOPOLOGY [Ch. 7
is a typical m-simplex of Mv In G.19), let the sign be + or —
according as (j0, . . . ,jm) is an even or an odd permutation of (г0, . . . ,
im). Then tf tj is oriented like sm (Exercise 31).
Another formulation for the simplexes {t} is useful. Let {sk} =
(s\, . . . , s% ) be the ^-simplexes of ? (к = О, ... , ж), and let p1· be the
barycenter of sk. A partial ordering of the simplexes {s} was defined
(Art. 4-5) such that s% < sk<^>s^ is a boundary face of sk. Let this
ordering be carried over to the barycenters of the simplexes {s}, so that
G.20) p\ < pfos*; < s*o(sfis a boundary face of sk),
and let
G.21) {?}=(?!,...,?«) (? = ?«*)
be the set of all the points pk, renumbered in the order (p\, . . . , p® ;
p\, . . . ,?^; . . . ; p?9... tPZJ.
Lemma 11. The unoriented ^-simplexes {t} of Mx consist of all
simplexes of the form
G.22) tk = qhQ... qhk such that qK < · · · < qhk.
Proof of Lemma. This follows from (A), since the unoriented faces
of tf mmmj consist of all subsets of the vertices on the right in G.19),
and a typical set of such vertices, in the relative order of their
occurrence, is also a typical subset (qh , . . . , qh ) with qh < · · · < qh .
(B) We remark that each ^-simplex of {t} which is on a ^-simplex of
{s} takes its orientation from the latter. We will presently specify
orientations for the other simplexes of {t}, which, for the present, we
take to be arbitrarily oriented.
(C) We will conventionally write the vertices of a simplex of {t} in
increasing order, so that
G.23) tk = ±qho . . . qhk => qho < - - - < qhk.
We refer to qh as the first vertex of tk and to qh eas its last vertex.
Lemma 12. Let S% be the set of simplexes of {t} with pk for last vertex.
Then sk = |SJ| (k = 0, . . . , m; i = 1, . . . , <xfc) (Exercise 33).
(D) The subsets Sk constitute a partition of the simplexes {t} (Art.
3-2E)). A dual partition, which we next study, is afforded by the
subsets Tf-* (k = 0, . . . , m; i = 1, . . . , ??), where Tf ~k is the set of
simplexes of {t} with p\ for first, instead of last, vertex.
Lemma 13. The set Tf is the star ofp$onM1{i = l,..., a0).
Proof of Lemma. Since its superscript is 0, p*· is the first vertex of
each simplex of Mx with which it is incident. Hence the lemma is true.

Art. 7-7] MANIFOLDS 169
(?) Let ?™ = \T™\. Since Ш is a homology manifold, we see from
Arts. 7-5D), 7-6D), and Theorem 4 that /Jf = ff'1 - ?™
is a homology (ra — l)-sphere, and that ?™ is a homology га-cell. The
set {rm} = (?™, ...,t™) will be the m-cells of a cell m-complex {?}
covering SR, which we proceed to define.
Lemma 14. Each outer boundary ??_1 = if — rf is a homology
(m — l)-manifold.
Proof of Lemma. The case m = 1 being almost trivial (Exercise
34), we assume m > 1. Let ?™-1be the star of a pointy e ?™-1 under
Fig. 7-2. Star neighborhood on /5f *.
a triangulation of a neighborhood of p, with ? as a vertex. It is to be
shown that Ът~2 = ?™-1 — ?7?'1 is a homology (m — 2)-sphere. Let
?™-1 be the cone complex with center p® and outer boundary 33ш-2.
(F) Then Bm~2 = ??1 ? ??1 is a triangulation of 93m~2 (Fig. 7-2).
The union ?™'1 = ?^ U ?^-1 is a homology (w - l)-sphere. For,
Щ~х is the surrounding complex of the midpoint q of p^p in a certain
triangulation ?171 covering a neighborhood of q on the homology m-
manifold Ж, ?™ being defined as the cone complex with center q and
outer boundary B™~x.
Since ?™_1 is a cone complex (j = 0, 1), a i-cycle Zk on Бш~2
bounds a chain С]+1 on Sf_1 (Theorem 3). The difference Zk+X =
Gk+i _ Ck+i is a (i + i)-Cycle on B™'1. Suppose к < m — 2, so that
S/H-iW-1) = 0 and therefore Z^1 bounds a chain C*+2 on Б™. Let

170 INTRODUCTORY TOPOLOGY [Ch. 7
C\+2 be the subchain of Gk+2 consisting of all terms of Ck+2 involving
simplexes incident with p, and let Gk+2 = Ck+2 — G\+2. Then
G.24) dCl+2 = dCk+2 - dCk0+2 = Zk+1 - BCk+2 = Ck+1 - Ck+1 - dCk0+2.
Hence the chain Gk+1 = Gk0+1 + dCk+2 is A) on ?™-1 by definition and
B) ???? because it equals Gk+1 - dCk+2. It is therefore on Bm~2 =
??~? ? ??. Now ЭС/С+! = dCk0+1 + ддСк+2 = dCk+1 = Zk. Hence,
since Zk ~ 0 was an arbitrary &-cycle on Bm~2,
Ьк(Вт~2) = 0 0<k<m-2,
G,2o) bo(Bm~2)is free cyclic·
Now suppose к = m — 2. Since Hm_1(Bm-1) is free cyclic, Zk+1 =
Zm~x is a multiple of the chain Z™'1 of Bm-\ But С™ - C™'1 =
???-г implies that Gf ~x = ???~\ where DJ1'1 is the chain of Zf_1
(j = 0, 1). Hence Zm~2 = dGf'1 = ????'1. Thus each (m - 2)-
cycle Zw-2 on Bm~2 is a multiple of the chain of Bm~2, so that
%m_2(Bm-2) is free cyclic. This, with G.25), implies that Bm~2 is a
homology (m — 2)-sphere, completing the proof of Lemma 14.
Theorem 13. In the notation of (D), let
G.26) rf-fc = \T™~k\ (k = 0, . . . , m; i = 1, . . . , afc).
Then A) rf ~k is a homology (m — A;)-cell, and B) the set {?} of all the
cells rf-* is a subdivision of Ш into a cell complex.
Proof. If ?] < pk, then each simplex of Tf~k is a boundary face of
some simplex of Tf~h.
(G) Hencep\ < p) => rf~k с ff ~h - rf~h. We then say that Tf ~*
is a face of rf ~h.
Let ? be the carrier in Mx of a point ? e 5W. If pf is the first vertex
of i, then ? g т11~к. ?? pk is any other vertex of Mv then ? ? rf~k.
(?) Hence each point of Ш is on exactly one of the elements {?}.
The elements ff~h and fj'~k intersect if and only if there exists a pde
such that ?\ < pde and pk < pd, which is true only if s] and sk are faces
of sd. Let sd be the lowest-dimensional simplex with this property.
(/) Then fd = T™~h ? fjl~k, and rd is the highest-dimensional ? on
f,- ? fy (Exercise 35).
By (E), each ?™ is a homology cell. So is each rf-1, since its outer
boundary 33J1-2 is a homology (m — 2)-sphere on some /3™_1 =
-m _ Tm (Lemma ?4)? gut ?™-1 is a homology (m — l)-manifold and
can play the role of Ш in (E) and Lemma 14, read with m — 1 replacing
m. This means that each ?™~2 is a homology cell.
(J) Continuing thus, inductively, we conclude that each rf ~k is a
homology (m — &)-cell (k = 0, . . . , m\ i = 1, . . . , afc).
Properties (G), (H), (I), and (J) characterize {?} as a subdivision of
Ш into a cell complex.

Art. 7-8]
MANIFOLDS
171
EXERCISES
31. (a) For the special case m = 2 and sm = p0p1p2> sketch the six simplexes
tf, j- of G.19), showing their orientations, (b) Show, for the general case, that
each tV1 . is oriented like sm.
32. Illustrate the numbering of the vertices in G.21) for the special case of
Exercise 31.
33. Prove Lemma 12. Suggestion: Show that each simplex of {t} is a subset of
the carrier sk ? {s} of its last vertex.
34. Prove the case m = 1 of Lemma 14.
35. Establish Statement (/).
7-8. The Poincare Duality Theorem
Consider now the triangulation {s} = ? and the dual cell complex
{?} = ? covering the manifold Ш = \M\ = \T\. We remark that
Mx = {t} is a common subdivision of ? and T, and that ? and T, by
Art. 7-7(D) and Theorem 13, are obtainable by partitioning the
simplexes {t} A) into the stars Sk and letting sk = \Sk\, then B) into the
stars Tf~k and lettingrf~k = \Tf~k\ {k = 0, . . . , m\ г = 1, . . . , ??).
Lemma 15. The point p\ is the intersection
G.27) V\ = ?* ? тГк (к = 0, . . . , m; г = 1, . . . , ?,).
Proof of Lemma. By definition, Sk is the set of simplexes of ?? with
pk for last vertex and Tf ~k is the set of simplexes of ? with p\ for first
vertex. Hence p\ is the only point common to \8\\ and \Tf ~k\ (see
Fig. 7-3).
We next assign special orientations to each cell rf ~k, which is
possible because Tf ~k is an orientable (m — &)-pseudomanifold with
boundary. Consider three simplexes
(a) t%=tVlV\i...VkikeSki (i = ik),
G.28) (b) trk = VPikPt\ · · · PZ e TT~\
(c) r^i-.^el,
where A) tk is an arbitrary ^-simplex of Sk, with ? chosen to make the
orientation like that of Sk, B) t™~k is an arbitrary (m — &)-simplex of
Tf ~k, with orientation yet to be specified, and C) tm is determined by
tkQ and tf~k, and ? is chosen to give tm the positive orientation of Ш.
(A) The value ?, hence the positive orientation of tf~k shall be
determined, for each tf ~k, by the requirement ??? = 1.
Lemma 16. (a) The convention (A) specifies a unique orientation
for each tf ~k on a Tf ~k. (b) These orientations are coherent on each
Tf~k (k = 0, . . . , w; г = 1, . . . , «*).

172 INTRODUCTORY TOPOLOGY [Ch. 7
Proof of Lemma. The vertices in G.28) are the barycenters of
correspondingly indexed simplexes of {s} such that
G-29) <<<<···<<<···< s?m-
All simplexes of S\ and of Tf k are obtained as sk = sk is held fixed
and the other s's range over all possibilities subject to G.29). As the
Fig. 7-3. Dual subdivisions of an M2.
first к of the s's in G.29) are thus varied, the coherently oriented k-
simplexes of Sk on sk are obtained. Each such variation entails
corresponding changes in the vertices of tk and of tm, but not in those of
tf~k. A change which reverses the sign of ? likewise reverses that of ?,
hence leaves ? fixed (Exercise 38). Thus all determinations of ? for
each t™~k are equivalent, proving Lemma 16, Part (a). Now let the
first к + 1 of the s's be held fixed and the last m — к allowed to vary,
subject to G.29). This keeps s% fixed and yields A) the set of coherently

Art. 7-8] MANIFOLDS 173
oriented m-simplexes tm incident with sk and B) the set of all (m — k)-
simplexes t™~k of Tf~k. A variation of the s's which changes the sign
of ? does the same for ?. Two selections of the s's yielding adjacent
simplexes tm likewise yield adjacent simplexes t™ ~k, and Part (b) of the
lemma is a consequence (Exercise 38).
(B) As a corollary to Lemma 16, the convention (A) specifies a
positive orientation for each rf~k, thus orienting the cell complex
? = {?}. The m-cells {тш} are coherently oriented.
Lemma 17. If sk_1 and sk have incidence number ??;_1, then Tf-k+1
and rf ~k have incidence number
G.30) n™~k = (-\)кект\
Proof of Lemma. Consider G.28) for the stars Sk and Tf~k covering
sk and rf ~k, and suppose the notation such that, for the stars covering
s*-1 and rf-*+1,
(a) tl1=^Vl...V1t;\eSf-\
G.31) (b) trk+1 = n'vX\ · · · К е T™~k+\
(с) Г = ??? ...р?п (identical with G.28c)).
Then the boundary chain dtk contains the term ( — lOcfp°o . . . ?^ =
( —l)*??'?o_1. Thus the incidence number of ^ and t^1 is ? =
( — 1)}€??' = ?^·_1 (since t^'1 and tk are oriented like 8к~х and sk).
Similarly, since A) the incidence numbers of Tjl-k+1 and rf ~k are the
same as those of ^~к+1 and t™~k and B) dt™~k+1 = ц<р\ . . . pfn +
• · - = r\r\t™~k + · · · , we have r\™~k = ??'. But the orientation
convention (A) implies ??? = ?'?'?, hence ??' = ??' (since each factor is
+ 1 or —1), and hence G.30) holds.
Theorem 14 (Poincare Duality Theorem). Let Ш be a compact
orientable homology m-manifold. Then
G.32) %к(Щ ъ 9)m~k(W) (k = 0, 1, . . . , m),
that is, the homology and cohomology groups of Ш in complementary
dimensions are isomorphic.
Proof. For the purpose of proving this theorem, we interpret ekfl
as the coefficient of s1-'1 in dsk and ??™-* as the coefficient of Tf~k+1 in
*r?-*.
(C) Lemma 17 is then equivalent to the statement that s* _1 has the
same coefficient in the boundary chain of (— \)ksk as Tf~k+1 has in the
coboundary chain of rf~k.
The simplex chain groups of ? = {s} are the integral modules
dk = [5*5 . . . , skk] (see Art. A-4) and the cell chain groups of ? = {?}

174 INTRODUCTORY TOPOLOGY [Ch. 7
are the integral modules J)w-fc = \r™~k, . . . , r™~k]. The
correspondence s\^Tf~k defines an obvious isomorphism between the
chain groups (?fc and Ът~к of complementary dimensions for each
к = 0, . . . , т.
The homomorphisms д : Сл -> G^ and ? : Ът-к -> D™-**1 lead in
precisely the same way to ЬАЩ апA Sm_fc(?7)· That is, the kernels of
д and ? are the &-cycle group !^k{M) and the (m — &)-cocycle group
3m~k(T), respectively. The image chains in (S^^ and J)w-fc+i
constitute the bounding &-cycle group ^к_г(М) and the cobounding co-
cycle group 3rw|-*+1(!ZT). By definition,
G.33) ым) = 3*W/S*W Sw~*m = З^т/М?7).
Since
G.34) Эв? = 2 fijT1^1 ^?™"* = (-l)fc ? еЙ-1^**1,
the isomorphism defined by s{ —> rf~j induces an isomorphism between
Ьк(М) and $r>m-k{T). This, with Theorem 12, implies Theorem 14.
Corollary 1. The Betti numbers of Ш satisfy the duality relations
?? = ??? = 1» ?*: = ???-k (k = 0, 1, . . . , m), and its kth torsion
coefficients are the same as its (m — к — l)th torsion coefficients.
This is the form in which Poincare duality was generally formulated
before cohomology theory was developed. It follows from Chapter 5,
Theorem 17, Chapter 6, Theorem 1, and Theorem 14 above (Exercise
39).
Corollary 2. If Щ is a compact homology m-manifold, orientable
or not, its connectivity numbers satisfy the duality relations ?0 =
??? = 1,?, = ???-,? = 0,1,...,?>?).
This is the result of mod 2 considerations strictly analogous to those
leading to Theorem 14.
Corollary 3. The Euler-Poincare characteristic JV(9Jl) of an odd-
dimensional compact homology manifold Ш is zero.
This follows from N(M) = ?™=0(-1O^ and Corollary 2, since ??
and /?w_fc appear in the summation with opposite signs.
Theorem 15. If Ш is a compact 3-manifold, then all its homology
groups are determined by ЬЛЩ and the orientability class of Ш.
Proof. Suppose Ш orientable. Then ?0 = ?3 = 1. Hence ?2 = ??
since N(m) = ?0 — ?1 + ? 2 — ? г = °· Since there are no °Ш torsion
coefficients, there are no second torsion coefficients (Theorem 14,
Corollary 1), a fact also known from the orientability. There may be
first torsion coefficients. If Ш is non-orientable, then A) ?0 = 1,
?3 = о, B) there is a second torsion coefficient equal to 2, and C)
??-??+?*~?* = 0. Hence ?2 = ?1 - 1.

Art. 7-9] MANIFOLDS 175
Corollary. If Ш is a non-orientable closed 3-manifold, then
§1(SR) is infinite.
For, ?? = ?2 + 1 > 0.
EXERCISES
36. Work through the proof of Lemma 16 in detail, with illustrative diagrams,
in the case where ? consists of the boundary faces of a 2-simplex.
37. Continue the special case of Exercise 36 through the proof of Lemma 17.
38. Verify the statements in the proof of Lemma 16. The form p. p. . ...
Pj j used in Art. 7-7 may prove useful in this connection.
39. Give the details of the proof of Corollary 1 to Theorem 14.
40. Express the first and second Betti numbers of a compact orientable
homology 4-manifold in terms of its Euler-Poincare characteristic.
41. Using methods like those in the proof, find out whether ?>?(?), $>>2(M),
and the orientability class of 9Л determine all its homology groups, if 9Л is a
compact homology 5-manifold.
42. Express the characteristic N(M) of an even-dimensional compact
homology manifold as simply as possible in terms of its connectivity numbers.
43. Show that a connected finite simplicial 3-complex ? is a triangulated
3-manifold if and only if the surrounding complex of each of its vertices is a
2-sphere.
44. The 3-dimensional torus T3 is obtained from the solid cube \xt\ < 1
(i = 1, 2, 3) by identifying pairs of boundary points symmetric to each other in
a coordinate plane, (a) Show that T3 is a 3-manifold and (b) find its homology
groups.
7-9. Relative Homology
The theory briefly discussed in this section and the next was
developed originally by Solomon Lefschetz. For an account by him of
some of its highlights, see [L2], pp. 195n\
Relative homology theory is obtained from the absolute homology
theory already presented in Chapters 5 through 7, by suppressing the
simplexes of a subcomplex L of a complex K. Absolute homology is
then the case L = 0 of relative homology. We discuss only the case К
finite (o \K\ compact).
Since К and L c= К are complexes, they are closed sets of simplexes
(Art. 4-6(Б)) and К — L is an open set. This implies (Chapter 4,
Lemma 4) that \K\ and \L\ are closed and \K — L\ is open.
If Ck is a i-chain on K, then KGk will denote the chain obtained from
Ck by changing to 0 the coefficient of each simplex of L appearing in
Cfc, a procedure described as suppressing the simplexes of L.
Lemma 18. The operator ? is, for each к e @, 1, . . . , m), a homo-
morphism ? : (?fc -> (?fc (Exercise 47).

176 INTRODUCTORY TOPOLOGY [Ch. 7
Two ^-chains C\ and G\ are equal mod L if XG\ = XG\\ that is, if
they are the same except for their subchains on L. Symbolically
G.35) C\ = C\ mod L о X(Ck - Ck) = 0.
(A) We refer to /K?fc = (?k(K, L) as the group of ^-chains mod L or
the &-chain group mod L. We next define a boundary operator dL
mod L, a &-cycle group mod L, and a bounding &-cycle group mod L,
as follows:
A) dL = Xd so that the mod L boundary of a &-chain Ck is obtained
from dCk by suppressing the simplexes on L.
B) Ck is a &-cycle mod L^>dLCk = 0.
C) Ck ~ 0 mod L^> 3 C*+1 such that dLCk+1 = XCk.
(B) The mapping dL : (?fc(if, i>) -> (^^(if, i>) is a homomorphism
(Exercise 48). Its kernel is the relative &-cycle group ^k(K, L), or the
&-cycle group mod L. The image chains under the mapping dL :
(?л+1 -> (?л constitute the group 5u(^> L) of bounding ^-cycles mod L.
The chains with non-zero multiples which bound mod L are the
boundary divisors mod L, and constitute a group T)k(K, L). These
groups satisfy the inclusions
G.36) %t(K, L) с Ък(К, L) с 3k{K, L) с Ць(К, L).
The relative (or mod L) homology groups, torsion groups, and Betti
groups are, as in the absolute case (Art. 5-2 and Eq. E.62)):
<bk(K, L) = 3k(K, L)l%k(K, L),
G.37) %k(K, L) = X>k(K, L)l%k(K, L),
S8k(K, L) = $k(K, L)j%k(K, L) = 3k(K, ЩЪк(К, L),
and the relative (or mod L) Betti and torsion numbers fik(K, L) and
rk(K, L) (i = 1, . . . , pk) are the Betti and torsion numbers of
ЫК, L).
(C) The following concepts and results directly generalize the
corresponding parts of the absolute homology theory:
A) ?0(?, L) is the number of components of \K\ which do not intersect
\L\. Hence ?0(?, L) = 0 for \K\ connected and L not vacuous
(Exercise 49).
B) If К is an m-pseudomanifold with boundary L, then (Exercise 50)
(a) К orientable => fim(K, L) = 1 and 3 no тт~1(К, L),
(b) К non-orientable => fim(K, L) = 0 and 3 one т"^, Ц equal
to 2.

Art. 7-10]
MANIFOLDS
177
C) The characteristic of К mod L is
m m
G.38) N(K, L) = J (-1)Ч(Я, ?) = 2 (~1)%(K, L),
fc=0 fc=0
where cck(K, L) is the number of k-simplexes of К — L.
Theorem 16. The relative homology group §fc(if, L) is a topological
invariant of \K\ and |i|.
Outline of proof. A subpolyhedron Л of a topological polyhedron
? means a subspace ? <= ? such that a subcomplex i> of some
triangulation К of ? = |if | is a triangulation of ?. The complexes
(K, L) are a covering pair of (?, ?).
(D) The work of Chapter 6 can be adapted (Exercise 53) to the
proof of Theorem 16. In the adaptation, singular simplexes are used
with the restriction that a simplex intersecting ? shall be entirely on
?. This permits a topologically invariant definition of singular relative
homology groups §fc(Il, ?). Simplicial approximations and
deformations can be used as in Chapter 6, with the restriction that under each
deformation no point moves from ? to ? — ?, or from ? — ? to ?,
to establish that each singular homology class of ? mod ? contains
exactly one simplicial homology class of К mod L.
EXERCISES
45. Let K2 be the set of boundary faces of a 3-simplex, and let L1 be the set of
boundary faces of one of the 2-simplexes of K2. Find all the homology groups and
the characteristic of K2 mod L1.
46. Let M2 be a triangulation of a 2-manifold with boundary, and let K1 be
the subcomplex of M2 covering its boundary. Find the homology groups of M2
mod K1. The result will be a generalization of Theorem 7 of Chapter 5.
47. Prove Lemma 18.
48. Show that Bl = Xd is a homomorphism for each chain group (?fc or
f$Ljc(K, L). Verify the other statements in (B).
49.* Prove Part 1 of (C).
50.* Prove Part 2 of (C).
51.* Establish the second equation in G.38), the first being a matter of
definition.
52. Give an appropriate definition of a relative A-pseudomanifold; that is, of
the statement that К is a pseudomanifold mod L.
53.* Develop the details of the outlined proof of Theorem 16.
7-10. The Lefschetz Duality Theorem
We continue with the notation of Art. 7-9. Our next object is to
find a simple expression for the singular homology groups %>к(П — Л).
* This problem calls for the writing of a short paper adapting the relevant parts of
t the absolute theory.

178 INTRODUCTORY TOPOLOGY [Ch. 7
This problem offers some novel features because ? — ? is generally
not compact, hence cannot be triangulated into a finite (closed)
complex.
(A) A covering pair (K, L) of (?, ?) is called normal if each
simplex of К with all its vertices on Л belongs to L (Exercise 54).
Given a normal covering pair (K, L), let St(L) be the union of the
stars on К of the vertices on L.
The subcomplex (Exercise 55) ? = M{K) = К - St(L) is an
approximating complex to ? — ?. Let R = St(L) — L. Then R is
the set of all simplexes of К each with at least one vertex on Л and at
least one on ? — ?.
Since the structure of the homology groups of a space is the same as
for all homeomorphs thereof, we are justified in letting ? = \K\,
where К is a linear simplicial complex in a euclidean space. We do
this, as usual, for the sake of having a metric and linearity.
Each ^-simplex pk e R can be written thus:
G.39) pk = p0 . . . Pj_iqj ...qk = A'/-'·,
where
(a) A' = p0 . . . Vj_x g L,
G.40)
(b) ?k-i = qj...qkeM.
(B) Each point ? on pk is on a unique segment a = ab, where
a g A5'-1, b g ??€~? (Exercise 56). Let (u0> ...,%) be the barycentric
coordinates of ? relative to (p0, . . . , Pj_v qjf . . . , qk). Then ? =
u0 + · · · + u^_x is a parameter, on each a = ab, which increases from
0 at b to 1 at a. We will denote the point on a with parametric value ?
by ?(?, ?).
Lemma 19. Let {a} be the set of all segments ab defined by (B), as pk
ranges over all the simplexes of R. Then each point on R is uniquely
represented as a point ?(?, ?), a e {a} and 0 < ? < 1.
Proof of Lemma. Since (K, L) is a normal covering of (П, Л), а
point ? of ? is on \L\ or on \M\ or on a simplex with some vertices on L
and some on M. In the last case, it is on a unique pk e R and the
lemma follows from (B).
Theorem 17. If ? = ? (?) is an approximating complex to
? — ?, then
G.41) $fc(II - ?) ^ §k(M) and §*(? - ?) ^ ?*(?).
Proof. The mapping ft defined on ? — Л by
iq^q (qe\M\)
G.42) ft :
1 ?(?, ?) -+?(?, A - t)x) @ < t < 1)

Art. 7-10] MANIFOLDS 179
is a deformation carrying ? — ? onto \M\. Theorem 17 now follows
from Art. 6-ЩЕ).
(C) We refer to ? as a (relative) homology m-manifold mod ? if
? — ? is connected and the local homology groups at each point of
? — ? are isomorphic to those of an m-sphere.
Theorem 18 (Lefschetz Duality Theorem). If ? is a relative
orientable homology m-manifold mod A, then
G.43) &(?, ?) ~ §w-fc(IT - ?) (* = 0, 1, . . . , m).
Proof. Note that this generalizes the Poincare duality theorem,
to which it reduces when ? = 0.
Let (K, L) be a normal covering pair of (M, A). Let (Kl9 Lx) be the
first barycentric subdivisions of (K, L), and let
G.44) Мг = Кг- StiLy) ? = К - St(L)
be the approximating complexes to ? — Л defined by Kx and К
respectively.
(D) We use the notation of Arts. 7-7 and 7-8. The dual cells {?}
are formed only when the first vertex (Art. 7-7(D)) is on К — L. The
set T(M) of all such dual cells is a closed cellular complex with the
approximating complex ?? for a subdivision (Exercise 58).
It follows from Theorems 12 and 17 that %m-k(Ti - ?) ^
%>m-k(T(M)). The relative homology groups $fc(n, A) are determined
by the incidence numbers Sjf1, where both ^_1 and sk are on К — L,
and these are related to the incidence numbers of T(M) as in Lemma
17. Hence, the arguments used in proving Theorem 14 also establish
the present theorem.
Corollary. The relative Betti numbers satisfy the relations
G.45) /yn, A) = /?w_fc(II - A) (* = 0, . . . , m)
and the kth torsion coefficients of ? mod A equal the (?? — k — l)th
torsion coefficients of ? — A.
EXERCISES
54. (a) Give an example, for a circle ? on a sphere ?, of a covering pair
(K, L) which is not normal, (b) Show, in general, that the first barycentric
subdivision of an arbitrary covering pair is a normal covering pair.
55. Show that M(K) is a complex (that is, a closed set of simplexes) in (A).
56. Prove (B). Suggestion: Show that a and b are necessarily the points
(vo, · · · , v0_l9 0, . . . , 0) and @, . . ., 0, wj9. . ., wk), if vh = uh/(u0 -\ + щ_г)
{h = 0 j - 1) and Wi = и^щ + · · · + uk) (i = j k).

180 INTRODUCTORY TOPOLOGY [Ch. 7
57. Let T2 be the torus with the polygonal representation ???~1?~1. For the
case ? = ?2, ? = A, find all the homology groups involved in Theorem 18, and
verify G.43).
58. Verify Statement (D).
7-11. The Alexander Duality Theorem and Consequences
Theorem 19 (Alexander Duality Theorem). Let ? be a
topological subpolyhedron of an m-sphere Sm, where ? ? 0 or Sm. If
m = 1, then
G.46) ?0(?) = ?^1 - ?).
If m > 1, then
(a) ?0(?) = 1 + ?^?" - ?),
G.47) (b) fi0(Sm - A) = 1 + ?^?),
(с) /уЛ) = ^h^OS - Л) (* = 1, . . . , т - 2).
Proof. Note first the symmetry of these relations in Л and Sm — Л.
Case 1 (m = 1). In this case, S1 is a circle, Л is a collection of
disjoint closed arcs and isolated points, and G.46) formulates the
obvious fact that the number of components of Л equals the number
of components of S1 — A.
Case 2 {m > 1). Let Ck be a singular &-cycle of Sm mod Л, where
1 < к < m — 1. This implies that B1c~1 = dCk is a singular cycle on
?. If C7c~0mod Л, there exists a (k + l)-chain Ck+1 such that
3Ck+1 =Ck - Dk, where Dk is a chain on ?. But Ck -Pisa cycle,
so dDk = dCk = Bk-\
(A) Hence Ck ~ 0 mod ? implies Bk~l = dCk ~ 0 on ?. It follows
that the boundary operator д induces a homomorphism d* of %k(Sm, A)
into §fc_1(A). For A) д maps ^-cycles mod Л on Sm into absolute
(k - l)-cycles on Л; B) d(G\ + G\) = dC\ + dCk; and C) if C\~ Ck
mod Л, where G\ is a &-cycle mod Л (г = 1, 2), then d{G\ — G\) ~ 0
on ?, as we have just seen.
Again let Ck be a &-cycle mod ?, suppose i?fc_1 = dCk bounds on ?,
and let Dk be a k-chain on ? with dDk = Bk~K Then Ck - Dk is a
?-cycle on Sm. Since к < ni, Ck — Dk bounds some chain Ck+1 on Sm.
Since BlC^1 = X(Ck - Dk) = lCk, it follows that Ck ~ 0 mod Л.
(Я) Letting C7c = Cf - C*, with C\ a &-cycle mod Л (i = 1, 2), we
deduce from the preceding paragraph that dC\ ~ dC\ on Л implies
G\ ~ G\ on Sm mod A. Hence no two different elements of 5)fc($m, Л)
are mapped by d* onto the same element of §?_?(?); that is, 9* is
one-to-one.

Art. 7-11] MANIFOLDS 181
If Bk_1 is a cycle on Л then it bounds on Sm, hence there exists a
chain Ck on Sm with dCk = B*-1.
(C) It follows that d* is onto. But a homomorphism which is one-
to-one and onto is an isomorphism. Hence
G.48) $k(Sm, A) *, $M(A) (k = 2, . . . , m - 1).
But, by Theorem 18 for ? = Sm, this implies
G.49) $M(A) ^ §-*(?- - Л).
Hence, by Chapter 5, Theorem 17, which, as shown by G.41), applies
to ? — Л even though the latter is not a complex, G.47c) follows
(Exercise 59).
Corollary 1. The kth torsion coefficients of Л equal the {m — к — l)th
torsion coefficients of Sm — A (k = 1, 2, . . . , m — 2) (Exercise 60).
It remains to establish G.47a,b). Let Zf_1, . . . , Z™_1 be a set of
homology-independent (m — l)-cycles on Л, where ? = ???_1(?).
Since /???_?(?™) = 0, there exist chains Cf, . . . , C™ on Sm with BCf =
Zf'1 (i = 1, . . . , ?). Since §w(?w) is free cyclic, there is a cycle Zw
on Sm whose integral multiples are representatives of the respective
mth homology classes of Sm.
Lemma 20. The chains {Cm} = (C™, . . . , C%), where
?
G.50) C™= Zm -?Cf,
i = l
are a maximal set of homology-independent га-cycles of Sm mod ?.
Proof of Lemma. The {Cm} are га-cycles mod ?, since their
boundaries are all on ?. If ??=0 afif ~ 0 mod ?, there is on Sm an (m +1)-
chain 0+1 such that ЭО+1 = Dm - If=0 a^f for some Dw on ?.
But, since dCm+1 is an га-cycle, Dm must have the same boundary as
S?_0a<C?; that is, &DW = If=1(a, - a0)Zf-\ Since (Zf"\ . . . , Z^)
is a homology basis on ?, this implies at = a0 (i = 1, . . . , ?), hence
??=0 а/З™ = a0 ??=0 Gf = a0Zm. But this is null-homologous only for
a0 = 0. Hence ??=0 afif ~ 0 mod ? => ?? = 0 (i = 0, . . . ', ?) so that
{Cm} is a set of homology-independent га-cycles mod ?. It remains to
establish that an arbitrary m-cycle Cm mod ? is homology-dependent
on them. Now dCm = Z™ is an (m — l)-cycle on ?. Hence, for
some numbers (a, av . . . , ??), not all zero, aZm~x + ??=1 а^~1 ~ 0
on Л. Hence there is on Л a chain Dm such that dDm = aZm~x +
???=1 ciiZ?-1 = B(aCm + ???=1 afif). It follows that Dm - (aCm +
??=1 ???) is a cycle, hence that Dm - (aCm + ??=1 afif) ~ 6ZW for
some 6. Using G.50), and noting that Dm is on ?, we deduce that
aCm - ЬС^ + ?$=1 (a, - b)Cf ~ 0 mod ?. Since (a, al9 . . . , ??) are
not all zero, at least one of the numbers (a, b, ai — b) (i = 1, . . . , ?)

182 INTRODUCTORY TOPOLOGY [Ch. 7
must differ from zero. Hence (Cm, C™, . . . , C™) are homology-
dependent, and Lemma 20 is proved. Relation G.47b) now follows
from Lemma 20 and G.45). Relation G.47a) is suggested as an
exercise (Exercise 63).
Corollary 2. The Alexander duality theorem holds with A) Sm
replaced by euclidean m-space Em and B) the right sides of G.46) and
G.47a) diminished by 1. Also, Corollary 1 holds with Em in place of Sm
(Exercise 64).
Corollary 3. The Alexander duality theorem holds with Betti
numbers ? replaced by connectivity numbers ?. After this replacement,
Sm can be replaced by Em9 with the corresponding other changes listed
in Corollary 2.
For, the connectivity numbers are merely the mod 2 case of the
Betti numbers in relative as well as absolute homology theory.
The Alexander duality theorem has been one of the most important
bases for the further study of manifolds. We mention one or two of its
consequences.
Theorem 20 (Jordan-Brouwer Theorem). Let #m_1 be a
topological (m — l)-sphere in a euclidean m-space Em (or on an m-sphere
Sm). Then S™-1 separates Em (or Sm) into exactly two regions.
For the case where $m_1 is a topological subpolyhedron of Rm =
Em or Sm; that is, where $m_1 is a subcomplex of some triangulation of
Rm, this follows from G.47b), which becomes
G.51) ?0(?? - S™-1) = 1 + ft^OS"-1) = 2.
We omit the case where $m_1 is not assumed to be a subpolyhedron.
Corollary (Jordan Theorem). A simple closed curve separates
the plane (or the 2-sphere) into exactly two regions.
This is the case m = 2 of Theorem 20.
Theorem 21. No topological subpolyhedron of Rm = Em or Sm of
dimension less than m — 1 can separate Rm.
For, dim ? < m - 1 => ?^?) = 0 => ?0(?? - A) = 1.
Theorem 22. No non-orientable closed (m — l)-manifold ? can be
a topological subcomplex of Rm = Em or Sm.
Proof. For any complex, the 0th Betti number equals the 0th
connectivity number. Hence, by Theorem 19 and its Corollary 2, if ?
is assumed to be a subcomplex of Rm,
G.52) ?0(?" -?) = ?0(?™ -M) = l+ /t-iW = 1 + ?^?).
But ??-^?) = 0 and /?w-iW = l> which contradicts G.52) and
establishes the result.

Art. 7-11] MANIFOLDS 183
Theorem 23. A connected topological 1-complex L on the plane E2
or the sphere S2 separates E2 or S2 into 1 + ??(?) = ? + 1 parts,
where ? is the cyclomatic number.
Further discussion of duality in E2 is suggested in Exercise 65.
Consider the Alexander duality theorem in Es, where A = \K\ and
К is a linear simplicial complex of dimension <3. Let Al9 . . . , A^
be the connected components of A. Suppose it possible* to enclose
each Лг· inside a topological 2-sphere $f which excludes the Af (j ? i).
If Z\ is a singular 2-cycle constituting a base for §2(^f )> then ^ can be
shown that (Z\, . . . , Z* ) is a homology base for Ez — A in dimension
2; that is, each singular 2-cycle ? on Es — A is homologous to a
linear combination of the Z\ (i = 1, . . . , /?0). This is the meaning, for
such a A, of the relation (Theorem 19, Corollaries 2 and 3) ?0(?) =
?,?? - ?).
Continuing intuitively, if A <= Ez is topologically a closed orient-
able 2-manifold, hence equivalent to a sphere with handles, then
?2(?) = 1 and the Alexander duality theorem yields ?0(?3 — A) = 2,
meaning that A separates Es into two regions. If A is, for example,
topologically the union of /?2(A) disjoint closed orientable 2-manifolds,
then A separates Ez into ?0(?3 — ?) = 1 + /?2(?) regions.
We have just commented on the significance of relations G.47ab)
interpreted in Es. Relations G.47c) reduce to
G.53) ?,(?) = ??(?* - ?).
Suppose, for example, that A is a connected linear 1-complex in Es,
and let (s\, . . . , s^ ) be a set of 1-simplexes whose removal will reduce
A to a tree. Imagine a set (Tj, . . . , ??? ) of small circles, where ?]
links s] (i = 1, . . . , /?х), in the sense that the plane disk bounded by
T} intersects s] in a single point and does not intersect A elsewhere.
It can then be shown that cycles (Z{, . . . , ??? ) on (?}, . . . , ??? )
constitute a first homology basis for E* — A, just as a set of cycles
(C\, . . . , CJ ) on A is a first homology basis for A if s] enters into G],
but not into G) for j ? i. The dual first homology bases (C{, . . . , Cj )
and (Z{, . . . , ??? ) give a geometric insight into the relation G.53).
The foregoing intuitive discussion of the geometric significance of
duality in the lower dimensions barely touches upon a well-developed
duality theory, which is beyond the scope of this book.
EXERCISES
59. Give the details of the last line of the proof of G.47c).
60. Prove Corollary 1 to Theorem 19.
* In some cases, this is impossible; for example, if ? consists of two linked circles.

184
INTRODUCTORY TOPOLOGY
[Ch. 7
61. Follow through the proof of the Alexander duality theorem, using
numerical values for the /Ts, in the special case where m = 2 and ? is homeo-
morphic to the edges and vertices of a tetrahedron.
62. Let M2 be a sphere with three handles in E3. Give the Betti numbers of
M2 (Chapter 5, Theorem 7), than obtain the Betti numbers of E3 - M2 by
duality.
63. Establish Relation G.47a).
64. Prove Corollary 2. Suggestion: Use the fact that Em is, topologically, the
punctured m-sphere; that is, Sm — q. Hence Em — ? is Sm — A — q,
topologically, forg?^m -?.
65. Discuss the geometric significance of the Alexander duality theorem,
where ? = \K2\ and K2 is a linear simplicial 2-complex in the plane E2. Illustrate
your discussion with examples in which ?0(?) > 1 and ??(?) > 1 and in which you
indicate homology bases for E2 — A.
66. Let M2 be a Moebius strip (Art. 2-2(D)) with a handle in E3. Find the
Betti and connectivity numbers of M2 (Chapter 5, Theorem 7), then obtain the
Betti and connectivity numbers of E3 — M2 by duality.

8
The Fundamental Group;
Covering Surfaces
In Art. 6-10, homotopy classes of mappings were defined, and
some basic results stated with regard to corresponding homomorph-
isms of homology groups. In this chapter, a special category of
homotopy classes will be developed, leading to the so-called
fundamental group, or Poincare group, of a complex, which sometimes
enables us to distinguish between some topologically different spaces
agreeing in their homology properties.
8-1. Paths and Path Products
In the plane E2, with a coordinate system {x, y), a curve can be
parametrically defined in the form ? =f(s), у = g(s), subject to
suitable restrictions on the functions. In topology, the concept of
curve is generalized as follows.
Let ? be a topological space. A (continuous) curve or path on ?
is defined by a continuous mapping /: [a, b] -> T, where [a, b] is a
segment a < s < b on an s-axis. Although we will refer to the path
? = f([a, b]) or ? : ? = f(s), s e [a, b], a path is not merely an image
f([a, b]) but consists of such an image together with a defining
mapping/. To emphasize this, we sometimes refer to a path (?,/). The
initial and terminal points of ? are f(a) and f(b). We say that ? joins
its initial point to its terminal point, and we refer to ? as a path from
?«) to/F).
(^4) A path (?,/) equals a path G7-, g), where/ has domain [a, b] and
g has domain [c, d], if A) f(a) = g(c), f(b) = g(d), and B) there exists
a homeomorphism h : [a, b] -> [c, d] such that / = gh (see Art.
з-1(Л)·
Condition A) means that equal paths have the same initial and
terminal points. Two paths can coincide as point sets, agree in initial
and terminal points, and still be unequal (Exercise 2).
(B) If two paths are equal, we refer to them as the same path;
185

186
INTRODUCTORY TOPOLOGY
[Ch. 8
otherwise as different paths. A given path ? can be defined by a
mapping f with an arbitrary preassigned range [a, b] (Exercise 5).
Now let G7-?,/?) and. G7-2,/2) be two paths such that the terminal
point of ?? is the initial point of 7r2, and let their defining mappings
fx and/2 have ranges [a, b] and [b, c], whose union is a segment [a, c].
The product ???2 is the path defined by
(Ms) a <s <b,
(8.1) ?,?2 :f(s) =
[f2(s) b <s <c.
Intuitively, ???2 is ?? followed by 7r2.
(C) Similarly, the product ?1?2 . . . ?? is defined (Exercise 6) if
the terminal point of 7гг is the initial point of
??+1 (* — 1, . . . , 72- — 1).
(D) In particular, the segment ?0?? = ?(?0, ??) in En as defined
in relations D.4) is a path, and a broken line pQpx . . . pm is also a path,
namely the product ???2 . . . ттт where 7гг = ??-^?? (г = 1, . . . , m).
Given a path ? с: Т, the selection of a particular defining mapping
/ : [a, b] —> ? constitutes a parametrization of 77, with s as parameter.
The parameter is a coordinate system, in the sense that a unique point
on ? is specified by each number s e [a, b]. A point on ? has as many
parametric values as it has images under/-1 (Art. 3-ЦЯ)). A point
is simple or multiple, according as it has one or more parametric values,
the number of such values being its multiplicity. If each point of ?
is simple, ? is a simple path, or an arc.
The subset of ? covered by a path ? may be a single point, or it
may, for example, be the closure of an open subset of a euclidean
space En (n > 1). It is then a space-filling or Peano curve. One
property of a Peano curve ? in En is that it has an everywhere-dense*
set of points, each of multiplicity at least ? + 1 (Exercise 8). Thus
a Peano curve in E2 has an everywhere-dense set of at least triple
points. See [K2, pp. 101-103] for an example of such a curve covering
the interior and boundary of a triangle.
(E) A curve, or path, ? is closed if its initial and terminal points
coincide. If, except for this coincidence, each point of ? is simple,
then ? is a simple closed curve or path. (See also Exercise 7.)
EXERCISES
1. Show that ? = s, у = s @ < s < 1); ? = 4 + 2s, у = 4 + 2s (-2 <
s < —1.5); and ? = t2, у = t2 @ < t < 1) all define equal curves.
2. Show that ? = cos t, у = sin t @ < t < 2?), and ? = cos 2t, у = sin 2t
@ < t < 2tt) define unequal curves, even though they are equal as point sets and
have the same initial and terminal points.
* A set of points of ? is everywhere dense if each point of ? is a limit point of the set.

Art.8-2] THE FUNDAMENTAL GROUP; COVERING SURFACES 187
3. Define two unequal curves from ? = 0 to ? = 1 on the x-axis, each of
which, as a point set, coincides with the segment [0, 1].
4. Show that the broken lines abcdb and abdcb generally represent unequal
curves, though they coincide as point sets and agree in their initial and terminal
points. Under what conditions are they equal?
5. Prove the italicized part of (B).
6. Write out the definition of тт^пг · · · ?? ?*1 accordance with Statement
(C).
7. Define closed and simple closed curves in terms of mappings of the unit
circle ? = cos 2nt, у = sin 2nt @ < t < 1). Include a definition of equal closed
curves.
8. With the aid of Brouwer dimension theory, show that a Peano curve
covering an ? -dimensional space must possess points of multiplicity ? + 1.
9. Suppose ? = f(t), у = g(t) defines a Peano curve in the (x, y)-p\ane. Show
that ? = f(t), у = g(t), ? = t defines a simple path in (x, y, z) -space.
8-2. The Fundamental Group
A topological space ? is arcwise connected if each two of its points
can be joined by a curve on ? (Exercise 10). Throughout this chapter,
? denotes an arcwise-connected topological space.
A singular 1-simplex (Art. 6-1 with к = 1) differs from a path in
the definitions of equality (compare Art. 6-l(B)) (Exercise 11) and
also in the algebraic processes applied to them. Singular 1-simplexes
were additively combined into 1-chains, including 1-cycles; and
homology classes of 1-cycles were used as elements of the first
homology group §1(?). Paths, on the other hand, are combined as
products; and homotopy classes of closed paths will be used as
elements of the fundamental group, which, unlike §?(?), is generally
not commutative.
Except where otherwise stated, each path ? =f(t) will be defined
by a mapping/ : ? -+?, where X = [0, 1] : 0 < ? < 1.
(A) Let (p, q) be two points, not necessarily distinct, on ?, and let
? = U(p, q) be the set of all paths on ? from ? to q. Two such paths,
7г0 and ?? are homotopic in ?, a relation symbolized by
(8.2) tti^tto in U(p,q),
if there exists a deformation (Art. 6-10) ft (t e ? : 0 <t < 1) such
that/j defines a path ?^?. As the notation implies, /0 and/x define
the given paths. We describe such an ft as a deformation in ?, and
we say that it leaves (p, q) fixed.
Otherwise expressed, Relation (8.2) means that there exists a
continuous mapping / of the unit rectangle
(8.3)
? ? ? :@ <x <l, 0 <t <l)

188
such that
(8.4)
INTRODUCTORY TOPOLOGY
[Ch.8
(a) f@,t)=p&ndf(l,t) = q @ < t < 1),
(b) Ъ=Мх)=/(х,Л (j =0,1).
(В) Using the notation of Fig. 8-1, one can describe/ : ? ? ? -> ?
as a mapping which maps the edge а0аг onto the point ? and Ъф1
t
fa
(a)
ax @,1)
??
a0@,0)
X=X0
(b)
MU)
M1.0)
¦> *
Fig. 8-1. The relationship ?? ^ 7r0.
onto q, so that ? and q are the common initial and terminal points
of all of the paths ?? @ <t < 1), defined by ft : X -> ? where
Lemma 1. The relationship of homotopy in ? is an equivalence
relationship among elements of ? (Exercise 12).
(C) Let p0 be an arbitrary fixed point of ?, and let ?0 = U(p0, p0)
be the set of all paths each having p0 for both initial and terminal

Art.8-2] THE FUNDAMENTAL GROUP; COVERING SURFACES 189
point. We will use the symbol ^ to stand for the relationship of
homotopy in ?0; that is,
(8.5) y0 ~ yx means y0 i^± yx in ?0.
The homotopy class (see Exercise 12) of y0 in ?0 will be denoted by
(8.6) Г0(у0) = {у|у^у0}.
A deformation in ? 0 is illustrated by a modification of Fig. 8-1 in
which ? = q = pQ.
Attention is temporarily confined to the class ?0 and the
corresponding homotopy relation ~ because the path products yxy2 and
y2yx are always defined if yx and y2 are in ?0. Note that a deformation
of y0 = yxy2 in ?0 need not leave fixed the terminal point of yv
which is the initial point of y2, although it must leave fixed the initial
point of yx and the terminal point of y2, both at p0, these being the
end points of y0.
Theorem 1. The homotopy classes {?0} = {Г0(у0) | y0 e ?0} are
the elements of a group ?0 whose product operation is defined by
(8-7) Г0(У1)Г0(у2) = Г0(У1у2).
The unit element of the group is ?0(?) where ? is the singular path
defined by / : X -> p0.
Proof. Note first that ?0(?) consists of those closed curves in ?0
which can be deformed, in ?0, into the point p0.
(D) The product of two homotopy classes, as formulated in (8.7), is
independent of the representatives ?? and y2 used in the formulation.
That is, y' ~ yi (j —- 1, 2) implies у'ху'ъ^ УхУъ (Exercise 13).
(?) For each у е П0, we have ?0(?)?0(?) = ?0(?)?0(?) = ?0(?)
(Exercise 14).
(F) For each у е П0, we have Г0(у-г)Г0(у) = ?0(?) (Exercise 15).
The elements of ?0(?) are said to be null-nomotopic.
(G) If 7j еП0 (j = 1, 2, 3), then Г0(У1)Г0(Г2Гз) = Г0(Г1Г2)Г0(у3).
This follows from the fact that multiplication of paths is associative.
The four group axioms (Art. A-l) are verified by Definition (8.7),
which is justified by (D), together with results (E), (F), and (G).
(H) The group Ф0 will be denoted, more explicitly, by ?(?, ?0)
and will be called the fundamental group of ? with fixed point p0.
Theorem 2. The groups ?, = ?(?, pj) (j = 0, 1) are isomorphic
for each pair of points (p0, px) on ?.
Proof. Let ? be a path from p0 to pv Such paths exist, because
? is arcwise connected. If yx e U1 = U(pv px) (see (^4)), then y0 =
ттухтт-х is obviously an element of П0.

190 INTRODUCTORY TOPOLOGY [Ch. 8
Lemma 2. If ?[ is homotopic to yx in Uv then y'Q = ??[?~? is
homotopic to ?0 = ????' in ?0 (Exercise 16).
It follows that ? induces a mapping ?* : ?? —> ?0.
Lemma 3. The mapping 77* is a homomorphism.
Proof of Lemma. This means that 7г*(Г1(у1) · Тх{у[)) = ??*?1(?1) ·
7?*?1G?), which is equivalent to 77*(Г1G1у{)) = Г^тту^-1) · Г^тгу^тг-1).
But the latter product, by definition of Ф0, is ?0G???7?~?7?7?7?~1) =
^?(?7?7???^ and this equals ^(Г^у^)) by definition of ?*.
Fig. 8-2. Dependence of 77* on 77.
Lemma 4. The mapping 77* has an inverse which is a homomorphism
(Exercise 17).
It follows from Lemmas 3 and 4 that ?* is an isomorphism. This
completes the proof of Theorem 2.
(/) In general, the isomorphism ?* depends on the path ? which
induces it. This can be seen intuitively, in Fig. 8-2, where ???~? is
not homotopic in ?0 to ?^?^1.
(J) Theorem 2 means that the structure of the fundamental group
of ? with fixed point p0 is independent of p0. This justifies the use
of the term fundamental group ?(?) of ?, also called its Poincare
group, with the understanding that ?(?) is determined up to an
isomorphism and is exemplified by ?(?, p0) for any choice of p0 e ?.
Theorem 3. The fundamental group ?(?) is topologically invariant.
Proof. We begin with the following lemma.
Lemma 5. A continuous mapping of ? into a space ?' induces a
homomorphism of ?(?) into ?(?') (Exercise 18).
By Lemma 5, a homeomorphism between ? and ?' induces both
a homomorphism of ?(?) into ?(?') and an inverse homomorphism

Art.8-2] THE FUNDAMENTAL GROUP; COVERING SURFACES 191
of ?(?') into ?(?). But a homomorphism with a homomorphism
for inverse is an isomorphism, and Theorem 3 is proved.
Theorem 4. In general ?(?) is not abelian.
Fig. 8-3. The non-abelian nature of ?.
Proof. This means that, in general, yxy2 and у2У\are n°t nomotopic
in П0. This fact is illustrated by the two diagrams of Fig. 8-3, where
У1У2 ^ У> У2У1 ^ у' but ? and ?' are not homotopic in H(p0, p0).
(K) A space ? is said to be simply connected if ?(?) consists of
the identity element alone. It is said that a closed curve у on ? can
be shrunk to a point or can be spanned (by a 2-cell) on ? if there

192 INTRODUCTORY TOPOLOGY [Ch. 8
exists a continuous mapping / : D —> ?, where D is the region x2 +
y2 < 1 on an (x, ?/)-plane, such that ? is defined by the restriction of/
to the unit circle ? : ?2 + У2 = 1 (see Exercise 7 above).
As a parameter t decreases from 1 to 0, the circle Kt : x2 + y2 = t2
shrinks from к to the origin. The curve yt defined by /1 Kt is
correspondingly deformed from ? = f (к) into the point ?0 =/(?0) =f@, 0),
the image of the origin. During the deformation, yt =f(Kt) sweeps
out the topological disk ? = f(D), which is the closure of the 2-cell
? =/(?).
Theorem 5. The space ? is simply connected if and only if each
closed curve on ? can be shrunk to a point (or can be spanned).
Proof. Suppose a closed curve ? can be shrunk to a point, and let
p0 be a point on ?. Then ? is null-homotopic (see (F)). This can be
seen with the aid of the mapping / of Statement (K), if the circles Kt
are replaced by a family of circles ?[, where ?{ is the circle on D of
radius ? ? [0, 1] tangent to к at a point q such that/(g) = p0 (Exercise
19). This shows that ?(?, pQ) reduces to the identity if each closed
curve through p0 can be spanned. Conversely, suppose that ?(?, ?0)
reduces to the identity, and ? is given on ?. By Theorem 2, we may
assume that ? e ?0, hence can be deformed in ?0 into ? (see Theorem
1). This implies the existence of a continuous mapping g of the
rectangle in Fig. 8-1, where g maps the three edges а0аг, Ь0Ьг, and X0 —
a0b0 onto p0 and where g : Хг —> ? defines ?. The proof is easy to
complete (Exercise 20).
EXERCISES
10. Show that the property of being arcwise connected is a topological
property.
11. Using the above notation, consider a continuous mapping / : X —> ?.
Show that/ and the mapping defined by px =f(x2), ? ? ?, define equal paths
on ? but generally define unequal singular 1-simplexes.
12. Prove Lemma 1. Note that homotopy in ? is a restriction of the more
general homotopy used to define classes of mappings in Art. 6-10, since the latter
has no restriction regarding a particular point p0.
13. Prove (D). Suggestion: Define a deformation in ?0 of ???2 into ?^?^ in
terms of deformations of y; into ?', (j = 1, 2).
14. Prove (E).
15. Prove (F). Suggestion: Show that ??~? can be so deformed into p0 that
each deformation path is on ?.
16. Prove Lemma 2. Suggestion: From a deformation of ?'? into ?? in Иг
deduce a corresponding deformation of y'Q into ?0 which leaves each point of
? and ?-1 fixed.
17. Prove Lemma 4, showing that ?*-1 is induced by ?-1.
18. Prove Lemma 5. Note that a similar result was proved for homology
groups.

Art.8-3] THE FUNDAMENTAL GROUP; COVERING SURFACES 193
19. In the proof of Theorem 5, complete the demonstration that ? is null-
homotopic.
20. Complete the proof of Theorem 5. Suggestion: Make use of the mapping
g0 of the rectangle onto a unit disk defined by x' = t cos 2??, у' = t sin 2??
@ < ? < 1, 0 < t < 1).
21. Show that a convex subset of euclidean ?i-space is simply connected.
22. If X is a unit interval, show that ?(? ? ?) ъ ?(?).
23. Show that a solid torus (that is, the closure of the region bounded by a
torus) has the same fundamental group as a circle.
8-3. Relation Between ?(?) and $?(?)
To simplify our discussion, we restrict attention to the case where
? = \K\ is a topological polyhedron, with oriented triangulation ?.
We will work with a rectilinear realization К of Я in a euclidean space
E, and with the polyhedron ? = \K\, homeomorphic to ?. The
replacement of ? by ? will involve no loss of generality, for present
purposes, since Ф(Р) ъ ?(?) by Theorem 3.
Let {s} = {sv . . . , 5? ) be the 1-simplexes of K, and let st be
defined as image st = А*(Х0) °f a un^ segment X0 : @ < ? < 1)
under a linear homeomorphism ?? : X0 -> i/. But ?? also defines a
path ei9 which we will call the path associated with s{. We can write
(8.8) (a) 8i = ?€(?0) (b) e, = ??(?0) (i = 1, . . . , ??).
The distinction between the simplexes s{ and the paths ei9 which
we will call elementary edge paths, is explained in Art. 8-2 above.
Let ?? : X0 —> ? denote the mapping defined by ?^?) = ЯгA — x),
? e X0. Then ?? defines the oriented 1-simplex — st and the edge path
ег~х, that is,
(a) ~sz = ??(??)
(8.9) (??(?) = Хг(\ - ?); г=1,...,о1).
(b) ??1 = ??(?0)
(A) Hence, if et is the path associated with si9 then e^1 is the path
associated ivith —s^
An edge path in general is a path product
(8.10) ? = ?\\... e% (?j = +1 or -1 j = 1, . . . , n).
(?) ?? ? =3 and the three factors in (8.10) are the three edges
of a 2-simplex s2 e K, then ? is an elementary boundary path. Each
s2 has six elementary boundary paths, each determined by the
selection of an initial edge and its exponent.
(G) Let ? and q be the initial and terminal points, respectively,
of ? in (8.10). A) If ? contains a pair of successive factors of the form
ее-1, then ? ^ ?' in U(p, q)9 where ? is obtained from ? by omitting

???
Fig. 8-4. ^?^^?^^?— 4?·
194

Art.8-3] THE FUNDAMENTAL GROUP; COVERING SURFACES 195
the factor pair ее-1 in (8.10) (Exercise 24). B) If ? contains a factor
e, and e belongs to some elementary boundary path efg, then ? c^ ?" in
? B?, q), where ?" is obtained from ? by substituting the factor pair
g-1/-1 for e in (8.10) (Exercise 25).
(D) By an elementary homotopy operation we mean A) the dropping
of ее-1 as described in Part A) of (C) or B) the inverse operation of
inserting a factor pair ее-1 in (8.10) just after a factor/, provided the
terminal point of/ is the initial point of e or C) the substitution of
gr-y-i for e as described in Part B) of (C).
Lemma 6. By elementary homotopy operations, it is possible (a)
to replace two consecutive factors ab of an edge path by c_1 if abc is an
elementary boundary path; (b) to eliminate three successive factors
abc from an edge path if abc is an elementary boundary path; (c) to
insert three factors abc9 constituting a boundary path, immediately
after a factor e, provided the terminal point of e is the initial point of a
(Exercise 26).
Given two edge paths, ?? and ?0, we will say that ?? is combina-
torially homotopic on К to ?0, symbolized by
(8.11) *h~*7o>
if and only if it is possible to pass from ?? to ?0 by a sequence of
elementary homotopy operations.
The relationship (8.11) implies that ?? and ?0 have the same initial
point and the same terminal point, since none of the elementary
combinatorial operations alters either end point of an edge path.
Lemma 7. The combinatorial homotopy relation ~ is an
equivalence relation.
Proof of Lemma. It is obviously reflexive and transitive. It is
symmetric because the inverse of each elementary homotopy
operation can be effected by an elementary homotopy operation. This is
obvious for the first two operations in (D). In the case of E3), it
follows from Lemma 6(a).
Theorem 6. The relation (8.11) holds if and only if
(8.12) щ~щшЩр,д)9
where ? and q are the initial and terminal points of ?0.
Proof. By the definitions and Statement (C), (8.11) implies (8.12).
To establish the converse, we first recall that (8.12) means that there
exists a continuous mapping/ of the rectangle ? = a0b0b1a1 @ < ? <
1, 0 < t < 1) (Fig. 8-4) into ? such that ?0 =f{a0b0), ?? =f(a1b1),
? =f{aoai)> Я =/(^(A)· Since ?j is an edge path (j = 0, 1), there
exists a subdivision Kj of ap^ on whose 1-simplexes / is linear. Let

196 INTRODUCTORY TOPOLOGY [Ch. 8
K2 be a subdivision of ? with K\ and K\ as subcomplexes. By the
approximation theory in Chapter 6 (see especially Arts. 6-7 and 6-9),
there exists, for a sufficiently fine subdivision K% of K2, a mapping
? : ? -> ?, which A) is linear on each simplex of K„, B) agrees with
/on the boundary of p, and C) maps each simplex of Kl onto a simplex,
possibly of lower dimension, of К (see Fig. 8-4bc). It is possible to
pass from ?0 to ?? by a sequence of elementary homotopy operations
each involving edges and elementary boundary paths on ?(?^). The
argument can then be completed with the aid of (C) and (D). While
the proof is straightforward, its details would involve a somewhat
tedious adaptation of methods already used in earlier chapters.
Now let p0 be a fixed vertex of К and let Щ be the set of all edge
paths with p0 for both initial and terminal point. The combinatorial
homotopy class of a path ? in Щ, defined by ~ (see Lemma 7), will
be denoted by ?$(?). The vacuous edge-path in ?$ will be denoted
by ?. Then ?$(?) contains ее-1, if e has p0 for initial point, and also
abc, if abc is an elementary boundary path from p0 to p0.
Theorem 7. The combinatorial homotopy classes {Г$} = (Г^(^) |
? e Щ} are the elements of a group ?0, whose product operation is
defined by
(8.13) ?*(?)?*(?') = ?*(??').
The unit element is ?$(?) (Exercise 27).
Theorem 8. The group ?0 is isomorphic to ?(?, p0), hence to ?(?).
Proof. By Theorem 6, it suffices to show that Г0(у) contains an
edge path ?. This follows easily, with the aid of the material in Art.
6-9 applied to mappings of the unit circle into P. Figure 6-3 is
suggestive.
(E) The combinatorial fundamental group ?0 is defined in terms
of the edges, finite in number, of a triangulation К of P. This makes
it relatively easy to analyze, and its structure is the same as that of
the invariantly defined Ф(Р). These groups are related after the
fashion of the finitely computable simplicial homology groups and
the invariantly defined singular homology groups.
(F) Since ?0 clearly depends only on the 2-skeleton K2 of K, it
follows that ?(?) ^ ?(?) ^ ?(\?2\).
Consider a group © given by a set of generators (gly . . . , gn) and
defining relations (see Appendix). By © made abelian we will mean
the group ©a with generators (дг, . . . , gn) and with the defining
relations of © supplemented by the commutativity conditions gtg5 =
g^i, equivalent to g^^gf1 = I (i = 1, . . . , n; j = I, . . . , n).
(G) If © is already abelian, then ©a = ©. If not, then some
elements are equal in ©a which are unequal in ©.

Art.8-3] THE FUNDAMENTAL GROUP; COVERING SURFACES 197
Theorem 9. The first homology group §?(?) is isomorphic to its
fundamental group ?(?) made abelian.
Proof. It is, by Theorem 8 and Art. 6-6, sufficient to show that
the simplicial homology group ^(K) is isomorphic to ?% =(?0
made abelian.)
(H) The symbol for an edge path ? from p0 to p0, modified by a
prime, will denote the corresponding chain ?' = ?(?). Thus, using
(sv . . . , sa ) for the oriented 1-simplexes of К:
r
(8.14) ? : ? = ejj . . . ??->4' = ???% (?? = +1 or ~lJ = l> · · · > r).
3 = 1
This defines a mapping 99 of edge paths into simplicial 1-chains.
Lemma 8. If ?0 ~ ?? then 77O ~ 7^. The converse does not hold.
Proof of Lemma. The elementary operation of deleting a pair of
successive factors of the form ss'1 in the product form for ? does not
alter ?', nor does the inverse operation. The other elementary
operation (D3) applied to ? corresponds to the addition to ?' of the
boundary chain of a 2-simplex and therefore yields a cycle
homologous to ?'. This proves the first part of the lemma. We remark,
without proof, that if ? is a torus with a contour and ? is an oriented
path around the contour, then ? ~ 0 but ? is not null-homotopic.
This illustrates the second sentence of the lemma.
Lemma 9. The mapping ? induces a homomorphism ?' : ?0 ->
ЬЛК).
Proof of Lemma. We will denote the homology class of 77' by [?'].
By Lemma 8, ? induces a mapping ? : ?$(?) —> [?']. Since {?^?^ —>
Vo + Vi, by (8.14) and ? ~ ?0?? => ?' ~ щ + ?[, this mapping is a
homomorphism.
Lemma 10. The homomorphism ?' induces an isomorphism
Proof of Lemma. The lemma means that ?' maps two elements
?*(?) and rg(?) of ?0 onto the same element of ?>?(?) if and only if
?*(?) equals ?*(?) in ?«.
Since ^'(rSl^jrjl^)) = [^ + 77 2] = [77 2 + 77J, 99' maps equal
elements of Tg onto the same element of ?>?(?)-
(I) To show that ?'(??(?)) = ?'(?*(?)) implies ?*(^) = Г*(?) in ?«,
it is sufficient to show that ?'(??{?)) = 0 implies ?^(^) = / in ?*,
where 0 is the null element of ^г(К) (Exercise 29).
(J) The condition <?'(?$(?)) = 0 holds if and only if some path in
Г*(??) nas the property that each edge appears in it equally often
with both orientations. Thus the proof reduces to the following
lemma.

198
INTRODUCTORY TOPOLOGY
[Ch. 8
Lemma 11. If ? is a path in whose expression of the form (8.14)
each edge appears as often with exponent +1 as with exponent — 1,
then ? is in the identity element of Tg.
Proof of Lemma. To simplify the notation, let e5 = &!·* (j — 1,.. ., r),
so that
(8.15) n = e1...er = 8?1...8?r.
Let G be the linear graph composed of the edges appearing in ?,
and their vertices. If G is a tree (Art. 1-3), let e be the first edge in ?
whose terminal vertex is a terminal vertex of G (Art. l-3(B)). Then
e_1 immediately follows e in ?, and ее-1 can be eliminated to obtain
?' ~ ? where ? has two fewer edges than ?. Now G(tt'), if not
vacuous, is a tree. By repetitions, we deduce that ? is in the unit element
of ?0, hence of ?».
If G is not a tree, let e be the first edge of ? on a circuit of G. Then,
by hypothesis, e_1 is an edge s ?? ? and we can write
(8.16) ? = ъ1еъ2е~1ъ3.
Since e is on a circuit of G, G — s is connected. Hence there exists a
path 7г0 on G — ? from the terminal point of e to the initial point of e.
Then, by (C) and (D),
(8.17) ? ~ (????????????^????????^^).
The transposition of the last two parentheses yields the path
(8.18) (^етгоНтго^^зНтго^тго) ~ {^г^п^-^щ) = ?',
where ?' contains the elements e and e-1 each once less than ?. Also
?' and ? are in the same element of ?{>, though not necessarily in the
same element of ?0. Now G(n') = G(n) or G{tt') = G{tt) — s according
as e appears or does not appear in ?. In the former case, we repeat
the process until we arrive at a ?" in the same element of ?% as ? for
which CrGr") = G(tt) — «. Continuing thus, we finally arrive at a ?"
for which СгGг'") is a tree, and the lemma is proved.
EXERCISES
24. Prove Part A) of (C) by showing that ? can be deformed into ?' in H(p, ?
(see Art. S-2(A)) by a deformation affecting only the factor pair ее-1 and reducing
ее-1 to a point.
25. Prove Part B) of (<7).
26. Prove Lemma 6.
27. Prove Theorem 7, following the pattern of the proof of Theorem 1.

Art.8-4] THE FUNDAMENTAL GROUP; COVERING SURFACES 199
28. Which of the following surfaces are simply connected? Justify your
answers, (a) The curved surface of a double cone, (b) Euclidean ?-space, (c) An
annulus.
29. Deduce (I) from the fact that ?' is a homomorphism.
8-4. The Fundamental Groups of En and of a Circle
(A) It is a simple matter to show (Exercises 21 and 28b above)
that euclidean тг-space En is simply connected, using deformations
along line-segments (Exercise 30).
Theorem 10. The fundamental group of a simple closed curve к
is a free cyclic group.
Proof. We will establish Theorem 10 by a method intended to
lay a foundation for more general procedure.
(B) Let E7 = [0, 1] : @ < и < 1) be the unit segment on a w-axis.
Let g : U —> ? define a path ? = g(U) on ?, and let ? : ? —> ?' be a
continuous mapping. As a ready consequence of the definitions,
cpg : U -> ?' defines a path ? = cpg(U) on ?'. We will also write
? = 99(a) to symbolize this manner of defining a curve ? on ?' by a
mapping ? from another space ? containing a.
We will make use of (B) in proving Theorem 10, with ?' = ? and
? = ?1, the motive being to capitalize on the simplicity of the
deformation properties of E1.
As a consequence of Theorem 3 and Exercise 7 above, our proof
will lose no generality in being given for the case where к is the unit
circle x2 + y2 = 1 in an (x, i/)-plane E2. Let E1 be interpreted as a
?-axis, and let ? : ?1 —> ?2 be defined by
(8.19) cp(t) = (x, y) where ? = cos 27r?, у = sin 2??.
Lemma 12. An arbitrary path ? = g(U) on к can be expressed in
the form ? = 99(a), where a is a path on E1.
Proof of Lemma. The mapping 99 is of period 1 in t. It is a
local homeomorphism in the following sense. A) If ? cz E1 is a
segment of length ? < 1, then the restriction 99,. = 991 ? is a
homeomorphism of ? onto an arc ? = ??(?) = ?(?), of arc length 2??.
B) Given an arc ? с K of length 2?? < 2?, the inverse set ?~1(?)
consists of a set of disjoint arcs {?} = 99-1(y), and {?} is mapped onto
itself by a translation t' = t + m (m an integer). If ? e {?} = <p-1(y)
then ?~? is a homeomorphism and is a local inverse to ?.
Case 1. ? is on an arc ? cz K of arc length <2?. Let ? e ?~1(?)?
and let ? = ^(?) = <Р^гд(и). Then /? = ?(&) = ??~^{?), since
<?<?-1 is the identity on т.

200 INTRODUCTORY TOPOLOGY [Ch. 8
Case 2. The general case. By the uniform continuity of g on U,
we can express ? as a product ? = ?? . . . ?? where ?? = g(U\) (г =
1, ... , ?) is on an arc yt с K of length <27r. Then Case 1 applies to
each individual ?^
Hypothesis. For some j e 2, . . . , ? the product ?'-1 = ?? . . . ?5_?
can be expressed in the form ??_1 = ?(???~1) where ?5-1 is a path on E1.
Applying Case 1, let ?? = ^(?1), fulfilling the hypothesis for j = 2.
Let qj_1 be the terminal point of a5-1. Among the inverse images
?~1(??) let Tj be the one containing qs. Let ?,- = ?~1{??)- Then
?§ = ?{?.?). Since the initial point of a, is the terminal point of a5-1,
the product a5-1^ = aJ is defined. Since continuous mappings
commute with the path product operation, (p(<xj) = ?? . . . ?? and the
inductive argument establishing Lemma 12 is complete.
Corollary. If q0 e ?~1(??)9 where p0 is the initial point of ?,
then there is a unique ? = a(g0) satisfying Lemma 12 and having q0
as initial point.
(C) Theorem 10 will follow (see Theorems 2 and 3) if ?0(?, ?0) is
shown to be free cyclic (where?0 = ?@) =- A, 0)). Letll0 = U(p0, p0)
on к (Art. 8-2D)).
Lemma 13. Let ? be a path on E1 from t = 0 to t = ?? (an integer).
Then 99(a) e ?0. Conversely, if ? e ?0, then ? = 99(a) for some a on
E1 from t = 0 to t = m9 where m is an integer uniquely determined
by/3.
This follows easily from Lemma 12 and corollary.
Lemma 14. Let/M be a deformation of ? in ?0 and let ??? = /?(?)-
Let gu(oi) be the path on E1 from t = 0 such that ?? = срди(к)> Then
gu is a deformation of ? with fixed end points.
It is a routine procedure to verify that gu is a deformation with
the initial point of ? fixed at t = 0. Initially, the terminal point of ?
is fixed, by Lemma 13, at some integral point t = ??. Since the
terminal point varies continuously with и and is always an integral
point, it must remain fixed.
If ? is a closed path, the path ? ? ... ? (?? factors) is denoted by ?171.
Lemma 15. Let rw = [0, ??] : 0 < t < ??, and let ym = 99(rw).
Then A) ym = ?? and B) each ? e ?0 is homotopic in ?0 to ym for
a unique integer ??.
Proof of Lemma. Part A) holds because 99 is of period 1 in t. Part
B) follows from Lemma 14 and the fact that an arbitrary path on E1
from t = 0 to t = m can be deformed into rm on E1 with fixed end
points.
With Lemma 15, the proof of Theorem 10 is complete.

Art.8-5] THE FUNDAMENTAL GROUP; COVERING SURFACES 201
EXERCISES
30. Let fj : X -> En (j = 0, 1) define two paths щ (j = 0, 1), where X =
[0, 1] : 0 < ? < 1, and where Д@) = /2@) =ДA) =/2A)· Formulate a
deformation ft of 7г0 into ?? along line segments @ < t < 1).
31. Show that ? can be expressed as ?? . . . ?? as asserted in Case 2 under
Lemma 12.
32. Prove Lemma 13, with special attention to the uniqueness of m.
33. Write up the proof of Lemma 15 in greater detail.
8-5. The Fundamental Group of a Surface
Theorem 11. The fundamental group of a torus is free abelian
with two generators.
Proof. We use the same type of argument as in the case of Theorem
10.
Let E2 be the euclidean plane of a coordinate system (x, y) and
let Ez be euclidean 3-space with a cylindrical coordinate system
(r, ?, z). We lose no generality in proving the theorem for the special
torus ,
I 0 = 2??,
(8.20) ?2 : -j r = 2 + cos 2тту,
I 2 = Sin 2774/.
Let ? be the mapping of E2 into Es defined by ?(?, у) = (?, ?, ?),
where (?, ?, ?) are given by (8.20). Then ?2 = ?(?2).
The reader would do well to draw diagrams, showing E2, T2 с Е*,
the coordinate systems, and the various geometric objects in the
following proof.
Lemma 16. Let ? be a path on E2. Then ? = 99(a) is a path on
T2. Each path ? on T2 can be thus defined.
Proof of Lemma. The lemma and its proof are strictly analogous to
Lemma 12 and proof. We comment on some major points (Exercise 35).
(A) The mapping ? has period 1 in ? and in y. Its restriction to a
square region (x0 < ? < x0 + ?, y0 < у < y0 + ?), where 0 < ? < 1,
is a homeomorphism. Hence so is its restriction to a circular region
of diameter < 1.
(B) There exists a number d > 0 so small that each subset of T2 of
diameter <d is on the image under ? of some circular region on E2 of
diameter <1.
In the inductive argument analogous to that for Lemma 12, the
role of the ?5 is played by sub-arcs yi of 7 of diameter <d, and the role
of the ?^ is played by circular regions of diameter <1.

202 INTRODUCTORY TOPOLOGY [Ch. 8
Corollary. Given a path ? on T2 with initial point p0, let (x0, y0) e
?~1(?0). Then there is a unique path a0 on E2 with (x0, y0) for initial
point, such that ? = ?(?0). A translation of a0 by a vector with integral
components (m, n) takes a0 into the path ? with initial point (x0 + m,
y0 + n) such that ? = ?(??).
We prove Theorem 11 by showing that ?0{?2, p0) has the specified
structure where p0 = ?@, 0), the point with cylindric coordinates
(r0, ?0, z0) = C, 0, 0). Let ?0 be the class of closed paths on T2 from
p0 to p0.
Lemma 17. A path ? on T2 belongs to ? 0 if and only if there exists a
path ? on E2 such that
{initial point of ? is @, 0),
terminal point of ? is (m, n) where (ж, n) are integers.
The path ? is unique, given ? (Exercise 36).
Lemma 18. Let/, be a homotopic deformation of у in П0 and let gt(&)
be the path in E2 from @, 0) such that /Д/?) = <р(^(а)). Then gt is a
homotopic deformation of ? with both end points fixed (Exercise 37).
Lemma 19. Let aw be the linear path along the x-axis from @, 0) to
(m, 0) and let ?? be the linear path along the i/-axis from @, 0) to @, n).
Let a = ?(?1), b = ?(?1). Then A)
(8.22) am = <р(Лт)9 Ъп = ?(??)
and B) each path ? e ?0 is homotopic in ?0 to ambn for a unique pair of
integers (m, n).
Proof of Lemma. Part A) follows from the fact that ? is of period 1
in ? and in y. Part B) follows from Lemma 17, an obvious deformation
in E2, and the fact that ? maps a deformation on E2 into a deformation
on T2. If m = 0, aw is a null path, and similarly for ??, am, bn.
Theorem 11 now follows.
(C) Intuitively, the fact that ? ^ ambn in ?0 means that ? runs
around T2 m times in the sense of a, while it twists around it ? times in
the sense of b before returning to its starting point (Fig. 8-5). These
are the net numbers of times that ? thus runs and twists around T2.
Thus, m = 3 if ? makes five circuits in the positive sense of a, then
doubles back and makes two circuits in the negative sense of a.
Theorem 12. Let M2 be a compact surface, represented (see
Chapter 2) by one of the following polygonal symbols:
(a) M2 = ??'1 (?2 is a topological sphere);
(b) ?2 = ?,?,?-^?-? . . . AnBnA^B^KJ)xK^ . . . KrDrK^
(8.23) (sphere with h handles and r contours, h + r > 0);
(c) M2 = СгСг . . . C^K^K-1 . . . KrDrK~x (sphere with
q > 0 crosscaps and r > 0 contours).

Art.8-5] THE FUNDAMENTAL GROUP; COVERING SURFACES
203
(b)
Fig. 8-5. The relation ? ^ ambn where ? = ?(??) and m == — I, n = —2.
Then, in Case (a) and in Case (b) with h == 0, r = 1, Ж2 is simply-
connected. In Case (b) with h > 0 or with A = 0, r > 1, Ф(Ж2) is
isomorphic to a group with 2h + r generators (?<} ?„ df) (i = 1, . . . , A;
J = 1, . . . , r) and the single group relation
(8.24)
«10A ?;1 · · · «А^Ш ... ir = J.

204 INTRODUCTORY TOPOLOGY [Ch. 8
In Case (с), Ф(М2) is isomorphic to a group with q + r generators
(?v - - - > ?? ^?' · · · ' ^r) an^ the single group relation
(8.25) y\y\ . . . y& ...br = I.
Outline of Proof. The cases where M2· is simply connected offer no
difficulty.
Let M2 be triangulated into a complex so that the edges of a
polygonal representation P2 in the form (8.23) are covered by a sub-
complex, and let the complex K2 be represented as a triangulation of
P2 with boundary simplexes suitably identified. Let p0 be the common
initial and terminal point of the A's and B's in Case (b), where it is
also initial point of each K{. In Case (c). let p0 be similarly defined for
the C's and the K'a.
It suffices to show that the combinatorial fundamental group
?(?2, p0) has the specified structure. An edge path on M2 belongs to
?0 = U0(p0, p0) if and only if its representation on K2 has copies of p0
for initial and terminal point.
(D) Each homotopy class in ?0 contains an edge path on the
boundary of P2. For, any path in ?0 can first be deformed into an
edge path ? (Art. 8-3). Then, any part of ? joining a point qx on the
boundary of P2 to another such point q2 and composed of edges
interior to P2 can clearly be deformed into either of the two parts into
which q1 and q2 separate the polygonal boundary.
(E) Each edge path on the boundary of P2 from one copy of p0 to
another is a path product made up of factors each of which is one of
the following or its inverse:
A) in Case (8.23b), the edge path аг· along A{ or the path ?? along B{
or the path ?3· first along Kj, then Dp then Kf1;
B) in Case (8.23c), the path yi along Ci or the path 6i just described.
This yields Theorem 12, except for the group relationships.
The only null-homotopic paths on the boundary of P2 are those
which begin and end at the same copy of p0. From this fact it is easy
to deduce that (8.24) and (8.25) are group relations on which all group
relations depend, in Cases (8.23b) and (8.23c) respectively.
(F) For example, the fundamental group of the projective plane
has just two elements, and that of the Klein bottle has two generators
(?? ?2) with the defining relation Г^Г^ = /.
EXERCISES
34. Show that if ? is the square region 0 < ? < 1, 0 < у < 1 and (?, у) is
restricted to ?, then Eqs. (8.20) are a formulation of a definition of T2 as ? with
opposite edges identified.
35. Write out the proof of Lemma 16 and corollary as summarized.
36. Prove Lemma 17, which is analogous to Lemma 13.

Art.8-6] THE FUNDAMENTAL GROUP; COVERING SURFACES 205
37. Prove Lemma 18, analogous to Lemma 14.
38. With the aid of the representation suggested in Exercise 34 above, show
that аЪагЧг1 is null-homotopic.
39. Show (a) that the Betti numbers and torsion coefficients of a closed
3-manifold are determined by its fundamental group, (b) that the fundamental
group of a non- orient able closed 3-manifold is infinite.
40. Illustrate Exercise 39a for projective 3-space.
8-6. Covering Complexes
(^4) Let g : ? —> ? be a continuous mapping of a connected
topological polyhedron ? onto a connected topological polyhedron ?. The
pair (?, g) is a covering space of ? if A) for each ? e ?, the set {p} =
g~x{p) of its inverse images, called its covering set, is denumerable;
and B) letting
(8.26) {p} = (pv p2, . . .),
there exists a neighborhood ? ?? ? and neighborhoods iV^ of p{ (i =
1, 2, . . .) such that (a) g maps N{ homeomorphically onto N and (b)
each inverse image of a point on N is on one of the neighborhoods
{N} = (Nv N2, . . .). We call ? a covering space and g its projection.
Property (a) is a local homeomorphism property (Art. 8-4).
(B) Each pt ? {p} is said to cover or to be above p, which is called its
base point, and N{ is said to cover or be above N (i = 1, 2, . . .). If {p}
contains the same finite number к of points for each ? e ?, then (?, g)
is a &-fold covering of ?. If {p} is infinite for each ? e ?, then (?, g)
is an infinite covering of ?.
(C) As a first trivial example, an arbitrary ? is covered by itself,
? = ?, with an arbitrary self-homeomorphism (the identity for
example) as projection.
(D) The unit circle ? has an infinite covering (?1, ?), defined by
(8.19). The covering set of a point ? : (?, у) on ? is of the form {p} =
(t, t ± 1, t ± 2, . . .), since ? is of period 1 in t. We can use for
$(t + ??) an open interval of length J, center at (t + m).
(E) The torus T2 has an infinite covering (?2, ?), with ? defined by
(8.20). If px{x, y) is a point on E2 above ? on ?72, then {p} consists of
all the points (x + m, у + ?) as m, ? range over the integers. Circular
regions of radius d < 1 about each of these points can be used for {$},
since ? is of period 1 in ? and in y.
(F) Let ac be the unit circle of (8.19) and let ? be another circle
defined by
(8.27) * = ?{??) :
2??
? = COS
к
(*е2,3,4,...)
. 2???
у = sin —
к

206 INTRODUCTORY TOPOLOGY [Ch. 8
and let g : к -> к be defined by the condition g(<p(t)) = cp(t). Then
(к, д) is a &-fold covering of к (Exercise 41).
(G) Coverings of the torus analogous to the covering of the circle in
(F) are obtained as follows: Let T2 be defined as in (8.20), let f2 be
defined by f2 = <p(E2), where <p(x, y) = (r, ?, z) is defined by
(8.28) ?2 = ?{?2) :
2770;
m
2vy
r = 2 + cos
. 2тту
? = sin
wgA,2, . . .)
ne A,2,...)
and let g : T2 -> T2 be defined by g(<p(x, y)) = ?(?, у). Then (?2, #) is
an mn-fold covering of T2 (Exercise 42).
As another example, consider a manifold ? in the form of a sphere
with three handles. Let ? be covered with five layers of identical
manifolds Mv M2, . . . , M5, like layers of skin, so that the result
could be regarded as a laminated surface made up of six juxtaposed
membranes (M, Mv . . . , Мъ) which, for the sake of being specific, we
number from the inside layer outward. The juxtaposition of the layers
affords a natural homeomorphism among them. If ? is any point on
M, then pi will denote the corresponding point on Mi (i = 1, . . . , 5).
The mapping g : ? -> ? will be called the projection of the union M$ =
5
(J М{ of the Mi onto M. Now (Ж*, g) fails to satisfy the definition of
t = l
a covering space in that ? J is not connected. To correct this, first let
three closed cuts be made, as suggested by a, b, and с in Fig. 8-6,
through all five surfaces M{. Allowing surfaces to penetrate one
another, let the edges be matched as suggested in the figure to obtain
a covering (M, g). Thus, along b, the two edges of Mx are rejoined,
and so are those of M2, while Ms is joined to Ж5, Ж4 to Ms, and Mb to
Ж4 in the direction of the arrow cutting across b in the figure.
(H) The surface M, with the projection g, is a 5-fold covering
space of M.
Consider a closed path ? on ? from px to pv where no point on a
cut is above pv and let ? be the path on ? obtained by liftingf ? into
? so that рг on ?? is the initial point of ?. Then ? stays on Mx unless
it crosses a, in which case it shifts to M2. Hence, ? goes from рг to p2
or else from px to pv according as it crosses a an odd or an even
number of times.
Theorem 13. Given a topological polyhedron ? and a covering
space ?, there exist triangulations К and ? ?? U and ft such that the
f That is, finding ? so that ? = д{тг) (Exercise 43).

Art.8-6] THE FUNDAMENTAL GROUP; COVERING SURFACES 207
(b)
Fig. 8-6. A 5-fold covering space of the sphere with three handles.
mapping g induces a mapping of the j-simplexes of К onto the j-
simplexes of К (j = 0, 1, . . . , m).
Outline of Proof. Let К be so fine a triangulation of ? that each of
its simplexes is on a neighborhood N of the sort entering into Definition
(A). Then the inverses of the mappings g | j\^ define a triangulation
К of ? satisfying the theorem.
Corollary 1. If ? is a &-fold [infinite] covering space of ?, then g
maps exactly к [infinitely many] simplexes of R onto each simplex of K.

208 INTRODUCTORY TOPOLOGY [Ch. 8
Corollary 2. The Euler characteristic of ft is к times that of ? if ?
is a &-fold covering space of ?.
For, in the notation of Theorem 13, if a,· is the number of j-cells of
K, then olj = koij is the number of j-cells of K.
From the fact that the characteristic of a closed 2-manifold ? is
B — 2h for a sphere with h handles,
2 — q for a sphere with q crosscaps,
we deduce the following result.
Corollary 3. A k-fola covering space of a sphere with h handles is
a sphere with kh — к + 1 handles. A &-fold covering of a torus is a
torus (Exercise 45b).
Theorem 14. Projective m-space Pm can be doubly covered by the
га-sphere Sm.
Proof. One can represent Pm as Sm with antipodal pairs of points
(p, p') identified. The mapping g which maps each point ? of Sm onto
the identified pair (p, p') yields a double covering space (Sm, g).
Corollary. The only k-fola covering space (k > 1) of the
projective plane is a double covering by the sphere.
For, P2 has characteristic +1. A k-iold covering space of P2 is a
closed 2-manifold of characteristic k. The 2-sphere is the only closed
2-manifold of characteristic к > 1.
EXERCISES
41. Give a proof of (F).
42. Give a proof of (G).
43. Discuss the lifting of a curve into a covering complex. This involves writing
a short paper, using methods analogous to those used in connection with Lemma
12 and corollary, also Lemma 16 and corollary. Include a proof of Theorem 15
below.
44. Given the closed path ? of Fig. 8-6, discuss all paths obtainable by lifting
7? into M. If тгк is lifted onto ?k on M, under what conditions will ffk be closed?
45. (a) Prove that a covering space of a closed orientable m-pseudomanifold
is a closed orientable m-pseudomanifold. (b) Establish Corollary 3 to Theorem 13.
46. Show that a &-fold covering space (k > 1) of a sphere ? with h > 1
handles is not homeomorphic to M.
47. Show that a 2-sphere can have no &-fold covering (k > 1). Compare the
proof of the corollary to Theorem 14.
8-7. Fundamental Groups and Coverings
Let (?, g) be a covering of a topological polyhedron ?, and let
/ : X -> ? define a path
(8.30) 77 = f(X) X = [0, 1], 0 < ? < 1.

Art.8-7] THE FUNDAMENTAL GROUP; COVERING SURFACES 209
Then 77 covers, or is a covering path of,
(8.31) 77 = д(тт) = gf(X) (see Lemma 12),
which is the base path, or the projection, of 77.
Theorem 15. If 77 is a path on ? and ? on ? covers the initial point
? of 77, then there is a unique covering path of 77 with initial point p.
The proof is left to the reader (Exercise 43 above).
Lemma 20. Let (p) = pv p2, . . . be the covering set of ? on ? and
let 7?г be the path above 77 with initial point pt (i = 1, 2, . . .). Let дг be
the terminal point of тгг. Then дг ? q5 if г ? j.
Proof of Lemma. As a consequence of Theorem 15, ??-f1 is the only
path above ?^1 with дг for initial point. Hence qt = q5 would imply
Pi = Vj-
Corollary. A covering (П, g) is either &-fold for some к or is an
infinite covering.
For, the paths 77 г· (г = 1, 2, . . .) and щ1 establish a one-to-one
correspondence between the covering sets of ? and q if 77 goes from ?
to q.
Theorem 16. If, on ?, 7?! ^ 7?2 with fixed end points, then the
projections 77, = ?7G7,) (г = 1,2) are also homotopic with fixed end
points.
This follows easily, since a deformation rectangle f(X ? ?) for a
homotopy of 7?! into 77 2 with fixed end points projects into such a
rectangle gf(X x T) for ?? and 772.
Theorem 17. If, on ?, ?? ^ 772 with fixed end points, and G^, 772)
are paths above G7l3 772) with a common initial point p, then ?1 ^ 772
with fixed end points.
Outline of Proof. This theorem is analogous to Lemmas 14 and 18,
but the argument must be modified, unless we introduce metrics on ?
and ?. Let a deformation rectangle on ? for a deformation 770 —> ??
be triangulated into a singular complex K2 so fine that each of its
simplexes is on a neighborhood N satisfying Art. 8-6(Л). Then there
is a unique singular complex K2 such that g(K2) = K2 and \K2\ is a
deformation rectangle for a deformation of ?1 into 772 with fixed end
points (Exercise 48).
Theorem 18. Let Ф(ГЬ р) be the fundamental group of ft with a
fixed point p. The projection g : ft -> ? induces a homomorphism
9* : ?(?, ?) -> ?(?, ?), where ? = g(p).
Proof. In the first place, Theorem 16 implies that g takes homotopy
classes into homotopy classes. Secondly, g commutes with the path
product operation, since д(тгтг') = д(тг)д(тг') provided 7777' is defined.

210 INTRODUCTORY TOPOLOGY [Ch. 8
Theorem 19. The homomorphism g* is an isomorphism between
?(?, ?) and the image subgroup <7*?(?, ?) c ?(?, ?).
Proof. This follows from Theorems 16 and 17. The latter implies
that no two different homotopy elements of ?(?, ?) can map onto the
same element of ?(?, ?).
(A) For example, in the case of the 3-fold covering к of к in Art.
S-6(F) with k = 3, g* maps the free cyclic group ?(?, ?) onto the
elements T3w of ?(?, ?) where ? is a generator of ?(?, ?) and m = 0,
±1, ±2, ....
(B) In the case of the coverings (?1, ?) of к and (?2, ?) of T2 (Art.
8-6(F)(G)), the fundamental group of the covering space consists of
the unit element alone and g* maps it onto the unit element of the base
space.
For another example, see Exercise 49.
A general analysis of relations between the fundamental group of a
topological polyhedron ? and the possible covering spaces of ? would
lead us into group-theoretic considerations beyond the scope of this
book. It turns out that there is a one-to-one correspondence between
classes of possible covering spaces and classes of so-called conjugate
subgroups of ?(?). The problem of determining all k-iold covering
spaces of ? is thereby reducible to the group-theoretic problem of
finding all classes of conjugate subgroups with a so-called index k in
?(?).
(C) In particular, it can be shown that each connected topological
polyhedron ? admits a universal covering complex, ?, characterized
by the property of being simply connected. We found the universal
covering of A) the circle к by E1, B) the torus by E2, C) the projective
plane P2 by S2. With suitably defined projections, one can show that
the universal covering ? is a covering of each covering ? of ?, and
that a covering space of all coverings of ? must be its universal
covering. Further details are to be found in [S-T], for example.
EXERCISES
48. Write out the outlined proof of Theorem 17. This involves writing a
short paper.
49. Consider the 6-fold covering of T2 in Art. 8-6(?) with m = 3, ? = 2. If
[a], [6] are the generating homotopy classes of a, b for T2, show that g* maps
?(?2, ?0) onto the subgroup of ?(?2, ?0) generated by [a3], [62].

Bibliography
The following list makes no claim to completeness. Additional
references can be found in the works here listed. The symbols in
brackets are used in this book for reference purposes.
Aleksandrov, P. S. Combinatorial Topology. Vols. 1, 2, 3. Albany: Graylock
Press, 1956, 1957, 1960.
Alexandroff, Paul. Einfachste Grundbegriffe der Topologie. Berlin: Julius
Springer, 1932.
Alexandroff, Paul, and Hopf, Heinz. Topologie I. Berlin: Julius Springer,
1935.
Courant, Richard, and Bobbins, ?. ?. What is Mathematics? New York:
Oxford University Press, 1948. Chap. 5.
Hilbert, David, and Cohn-Vossen, S. Geometry and the Imagination. New
York: Chelsea Publishing Co., 1952. Chap. 6.
[Kx] Kelley, John L. General Topology. New York: D. Van Nostrand & Co., Inc.,
1955.
[K2] Kerekjarto, B. v. Vorlesungen uber Topologie I. Berlin: Julius Springer, 1923.
[Lx] Landau, Edmund. The Foundations of Analysis. New York: Chelsea Publishing
Co., 1951.
[L2] Lefschetz, Solomon. Introduction to Topology. Princeton: Princeton
University Press, 1949.
Pontryagin, L. S. Foundations of Combinatorial Topology. Albany: Graylock
Press, 1952.
[S-S] Schreier, O., and Sperner, E. Introduction to Modern Algebra and Matrix
Theory. New York: Chelsea Publishing Co., 1951.
[S-T] Seifert, H., and Threlfall, W. Lehrbuch der Topologie. New York: Chelsea
Publishing Co., 1957. Original ed.: Leipzig: Teubner, 1934.
[T] Thurston, H. A. The Number-System. London: Blackie & Son, 1956.
[V] Veblen, Oswald. Analysis Situs. 2d ed. Vol. V, part II. American
Mathematical Society Colloquium Publication, 1931.
211

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APPENDIX

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Group-Theoretic Background
The purpose of this appendix is to set forth the concepts and results
of group theory used in this book. We confine ourselves largely to
descriptive statements, omitting many important logical
considerations. A good modern approach to the subject for its own sake would
be quite different from the present assemblage of facts for reference
purposes. A few exercises are included.
A-l. Basic Terminology
A (multiplicative) group © is a set of elements {g} and a product
operation, called (group) multiplication, as follows: Given gx e © and
g2 e ©, there exists a product дгд2, subject to the following group axioms:
I. gxg2 is an element of ©.
II. If gi ? © (i = 1, 2, 3), then g1(g2gs) = (g1g2)gs· That is,
multiplication is associative.
III. There exists an element 1 e {g}, called the unit element, such
that gl = Ig = g for each g e ©.
IV. For each g e ©, there exists an element g-1 e ©, the inverse
of g, such that gg-1 = д~гд = 1.
Added conditions define special groups or classes of groups. In
particular, the important class of commutative (or abelian) groups
is defined by adding the following axiom of commutativity:
V. If дг ? © and g2 e ©, then gxg2 = g2gv
(A) The '"product" operation for a commutative group is
frequently written in the form of a "sum," and the following definition is
used. An (abelian) additive group 91 is a set of elements {a} and a sum
operation called (group) addition, as follows: Given ax e 91, a2 ? 91,
there exists a sum, ax + a2, subject to the following axioms for
abelian additive groups:
I. ax + a2 ? 91.
II. ?? + (a2 + аг) = (аг + a2) + as (Associativity).
III. There exists an element 0 e 91, called the zero or null element
such that a + 0 = a for each a e 91.
IV. For each a e 91, there exists an element —a e 91, the negative
of a, such that a + (—a) = 0. This relation is generally
expressed in the form a —a =0.
V. ax + a2 = a2 + av
215

216
APPENDIX
Axioms II justify such notation as
Я\ЯгЯъ = 9i(929s) = ЙМЪ^з»
(АЛ)
? + ? + ? = ?? + (a2 + as) = ? + ?? + аз·
By a recurrent definition, a meaning is attached to
Я1Я2 · · · 9m>
«1 + ? + ' ' ' + am·
In the special case of (A.2) where the elements gi} or the elements ai}
are all equal, one writes
(A.3)
gm = gg . . .g (m factors)
g~m = g'Y1 · · · <7_1 = (g™)'1 (m factors)
ma = ? + ? + ···+? (m terms)
—ma = (—a) + (—a) + · · · + (—a) = —a — a · · · — a (m terms)
(B) Examples which the reader may wish to verify: A) the integers
Z, with addition, but not with multiplication; B) the integers mod m,
Zm (see (H) below), with addition, but not with multiplication, where
m is a positive integer; C) the rational numbers Q or the reals R,
with addition, but not with multiplication; D) the non-zero rational
or real numbers, with multiplication, but not with addition; E) square
matrices of equal size whose elements belong to ? or to Zw or to Q or
to R, with matrix addition; F) square matrices of equal size having
non-zero determinants, with rational or with real elements, and with
matrix multiplication. This is the only one of the six examples
which is not commutative. The symbols ? and R are frequently
used below.
Permutations on a given finite set of objects, with the product РгР2
of two permutations defined as the permutation resulting from Px
followed by P2, constitute a group. The reader may verify the group
axioms and decide whether the group is commutative.
A set of generators of a multiplicative [additive] group © is a set
{y} cz © of elements of © such that each element of © is expressible
as a product [sum] of the elements {?} and their inverses [negatives].
(C) We confine attention, save where the contrary is explicitly stated,
to groups admitting a finite number of generators. Such groups are
described as finitely generated.
(A.2)
ш<
г = 1

GROUP-THEORETIC BACKGROUND
217
Let © be a multiplicative group with generators (gv . . . , gn).
Then, by definition, each element g e © can be expressed in the form
(A.4) g = g\\ . . . g% (?j = +1 or -1; i, e 1, . . . , n; j = 1, . . . , r).
In a product like that on the right of (A.4), one can eliminate
factors equal to the identity, and one can combine consecutive factors
with equal subscripts, using (A.3) and the easily proved relation
дшдп = дпц-п ^ я, G ?) to reduce (A.4) to the form
9 = 9T11·· -9Tkk°T l> where
(A.5)
к ? Jh-v mh G z> 9ih^1 (A = 1, . . . , k).
We refer to (A.5) as the reduced form of (A.4), relative to (gv . . . , gn).
Thus g = g3g3gs1g4949s1 reduces to g = g^gbs1·
The inverse of (A.5) is easily seen to be
(A.6) g-1 = gjkm«.. . дттК
(D) In the case of an abelian additive group © with ? generators
(9i> · · · j 9n)i an arbitrary element g = ?}=1 е$г. similarly reduces,
with the aid of permutations of terms and the property mg + ng =
(?? + n)g, to a unique reduced form relative to (gv . . . , gn), namely
?
(A.7) 9 = lmi9i KeZ).
i = l
(E) The free multiplicative group with ? generators (gv . . . , gn) is
the group consisting' of all elements expressible in the form (A.4),
two elements being equal if and only if they have identical reduced
forms (A.5) relative to (g1} . . . , gn). The free abelian additive group
with ? generators is defined in the same way, save that (A.4) is put
in additive form and its reduced form is (A.7) instead of (A.5).
A group © (multiplicative or additive) with generators (gv . . . , gn)
can be specified by supplementing the axioms by a set of group
relations, each of which asserts the equality of two elements g and h
whose reduced forms are not identical, thus:
(A.8) sr«i . . . g«* = g% . . . g% (subscripts from the set A, . . . , л)).
This relation can be written as
(A.9) C-0r'-C = 1
where the reduced form of the left side is not 1.

218 APPENDIX
(F) Thus the set of all group relations is equivalent to the set of all
representations of the unit element 1 or, in the additive case, of the zero
element 0. A set of group relations
(a) IT (9ijh)miih = 1 (J = I? 2' . . . , k) (multiplicative case), or
h = l
(АЛО) rs
(b) ? mijh9ijh = 0 (j=l,...,k) (additive case),
is independent if no one of them is a consequence of the others. If
one of them is a consequence of the others, it will be said to depend
on them.
For example, in the following set of relations, for an abelian
additive group,
(a) 2ft + 3ft - 2ft = 0,
(A.ll) (b) ft - 402 + 3gr8 + flr4 = 0,
(c) 8ft + ft + 2ft = 0, ^ ^ \m
each pair is independent, (but each relation depends on the other two.J)
(G) A set of group relations on which all other group relations for
E depend is a set of defining relations of ©. A group is determined
by a set of generators and defining relations.
A group with just one generator g is described as cyclic. It is
necessarily abelian. If there are no group relations, its elements are
l,g±1,g±z, . . . (multiplicative)
(A.12)
0, ±g, ±2g, . . . (additive)
and the group is called the free cyclic group. If there exist group
relations, then all such relations are easily seen to depend on a single
relation gm = 1 or mg = 0.
The elements are then
(A.13) 1, g, g\ g\ . . . , g^1 or 0, g,2g,...,(m- l)g,
and the group is described as cyclic of order m (Exercise 1).
Given a positive integer m, ? is said to be congruent to у modulo m
(abbreviated mod m), where ? and у are integers, if and only if
(A. 14) ? — у = km for some integer k.
This relationship is symbolized thus:
(A.15) ? = у modm.
A residue class mod m consists of all integers congruent mod m to a
given integer. Let the residue class mod m containing a number j
be denoted by J. ¦

GROUP-THEORETIC BACKGROUND
219
(?) There exist exactly m different residue classes mod m : 0, I,
2, . . . , m — 1. They constitute an abelian group with j + к = J + к
as the group operation. This group is denoted by Zw and called the
group of integers mod m. It is cyclic of order m (Exercise 2). The
additive group ? of the integers is free cyclic.
The statements in (H) illustrate that an element of a group may
be an entire equivalence class of objects. Such is the case in most
topological applications.
EXERCISES
1. Show that, if m is a positive integer, then
52 2n 2n
(A.16) A = em = cos— + г sin— (г2 = -1)
m m
generates a cyclic group of order m, with multiplication of complex numbers as
the group operation.
2. Establish the three statements in (H).
A-2. Homomorphisms and Isomorphisms
Just as we consider mappings of a set into or onto a set (Art. 3-1),
so we consider mappings of a group © into [onto] a group ©, restricted
by the requirement that the product or sum of two elements of the
first group shall map onto the product or sum of their images. Let
g e © denote the image of g e ©. The requirement just stated can
then be symbolized thus:
(A.17) gs = gxg% if gs = gxg2 or gs = g± + g2 if gs = g± + gz.
A mapping of © into © satisfying this condition is called a homo-
morphism of © into ©. A homomorphism is called an isomorphism
if it is onto (Chapter 3) and one-to-one. An isomorphism can be
described as a homomorphism with an inverse which is also a
homomorphism. A group © is isomorphic to a group © if it is possible to
establish an isomorphism of © onto ©. The relation of being
isomorphic is an equivalence relation. A group invariant is a property
which, if possessed by a group ©, is possessed by each group
isomorphic to ©. To say that two groups have the same structure is to say that
they are isomorphic.
A set of properties will be said to characterize a group © or to
determine its structure if each group sharing all these properties with
© is isomorphic to ©.
A subgroup of a group © is a group ©0 whose elements all belong to
© and whose group operation is the same as that of ©. We symbolize

220
APPENDIX
this relation by ©0 с ©. If the elements of ®0 are a proper
subset* of those of ®, then ®0 is a proper subgroup of ®.
Let ® be an additive group and 31 a subgroup of ®. Then B2 e ®
is congruent mod 31 to Bx e ©, symbolized by
(A.18) B2 = Бх mod 51,
if B2 — 51? hence also Bx — Б2, is an element of 31.
(Л) The relation (A.18) is an equivalence relation. The set of all
elements congruent mod 31 to В е ® will be denoted by 5 and is
called the coset of1 В in ® mod 9?. Thus 0 is the set of elements of 31.
Theorem 1. Let B19 B2 be two elements of an abelian additive
group ® with subgroup 31. If Сг e Bx and C2 e B2, then Сг + C2 e
Bx + B2. The cosets of ® mod 31 form an abelian group, with addition
defined by Бг + B2 = Bx + B2.
Proof. The relation Сг e Bt (i = 1, 2) means that there exist
elements Ai¦ e 31 such that Gi = Bt + A{. Hence Cx + C2 = Bx +
Б2 + ^! + Л2 g Бх + Б2, since Ax + Л2 e 31. This proves the
first part of the theorem and verifies Axiom I. The rest of the proof
is left to the reader (Exercise 7).
(B) The group defined in Theorem 1 is the factor group or quotient
group
(A.19) ® = ®/Jl.
It is sometimes called the difference group and denoted by ® — 31.
(C) Let ® and ®x be abelian additive groups and let In : ® —* ®x
be a homomorphism. The set 31 of all elements of ® which map onto
the unit element of ®x is a subgroup of ®, called the kernel of h.
The homomorphism h can be interpreted as an isomorphism between
©/31 and ?1 (Exercise 8).
Suppose that © is defined by a set of generators (Av . . . , Am)
and a set of relations
(A.20) ??(??...,?„) = 0 (i=l,...,r)
where ??(?1, . . . , Am) is a sum of multiples klAi {kt e Z) of the
generators Лг. Let 31 be the subgroup of ® generated by a given set
of elements (Xv . . . , Xn), and let the Z's be expressed in the form
(A.21) X. = di(Av . . . , An) (i=l,...,n),
which is possible, since the ^'s generate ®.
Theorem 2. The factor group ® = ®/9? is isomorphic to the
additive group defined by generators (Av . . . , A m) with relations
(A.20) and
(A.22) 6i(Av...,AJ = 0 (г= 1,...,л).
* That is, if there exists an element in E which is not in E0.

GROUP-THEORETIC BACKGROUND
221
Proof. In the first place, if ? e 9?, then ? is expressible in terms
of the X's, and if ? ? 9?, then ? is not expressible in terms of the
X's (Exercise 9). The proof can now be completed with the aid of @)
(Exercise 10).
Theorem 3. The adjunction to a set of defining relations (A.20)
for an abelian group © of a set of relations (A.22) yields a set of defining
relations for a group isomorphic to ®/9?, where 9? is the subgroup of
© generated by the elements Xt = 6i(Av . . . , Am) (i = 1, . . . , n)
(Exercise 11).
EXERCISES
3. (a) Show that the mapping ? —> Zm defined by j —>> у is a homomorphism
(see Art. A-\(H) for notation), (b) Define an isomorphism between Zw and the
group defined in Exercise 1.
4. Show that
ki
b = em (k, m. are integers; i2 = —1),
with multiplication as operation, generates a group isomorphic to Zw if and only
if к and m are relatively prime.
5. Show that the multiples of an integer m constitute a subgroup of Z, with
addition, but that the odd numbers do not. Determine whether the odd numbers,
with multiplication, are a group.
6. Prove the first and last sentences in (A).
7. Complete the proof of Theorem 1 by verifying Additive Group Axioms
II-V.
8. Establish Statement (C).
9. Establish the first statement in the proof of Theorem 2.
10. Carry out the suggested completion of the proof of Theorem 2.
11. Prove Theorem 3.
A-3. The Structure of Finitely Generated Abelian Groups
(A) An abelian additive group g is the (direct) product*
(A.23) g = & ? 52 ? - - - ? gw
of the additive groups %v . . . , gw if A) & <z g (i = 1, . . . , n),
B) g. ? gy = 0 (i ? j), C) each element of g except 0 can be
expressed in one and only one way in the form Ai + A{ + · · · + Ain
where Ai. e 5y (j = 1, . . . , n). The product of a given set of abelian
additive groups g^ (i = 1, . . . , n) is generated by the aggregate of
the generators of the groups ge and has for a set of denning relations
the aggregate of a collection of defining relations for the respective
groups fo.
* Sometimes called the sum 5=5i+S2 + #,,+ Sn·

222
APPENDIX
(B) The factor group g/g^ is isomorphic to gx ? · · · ? g^ ?
дг+? ? * · * ? ?? (Exercise 12).
Let g be an abelian additive group with a finite set (Av . . . , An)
of generators. A general form for an element of g is then
?
(A.24) A = 2 xAi (яг an integer).
i = l
A group relation can be expressed thus:
(A.25) 2 3? = ° (Уг Ф ° for some *)¦
г = 1
(C) The free abelian group g with ?г generators (Av . . . , ij is
isomorphic to the additive group of integral vectors* in En (see also
Exercise 13). A basis for g is a subset of its elements in terms of which
each of its elements is uniquely expressible.
Theorem 4. Each basis of g consists of exactly ? elements.
Proof. The set (Av . . . , An) is a particular basis. Suppose
(Bv . . . , Bm) is also a basis. Then (see Exercise 14 for an example)
?
(a) Bi= 2 xi}A} (i=l,...,m),
(A.26) i_1
m
(b) As = 2 yjkBk (j = 1, . . . , n).
fc = l
Suppose m > n. The rankf of the matrix (xtj) cannot exceed ? (the
number of its columns) and is therefore less than the number m of
equations in the set (A.26a). Therefore the right side of at least one
of these equations, say the one for which г = r, is linearly dependent,
with rational coefficients, on the right sides of the other equations
in the set. But this means that, for some integer к Ф 0, kBr equals
a linear expression in the other B's. The element A = kBr of g is
then expressible in two different ways in terms of the B's, namely as
kBr and as the linear expression just mentioned, contradicting the
definition of (Bv . . . , Bm) as a basis. Hence m < n. By symmetry,
? < m, and hence m = n.
Corollary. Two finitely generated free abelian groups are
isomorphic if and only if they have the same number of generators
(Exercise 15).
The reader unfamiliar with matrix theory should study, in some
standard text, the definition of the product ? ? of two matrices,
* That is, vectors whose components are integers.
f We assume a basic familiarity with linear equations, with the use of determinants
in solving them, and also with some of the fundamental properties of matrices.

GROUP-THEORETIC BACKGROUND
223
where the number of columns equals the number of rows. He should
note that matrix multiplication is not commutative (Exercise 16)
and that it is associative; that is, X(YZ) = (XY)Z.
A one-row matrix is a row vector and a one-column matrix is a
column vector. The elements are the components of the vector. If
X is a row vector {xx . . . xn) and ? is a column vector* {yx . . . yn)',
then XY = YIl=1xiyi, while YX is the ? ? ? matrix (y^Xj) (i =
1, . . . , n\ j = 1, . . . , n).
If (Av . . . , An) and (Bv . . . , Bn) are two bases for the free
additive (abelian) group with ? generators, then, by (A.26) with
m = n,
(A.27) Bt = | 2 xijVjkBk (i = 1, . . . , n).
k=lj=l
By definition of basis, the right side reduces to B{, so that, using the
Kronecker delta 6ik,
? (\ if i = k
(A.28) J *иУ* =<*<*= (< = 1, . . . , л; 4 = 1, . . . , л).
i=i [0 \ii ? к
But this implies that the product of the determinants \xi5\ · \y}k\ is
|E.;| = 1; so that \xu\ = \yjk\ = +1 or —1, since them's, them's, and
hence the determinants are integers.
A square matrix is unimodular if its determinant is +1 or —1.
Lemma 1. The matrix of a change of basis in a free additive abelian
group g with ? generators is a unimodular matrix of order n.
Conversely, any unimodular matrix of order ? is the matrix of a change of
basis in g.
The first part of this lemma has just been proved. The second part
presents no difficulty.
Let 9? be a subgroup of g. Let Nv . . . , Nm be a set of generators
of 9?, not necessarily independent, and consider their expressions
(A.29) Nt= 2><A (г = 1,...,т)
in terms of the ^'s.
(D) The matrix (еи) of m rows and ? columns uniquely determines
the N{, hence 9?, and hence the group
(A.30) g = g/ji.
The same g would be determined by any coefficient matrix
corresponding to different choices of generators for g and 9^.
* The symbol for a matrix, modified by a prime, will denote its transpose.

224
APPENDIX
This suggests the concept of an equivalence class of matrices,
two m ? ? matrices being equivalent if and only if each of them is
the matrix of a set of relations of the form (A.29) expressing some set
of m generators of У1 in terms of some basis of g.
Lemma 2. Two matrices (etj) and (e^) belong to the same equivalence
class, as just defined, if
(A.31) «„) = (yik)(eki)(xih)
where (угк) and (xJh) are unimodular matrices of orders m and ?
respectively.
Proof of Lemma. Lemma 2 is a consequence of Lemma 1 applied
A) to a change of basis
(A.32) A,= | xjhBh (j=i,...,w)
/i = l
in g and B) to a change of generators
m
(A.33) Mt=XyJ!fk (»=l,...,n)
k = l
in 5R. The matrix (e-h) is then the matrix of the expression of the
generators Mt of 9? in terms of the generators Bh of g.
(E) By multiplying (eu) on the right by a unimoduJar matrix of
order n, any of the following effects can be achieved: (a) the
multiplication of a column by —1, (b) the interchange of columns j and k,
(c) the replacement of column к by column к plus an integral multiple
of column j (j ? к) (Exercise 17). None of these operations affects
the rank of the matrix. Since we can also suppose (еи) multiplied on
the left by the identity matrix (??}) of order m without affecting the
result, we see that the operations mentioned are of the form (A.31),
and hence transform (ei}) into an equivalent matrix.
(F) Statement (E) holds for {ei3) multiplied on the left by a
unimodular matrix of order m and "columns" replaced by "rows"
throughout.
Theorem 5. Let ? be the rank of the matrix (еи) of equations
(A.29). Then (ea) is equivalent to an m ? ? matrix whose elements
are all zeros except for the first ? elements on the main diagonal, which
are positive integers (?1? ?2, . . . , ??) such that тг+1 divides ri (i =
l,...,y- 1).
Proof. If each of the eiS equals zero, then 9? contains only the null
element of g. Hence ^ = g and (etJ) is already in the specified form.
We turn to the case where at least one ei} is not zero.

GROUP-THEORETIC BACKGROUND
225
(G) In proving Theorem 5, we lose no generality by assuming for
(еа) any set of properties possessed by an equivalent matrix. We will
use this principle to justify some convenient assumptions.
Assumption I. en > 0 and en is a minimum among the absolute
values of the non-zero terms of (ei5).
Justification. Let egh be a term of minimum absolute value among
the non-zero terms of (e{j). If egh < 0, change signs throughout row g.
Next interchange rows 1 and g, then columns 1 and h. This leads to a
matrix equivalent to (еи) and satisfying Assumption I.
The greatest common divisor of a set of integers is the largest integer
which divides all of them.
Lemma 3. Either en divides each element eb- or there exists an
equivalent matrix (e'u) containing an element e'st such that 0 < e[t < en.
Case 1. The first row of (еи) contains an element elh not divisible by
€n. Let jen be the largest integral multiple, positive or negative, of en
such that elh — jen > 0. Then,
(A.34) elh = jen + e'lh where e'lh e A, 2, . . . , en — 1)
Hence, in this case, the replacement (see (E)) of column h by
(col h) — j(col 1) leads to an equivalent matrix containing a positive
element e[h < en.
Case 2. The first column contains an element ehl not divisible by en.
This case is treated like Case 1, with rows replacing columns.
Case 3. The first row and column are both made up of multiples of
elv but there exists an element est not divisible by en. By adding a
suitable positive or negative multiple of row 1 to row s, we obtain an
equivalent matrix in which the element in column 1 and row s is 0.
Let the new row s be added to row 1 to obtain a new row 1, in which
en is unchanged and a term of the form (est + &en), with к an integer,
appears in row 1. This reduces Case 3 to Case 2.
Assumption II. en divides every element e^.
Justification. If en does not divide all the eiP let Lemma 3 be
applied to obtain an equivalent matrix (е'и) with some positive
element less than en. By Assumption I, this element can be assumed
to be e'u. If (eij) does not satisfy Assumption II, there similarly exists
an equivalent matrix (elf) with en > e'u > e\x > 0. This process can
be repeated only a finite number of times, since the number of integers
between en and 0 is finite. Hence there exists a matrix (eff) equivalent
to (егД where e^ > 0 and efx divides every efp and the greatest
common divisor of the elements of (e*) is the same as that of (еи).
We are now ready to complete the proof of Theorem 5. By
subtracting suitable multiples of the first column from the other columns,

226
APPENDIX
then multiples of the first row from the other rows, let (e^) be reduced
to the form
0 0 ... 0
(A.35)
" e22 e23
0 e32 633
where en divides all the other elements.
By operations involving only columns 2 to ? and rows 2 to m,
following the procedure just used to reduce {еи) to (A.35), let (e^) be
next reduced to the form
(A.36)
«u
0
0
0
^22
0
0 .
0 .
esa
.. 0
.. 0
.. ея
Sn
0 0
"m3
where e22 is the greatest common divisor of the terms other than en in
(A.35). Hence e22 is divisible by elv
Continuing recurrently the process whereby (eu) was reduced first
to (A.35) and then to (A.36), we reduce (eu) to a matrix in which all
elements are zero, save the first ? elements on the main diagonal,
which are positive integers elv . . . , eyy where eu divides ei+1 i+1 (i =
1, . . . , ? — 1). The non-zero elements must be ? in number, since the
operations we have used (see (E) and (F)) preserve rank. By suitable
interchanges of rows and of columns, we can reverse the order of the
e's, then let eH = r.._i+v to obtain the form specified in Theorem 5,
which we will call the normal form of (ea). It will be shown (see proof
of Theorem 8 below) that the numbers (??5 . . . , т.,). called the
invariant factors of (егД are independent of the particular way in
which (ea) was reduced to normal form.
(H) By Lemma 2, (еи) can be transformed into normal form by
multiplying on the right by a unimodular matrix X = (xjh) of order ?
and on the left by a unimodular matrix ? = (yki) of order m, where
X and ? are the matrices of appropriate changes of generators in the
groups g and 5R respectively. Let (Bl9 . . . , Bn) be the new basis of g.

GROUP-THEORETIC BACKGROUND 227
Then, since the transformed matrix is the matrix of the expression for
the new generators of У1 in terms of the B's, and any generator equal to
zero can obviously be dropped, the new generators of У1 are (r1Bv . . . ,
rvBv).
Two elements
(A.37) A = f ?.,?{, ?=2 ?,??
г: = 1 г: = 1
of g belong to the same coset of g mod 9^ if and only if their difference,
defined as the first plus the negative of the second, belongs to 9^; that
is, if and only if, for some integers (Я15 . . . , ??),
(A.38) | (ft - а,)Я, = i ?,?,?,.
г=1 г=1
Since the B's are a basis, (A.38) holds if and only if
(A.39) A ~ Ki = V< {? = 1-··' ^
ft — а,- = 0 (j= ? + 1, . . . ,n).
Since the only essential property of the A's is that they are integers,
their precise values being irrelevant, we arrive at the following
statement.
Two elements В and A of g belong to the same coset of g mod У1
if and only if, in the notation of (A.37),
(a) a, = ft mod т,· (г = 1, . . . , у),
(A.40) Pt ?
(b) a, = ft (j = у + 1, . . . , n).
The t's are a sequence of positive integers each divisible by the
next. Hence ?, = 1 implies rj+1 = · · · = ?? = 1. This means that
there exists a number ? @ < ? < ?) such that
тг > 1 (г = 1, . . . , ?),
(?.41) r
tj = ! 0' = ? + !> · · · > 7)·
Every pair of integers a, /8 satisfies ol = ? mod 1. Since the last
? — ? conditions in (A.40a) accordingly impose no restriction on the
oc's and fts, (A.40) is equivalent to
A af = ft mod ?, (? = 1, . . . , ? < у),
<*i = ft U= у + 1, · . · , w).
This proves the following result.
Lemma 4. Let the symbol for an element С of gf, modified by a bar,
denote the coset G_of С in g* mod % so that G is a typical element of
3f/5R. Then each С can be expressed in exactly one way in the form
(A.43) ? = ? uiEi + ? viSi (see (я) for the B's)
г = 1 i-y+1

228
APPENDIX
where
щ e Zr. (integers mod tJ (i = 1, . . . , p),
(A.44)
Vj eZ ' (integers).
Theorem 6. If g is an abelian group with a finite number of
generators Al9 . . . , Am9 then g is isomorphic to some factor group
? = ?/^> where g is a free abelian group.
Outline of Proof. In particular, let g be the free abelian group with
m generators Av . . . , Am. Let each element ??=1 aiAi of g be mapped
onto the element ?™=1 а{А{. The resulting map is a homomorphism of
g onto g. The elements of g which map onto the null element of g
constitute a subgroup 9? of g. The elements of g are images of the
cosets of g mod 9?, and g is isomorphic to g/5R (Exercise 18).
An element A of an additive group, abelian or not, is of order ?,
if ? is the smallest positive integer such that ? A = 0, the null element
of A. If no such integer exists, A is of infinite order. A similar
definition of order applies to multiplicative groups, with AT = 1
replacing ? A = 0.
Theorem 7. A finitely generated abelian group g is the product of
A) a number ? > 0 of free cyclic groups and B) a number ? > 0 of
finite cyclic groups of orders ??5 . . . , ?? where тг+1 divides т{ (г =
1, ... , ? — 1) and all the t's exceed 1.
This theorem is a consequence of Theorem 6 and Lemma 4, with
? = ? — ?.
(I) The number ? is the difference, /? = ? — у, between the number,
/г, of columns of (ег,·) and the rank, y, of (егу). It is called the Betti
number of g. The numbers тг are, as mentioned above, the invariant
factors of (e^) exceeding 1. They are called the torsion coefficients of
g. The terminology is motivated by the geometric significance of
these numbers in topological applications.
Theorem 8. Two finitely generated abelian groups are isomorphic
if and only if they have the same Betti numbers and torsion coefficients.
Proof. From Theorem 7 it follows that the Betti number and the
torsion coefficients determine the structure of a finitely generated
abelian group g. It remains to prove the following.
(J) The numbers ? and (??5 . . . , ??) of Theorem 7 are uniquely
determined by g; in other words, if g is expressed in a different way as
the product of ?' free cyclic groups and ? finite cyclic groups of orders
t[, . . . , r'p> where r'i+1 divides т\ (г = 1, . . . , ?' — 1) and ?'?> > 1, then
(?.45) (a) ?' = ? (b) ?' = ? (с) т\ = ?, (г = 1, . . . , ?).
For simplicity, we replace By+j hyCJ(j = l,..., ?) in the notation
of Lemma 4, and denote A) with (Gl5 . . . , 0?) a set of generators of

GROUP-THEORETIC BACKGROUND
229
the free cyclic groups of Theorem 7 and B) with (Bv . . . , Bp) a set of
generators of the cyclic groups of orders (??5 . . . , ??). We similarly
denote with (G[, . . . , 0'?>, B[, . . . , Bp>) the generators of a second
expression for g, as suggested in (J).
We now adapt an argument from Veblen's book ([V, pp. 129-130]).
(K) Condition (A.45) is equivalent to the following: For each
integer r, the number of generators in {G, B} = (Gl5 . . . , 6?, Bl9 . . . ,
Bp) each of order >r equals the number of generators in {C\ B'} =
(C'l9 . . . , C'p, B'l9 . . . , Bp.) each of order >r.
To verify (K), note first that ? and ?' are the numbers of generators
in [G, B) and {C, B'} respectively of orders > max (??5 ?[). The rest of
the argument is relegated to Exercise 19.
Assume the theorem false. Then {G\ B'} exists, as described above,
such that for some r, the number of generators in {G\ B'} of orders
>r is unequal to (and we assume it greater than) the number of
generators in {G, B} of orders greater than r. The respective generators
of orders exceeding r are as follows, for some к and k':
(A 46) (a) °19···' CP> Bv->Sk (Tk > r, rk+1 < r)
(b) C'v . . . , fy, S'v . . . , B'k, (r'k, > r, ??41 < r)
where ?' + к' > ? + к.
Because {В, С) are generators, we have
(A.47)
? ?
г=1 i=l
J = l
(А = 1,.
(t = 1, . .
..,л
.,*')
«м е ?.
wtt e ?,
¦ «?< G Z>
There are ?' + k' > ? + к equations in the set (A.47). Hence there
exists a set of ?' + к' integers (rl5 . . . , r^, sl5 . . . , %), not all zero,
such that, if we multiply the equations (A.47) by these respective
integers and add the results, the coefficients of (Gv ...,^, Bv ..., Bk)
all reduce to zero; that is,
(A.48) 2 '& + ? «Л = I (i v« + I ^щ)в}.
We can so select the numbers (rl5 . . . , ?y, sl5 . . . , s^) that 1 is the
largest integer dividing all of them. Since the order of Bj divides
?
k+l
(j = к + 1, . . . , ?),
(A.49) § fk+1rhC'h + | fk+1stB't = 0.
Since g is the direct product of the groups generated by the C'h, of
infinite order, and the B[, it follows that, in (A.49), fk+1rh = 0, hence
rh = 0 (A = 1, . . . , /?'); and тл+1^ = О mod f't since f/ is the order

230
APPENDIX
of B\(t = 1, . . . , k'). But this implies fk+1st = 0 modf^ > r, since f^
divides f't (t = 1, . . . , k'). Since all the rh's are zero, the s/s are not
all zero and they have 1 as highest common factor. Therefore (fk+1sl9
. . . , fk+1sk,) has fk+1 < r as highest common factor, contradicting the
set of relations fk+1st = 0 mod т'к> > r (t = 1, . . . , k'). Hence the
theorem is true.
EXERCISES
12. Prove (B). Suggestion: First discuss the cosets mod 5*of an element of 5y·
13. Show that a free abelian group with ? generators is the product of ? free
cyclic groups.
14. Show that ?? = 4AX - 6A2, В
abelian additive group generated by ?? and A2.
15. Prove the corollary to Theorem 4.
16. Find the product matrix XY and, if defined, YX if
3?? — 4A 2 is a basis for the free
(a) X
2
0
3
1
5
7
-2
4
0
1
1
1
(b) X
(c) X
110 0 0 0
0 0 1110
0 0 0 0 0 1
(using mod 2
operations);
17. Prove all parts of (E) and of (F).
18. Write out the outlined proof of Theorem 6.
19. Complete the proof of (K). Suggestion: Consider the implications as r
decreases from max (r1? rj[) to 1.
A-4. Integral Modules, Contravariant and Covariant
Components
The homology theory of a finite complex К is concerned with
certain factor groups of abelian groups. All these factor groups are

GROUP-THEORETIC BACKGROUND
231
defined (Chapter 5) in terms of the so-called chain groups (?л of К
(к = О, 1, 2, . . .), where &k is the free abelian group generated by the
oriented &-simplexes of K. The development of homology theory
involves changes of basis in (?fc. As background material, we discuss
some of the properties of a free abelian group with a specified initial
basis.
By the integral module [s] = [sv . . . , sa] we mean the free abelian
group with basis {s} = (sl5 . . . , sa) and with integral coefficients. The
particular basis {s} is called the initial basis of [s]. The situation is
much the same as when we work with a euclidean space R having a
specified rectangular cartesian coordinate system and introduce other
coordinate systems, skew or rectangular, by transformations of
coordinates. Indeed, we can interpret [s] as the additive group of
integral vectors in R*.
The scalar product (C · D) of two elements
(A.50) (a) 0 = 2 cfs^ (b) D = 2 b% (a\ 6* integers),
i = 1 г = 1
of the integral module [s] is the integer
(A.51) (G-D)= 2 ??.
i = l
If the initial basis is interpreted as an orthogonal set of unit vectors
in Ra, then (C · D) equals the product of the lengths of С and D by the
cosine of the angle between them. Thus (C · D) = 0 if and only if the
vectors С and D are perpendicular or if one of them is zero.
For our purposes, however, it is more relevant to note that the
scalar product (C · D) has important interpretations when С and D are
appropriate elements of certain chain groups (see Art. 5-9, for example).
(^4) The scalar product is obviously symmetric, (C · D) = (D · G),
and linear, (G · (Dx + D2)) = (C · DJ + (G · D2), (Сг +С2)- D =
(Ox · D) + (C2 · D).
According to Theorem 4, any basis for [s] has just ? elements. Let
Щ = (il5 . . . , ta) be an arbitrary basis for [s], and consider the
expression
(A.52) 0=2 c%
г = 1
for an arbitrary element С е [s] in terms of {t}. The coefficients
(c1, . . . , ca) are called the contravariant components of С relative to {t}.
They are merely the coordinates of the terminal point of С relative to
the basis {t}, when С is interpreted as a vector with initial point at the
origin. They are referred to as contravariant because of the way they
are transformed corresponding to a change of basis, in comparison

232 APPENDIX
with the transformation of the covariant components of G relative to
{?}, the latter being defined as the numbers
(A.53) Cj = (<7.*y) = (*,.<?).
Note the use of superscripts and subscripts to distinguish between
covariant and contravariant components.
Relative to the initial basis {s}, the contravariant and covariant
components of a vector
(A.54) G = b\ + · · · + b\
are the same, since (G · s{) = b\ In order to study their behavior
under changes of basis, consider a typical such change ti = ? aHsb
which is expressible also in the following matrix form:
? ? 1\ /4 ' ^
(A.55) | · I = A [ · I with inverse | · J = A'1
According to Lemma 1, A is unimodular.
The vector G can accordingly be formulated as follows:
(A.56) G = (b1. .. ba) I · = F1. .. ?^-1 ?
Comparing (A.52), we find (c1 . . . ca) = F1 . . . ba)A l or, using a prime
to indicate the transpose of a matrix, and using the fact that (AB)' =
B'A\
Л lbi\
(A.57) \= (A-1)
On the other hand, from (A.53), (A.55), and (A.54) we obtain Cj =
? au? or
? /?
(A.58) I · | = A I · J = A

GROUP-THEORETIC BACKGROUND
233
Thus the transformation carrying the covariant vectors relative to {s}
into those relative to {t} has exactly the same form as the
transformation {s} —> {t}, which justifies the term covariant. The contra-
variant vectors transform in accordance with the quite difFerent
formula (A.57).
It now follows that if {t} is carried into another basis as follows:
(A.59)
В
= ??
then the covariant and contravariant components (dl5 . . . , da) and
(d1, . . . , da) of the vector С appearing in (A.54) are given by
K\
= В
= ??
(A.60)
К
???
= (?-1) ? ? = (В-1) (A-1)
\dal \c*
From (A.58) and (A.57) we infer that
Ir \ IA
I \ I »
(A.61) I I = AA \
w w
(jB) It is obvious that a vector is determined by its covariant, as well
as by its contravariant, components relative to a given basis.
A-5. Dual Bases in a Module
A pair of bases {u} = (uv . . . , uj and {v} = (v^ . . . , va) for the
module [s] are dual to one another if their scalar products satisfy the
conditions
(A.62) (ut · v,) = du.
The elements иг and vi are called corresponding elements of {u} and {v}.

234
APPENDIX
Theorem 9. There exists a unique basis {v} dual to any given basis
{u} of [s].
Proof. Let the transformation {u} —* {s} be expressed by
1, . . . , ?).
a
Hence, if {и} and {v} are dual, (vt · sy) = ?^=1 bjh6ih = Ън.
(A) Conversely, the duality of {u} and {v} is implied by the
equations (v{ · Sj) = Ън (Exercise 20). But this is equivalent to the set
of equations
(A.65) v,= J 6УЛ (* = !>···>")>
J = l
which establishes the theorem and yields the following result.
Corollary. The basis {v} dual to {u} is obtained from the initial
basis {s} by the transformation {s} —> {v} whose matrix is the transpose
of the inverse of the matrix of the transformation {s} -> {u}. (Compare
(A.63).)
A geometric interpretation of dual bases is illuminating. Let the
initial basis {s} be regarded as a set of mutually orthogonal unit
vectors in Ra.
(B) Condition (A.62) means that A) the vector Vj is normal to the
(a — l)-plane Щ'1 determined by the vectors ut (i ? j), B) the length
of the vector vi is the reciprocal of the length of the projection of ui on
v^ and C) the angle between tij and vi is acute.
It is easy to verify, either geometrically or algebraically, that the
only self-dual bases of [s] are A) those of the form {sxsl9 . . . , ??#?}
where ег = +1 or —1 and B) transformations of them with
orthogonal matrices.
Consider an element
(A.66) S = аг8г + · · · + aasa
of the module [s]. Expressed in terms of {u}, it is
(A.67) S = i I afiihuh9
h = lj=l
so that its contravariant components relative to {a} are
(A.68) ch= J ajbih (A=l, . . . , ?).
3 = 1
(A.63) 8j = ? bihuh (j =
Л = 1
Then
(A.64) (vr8,)= L- J bjhuh) =

GROUP-THEORETIC BACKGROUND 235
Its covariant components relative to {v} are, by definition,
(A.69) dh=(S'Vh)=(f ajSj · i bkA) = | afiih = c\
This proves the following result.
Theorem 10. If {u} and {v} are dual bases of [s]9 then the contra-
variant components of s e [s] relative to {u} equal its covariant
components relative to {г;}.
Next consider a pair of modules [s] = [sv . . . , sa] and [w] =
[wv . . . , ???] and a pair of homomorphisms
(a) f:[8]-+[w],
(?·7°) /? r ? ?
(b) g:[w]-+[8].
Then / and g are called dual to each other if, for each pair of elements
S ? [s] and W ? [w], the following relationship holds between scalar
products:
(A.71) (f(S)-W) = (S-g(W)).
Theorem 11. Let {t}9 {u} be dual bases of [s] and let {x}9 {y} be dual
bases of [w]. Let/ and g be dual homomorphisms of [s] into [w] and of
[w] into [s]. Then if
(A.72) /(*<) = ? ?<? (i=l,...,oc),
3 = 1
it follows that
(A.73) д(у,) = 1*цщ (i=i,...,a
i = l
Proof. The covariant components of дг(г/у) relative to {?} are
(еи, . . . , eaj), since
(A.74) A7(%)-д = («л-17(У,))
= (ЛУ · У,) by (A.71)
= ?, ehk\Xk ' Vj) ~ Z, ehk°kj = ehj
k=l k=l
because {x} and {y} are dual bases. The covariant components of
?"=1 еищ relative to {t} are also (elh . . . , еаД since
= 6;
? еаиг - кI = ? euiui' h) = ? eiA
г=1 / г=1 г=1
Equations (A.74), (A.75), and Art. А-ЦВ) imply Eqs. (A.73) and also
imply the following result.
Corollary. The ith covariant component of д(у^ relative to {t}
equals the jth contravariant component of/(^) relative to {x}.

236
APPENDIX
(С) The material in this article and the preceding article can be
modified to the case where the integral coefficients are replaced by
integers mod 2.
EXERCISE
20. Prove Statement (A). See Art. А-ЦА).

Index of Symbols
Sets
{}0un=><=E 41
X (product) 45
Mappings
-> \ 43
~ 152 , 187
/_1 (exponent —1) . . . . 43
/ | A (vertical bar) .... 42
Spaces
X (product) 53
A (bar , for A <z S) . . . . 53
d (for A c £) 53
diam 57
dim 62
.E7 50
r(p0» >->Pk) 66
MPo> >- , Pk) 66
Ak 70
(£ , 93) 52
(S , p) 57
(S , {o}) 51
Complexes
< > (for simplexes) .... 75
afc 74
h (for surfaces) 33
K 74
51 76 , 78
M 111
P = \K\ 76
n = |#| 76
q (for surfaces) 33
r (for surfaces) 33
s 72
S 78
a 76
St 135 , 142
I 141
Groups
~ 96
~ 189
~ 195
*/> 124
« 153
~ 187
X 221
{Ak , Bk , Ck) 118
93fc 97 , 118 , 176
93* 132
pk 97 , 176
Ck 92
Ck 120
dk 92 , 118 , 176
(£fc 121
ak 124
Dfc 118 , 176
3>j? 132
d 123
d 92 , 94
h 176
d 121
dk 95 , 118 , 176
g* 132
S* 121
gfc 124
$>k 95 , 118 , 176
S*(n , p) 199
Sfc* 132
k 121
$>k 124
Ju 114
J* 115
Jk 120
N 118 , 177
237

238 INDEX OF SYMBOLS
Groups (Continued)
O 190
Q 216
R 216
Xk 97 , 118 , 176
X* 132
r\ 97 , 176
Z , Zm 216
3k 94 , 118 , 176
3 , * 132
3* 121
3k 124
I 141
Special Coefficient Groups
Z 95
Z2 95
Numerical
$ij 68
Implication
=><=<> 41

General Index
Abelian; see Group
Adjacent , 112
Affine transformation , 68
Aggregate , 41
Alexander duality theorem , 180
Approximating complex , 143
Approximating simplex , 143
Arc , 186
Arithmetical continuum , 48
Barycenter , 73 , 83
Barycentric coordinates , 70
Barycentric subdivision , 83
Base
Betti , 119
connectivity , 122
homology , 119
of a topology , 52
torsion , 119
Basis , 222
Betti
base , 119
group , 97
singular , 132
number , 7 , 97 , 228
relative (or mod L) , 176
Bolzano-Weierstrass theorem , 59
Boundary
chain , 92
complex , 100
face , 77
mod 2 , 121
of a chain , 94
of a set , 53
of a simplex , 72
of a surface , 15
point , 53
simplex , 72
Bounded , 57
Brouwer dimension , 62 , 136
Brouwer's fixed-point theorem , 137
Carrier , 133
Cell
abstract , 78
dimension of , 78
boundary of , 87
chain , 165
complex , 165
face of , 87
homology , 165
polyhedral , 87
Chain , 92
boundary of , 92 , 94
cell , 165
coboundary , 122
group , 92 , 95
of a complex , 112
singular , 130
equality of , 130
Characteristic , 109 , 118
mod L , 177
Circuit , 6
Class , 41
equivalence , 44
of mappings , 152
residue , 218
Closed interval , 48
Closed mapping , 55
Closed set , 53
Closed surface , 15
Closure , 53
Coboundary , 123
chain , 122
operator , 122
Cocycle , 123
Coherently oriented , 93 , 112
Cohomology
class , 124
group , 124
singular , 131
Cohomology-independent , 125
Collection , 41
Column vector , 223
Combinatorial equivalence , 83 , 88
Combinatorial homotopy , 195
Compact , 59
Compactum , 59
Complement , 42
Complex , 74
abstract , 77
denumerable , 77
dimension of , 77
finite , 74 , 77
realization of , 78
boundary , 100
cell , 165
cone , 154
239

240
GENERAL INDEX
Complex {Continued)
floor , 145
generalized , 78
infinite , 75
isomorphic , 80
linear , 77
of a chain , 142
oriented , 92
polyhedral , 87
subdivision of , 87
roof , 145
singular , 142
topological , 76
Component , 76 , 77
Components
contra variant , 231
covariant , 232
Composition of mappings , 43
Cone complex , 154
Congruent mod m , 218
Congruent mod 9t , 220
Connected , 76 , 77
arc wise , 187
simply , 191
Connected graph , 4
Connected surface , 15
Connectivity
base , 122
group , 121
number , 121
Continuity , 51
Continuous , 49 , 50 , 51
at a point , 54
on a space , 54
Continuous curve , 185
Continuum , 48
Converge , 54
Convex , 66
Coordinates
affine , 68
barycentric , 70
rectangular cartesian , 68
skew , 68
Coset , 220
Countable , 44
Cover , 205 , 209
Covering , 59
closed , 59
^-covering , 61
infinite , 205
fc-fold , 205
of mesh ^ , 61
open , 59
order of , 62
simple , 64
Covering complex , universal , 210
Covering pair , 177
Covering path , 209
Covering set , 205
Covering space , 205
Crosscap , 22
Curve
closed , 186
simple , 186
Peano , 186
Cycle , 94
group , 94 , 95
Cyclomatic number , 7
Decomposition , 45
Deformation , 152 , 187
homotopic , 151
path , 151
Denumerable , 43
Diagonal , 136
Diameter , 57
Dimension , 65
Brouwer , 62 , 136
number , 65 , 67
numerical , 67
Disjoint , 42
Distance
between two sets , 57
euclidean , 50
Domain , 42
Dual bases , 233
Dual homomorphisms , 235
Duality theorem
Alexander , 180
Lefschetz , 179
Poincare , 173
Edge , 19
free , 19
paired , 19
path , 193
Empty , 41
Equivalence
class , 44
combinatorial , 83 , 88
relation , 44
Euclidean space , 48 , 50
Euler-Poincare characteristic , 109 , 118
Euler's theorem , 12
Extension , 42
Face , 77
boundary , 77
of a simplex , 72
Factor group , 220
Fixed point , 136
Fixed-point theorem , 137
Floor complex , 145
Function; see Mapping
Fundamental group , 185 , 189 , 190 , 196

GENERAL INDEX
Genus , 109
Graph , 4
linear , 4
connected , 4
of a function , 51 , 136
of a relation , 45
Group
abelian , 215
addition , 215
additive , 215
axioms , 215
basis , 222
Betti , 97 , 176
relative (or mod L) , 176
singular , 132
bounding cycle , 95 , 166
mod L , 176
chain , 92 , 95
mod L , 176
singular , 131
cobounding cocycle , 166
cocycle , 166
cohomology , 124 , 166
commutative , 215
connectivity , 121
cycle , 94 , 95 , 165 , 166
mod 2 , 121
relative (or mod L) , 176
cyclic , 218
of order m , 218
difference , 220
factor , 220
finitely generated , 216
free abelian , 217
free cyclic , 218
free multiplicative , 217
fundamental , 185 , 187 , 189 , 190
combinatorial , 196
generators , 216
homology , 96 , 166
local , 159
relative (or mod L) , 176
singular , 132
invariant , 219
inverse element , 215
made abelian , 196
multiplication , 215
multiplicative , 215
Poincare; see Group , fundamental
product , direct , 221
quotient , 220
relations , 217
relative cycle , 176
structure , 219
torsion
relative (or mod L) , 176
singular , 132
unit element , 215
Half-space , 86 , 111
closed , 86
open , 86
Handle , 25
Heine-Borel theorem , 60
Homeomorphic , 54
Homeomorphism , 54 , 55
linear , 69
Homologous , 96
Homology
base , 119
cell , 165
class , 96
group , 96
singular , 132
relative , 175
singular , 131
Homology-independent , 97
Homomorphism , 219
dual , 235
Homotopic
chains , 152
complexes , 152
Homotopy , 152
class , 152 , 189 , 196
combinatorial , 195
Image , 42
Incidence , 77
matrix , 114
normal , 115
number , 114 , 165
relation , 77
Incident , 17 , 75
Independent
cohomology , 125
homology , 97
linearly , 71 , 97
Induced homeomorphism , 81
Induced mapping , 82
Induced orientation , 93
Induced topology , 53
Inner point , 53 , 72
Integral module , 231
basis , 231
initial , 231
Intersection , 41 , 42
Interval
closed , 48
half-open , 49
open , 48
Inverse image , 43
Inverse mapping , 43
Isomorphic complexes , 80
Isomorphism , 80 , 219
Jordan-Brouwer theorem , 182
Jordan theorem , 182

242
GENERAL INDEX
Kernel , 220
Klein bottle , 38
Kronecker delta , 68 , 223
Lebesgue number , 64
Lebesgue's lemma , 64
Lefschetz duality theorem , 179
Limit , 49 , 51 , 54
Limit point , 53
Linear complex , 77
Linear simplex , 76
Linear space , 66
Linearly independent , 71 , 97
Local homology group , 159
Manifold , 18 , 162
homology , 162
relative (or mod A) , 179
topological , 162
Map , 42
Mapping , 42
closed , 55
continuous , 49 , 50 , 54
extension of , 42
into , 42
inverse , 43
one-to-one , 43
onto , 42
open , 55
restriction of , 42
simplicial , 81 , 82
Matrix , normal incidence , 115
Mesh , 61 , 86
Metric , 56
space , 57
Module , integral , 231
dual bases , 233
Moebius strip , 25
Neighborhood , 52
spherical , 56
Non-orientable , 26
Non-orientable pseudomanifold , 112
Null , 41
Null-homotopic , 189
One-sided , 26
One-to-one correspondence , 43
One-to-one mapping , 43
Open interval , 48
Open mapping , 55
Open set , 49 , 51
Operator
boundary , 165
mod L , 176
coboundary , 165
subdivision , 141
Opposite faces , 73
Order , 75
of an element , 228
partial , 75
Orientable , 27
Orientable pseudomanifold , 112
Orientation
coherent , 93
induced , 93
like , 111
negative , 111
of E» , 111
positive , 111
similar , 111
unlike , 111
Oriented , coherently , 93 , 112
Oriented complex , 92
Oriented simplex , 90
singular , 129
Partition , 45
Path , 4 , 185
closed , 186
edge , 193
equality , 185
homotopy , 187
product , 186
simple , 186
closed , 186
Piecewise linear , 81
Plane , 70
Poincare duality theorem , 173
Polygonal symbol , 19
Polyhedral cell , 87
Polyhedral complex , 87
subdivision of , 87
Polyhedron , 75
simplicial , 77
topological , 76
Prism , 144
floor of , 144
roof of , 144
Prism complex , 145
Product , 43
cartesian , 45
projection of , 45
of mappings , 43
topological , 53
Projective plane , 21
Projective space , 155
Pseudomanifold
closed , 111
coherently oriented , 112
mod L , 177
non-orientable , 112
orientable , 112
relative (or mod L) , 177
topological , 163
with boundary , 111

GENERAL INDEX
Quotient group , 220
Range , 42
Realization , 78
Relation , 44
domain of , 44
equivalence , 44
graph of , 45
range of , 44
reflexive , 44
symmetric , 44
transitive , 44
Relations
defining , 218
dependent , 218
group , 217
independent , 218
Residue class , 218
Restriction , 42
Roof complex , 145
Row vector , 223
Segment , 48 , 66
Set , 41
closed , 53
open , 49 , 51
Simplex
abstract , 77
boundary face of , 77
face of , 77
boundary , 72
closed , 72
face of , 72
inner points of , 72
linear , 76
open , 72
oriented , 90
singular , 129
boundary chain of , 131
boundary faces of , 131
closed , 129
equality of , 129
faces of , 131
linearity on , 129
oriented , 129
topological , 76
Simplicial mapping , 81 , 82
Simplicial polyhedron , 77
Simplicial subdivision , 88
Simply connected , 191
Singleton , 42
Singular chain , 130
Singular cohomology , 131
Singular complex , 142
Singular homology , 131
Singular simplex , 129
closed , 129
Skeleton , 76 , 88
Space
compact , 59
euclidean , 48 , 50
oriented , 111
Hausdorff , 55
linear , 66
metric , 57
metric on , 57
metrizable , 57
of mappings , 61
projective , 155
topological , 51
Sperner's lemma , 134
Sphere , 21
homology , 164
topological , 154
unit , 154
with crosscaps , 21 , 29
with crosscaps and contours , 32
with handles , 29
with handles and contours , 32
Star , 135 , 142 , 154
closed , 154
center of , 154
outer boundary of , 154
Star-related , 142
Subcomplex , 75
Subdivision
barycentric , 83
minimal central , 88
operator , 141
simplicial , 88
Subgroup , 219
proper , 220
Subset , 42
proper , 42
Surface , 15 , 18
boundary of , 15
closed , 15
connected , 15
non-orientable , 26
one-sided , 26
orientable , 27
Surrounding complex , 159
Topological complex , 76
Topological polyhedron , 76
Topological property , 54
Topological simplex , 76
Topological space , 51
Topology , 52 , 54
base of , 52
discrete , 52
induced , 53 , 57
by a metric , 57
trivial , 52
Torsion
base , 119

244
GENERAL INDEX
Torsion (Continued)
coefficient , 97 , 116 , 228
group , 97
singular , 132
number , relative (or mod L) ,
176
Torus , 25
Tree , 6
Triangulation , 75 , 76
Triple of related chains , 147
Unbounded , 57
Unicursal , 4
Unimodular , 223
Union , 41 , 42
Unit sphere , 154
Vacuous , 41
Vector , 68 , 223
components , 223
Void , 41