homology lecture notes

8/10/2019 Homology Lecture Notes

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Lecture Notes on Homology Theory

Dr. Thomas Baird (illustrations by Nasser Heydari)

Winter 2014

Contents

1 Introduction 2

2 Review of Point-Set Topology 52.1 New spaces from old. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.2 Connectedness and Path-Connectedness . . . . . . . . . . . . . . . . . . . 82.3 Covers and Compactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.4 Metric spaces and the Lebesgue number lemma . . . . . . . . . . . . . . . 92.5 Hausdorff spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

3 Singular Homology 103.1 Simplices. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103.2 Chains, cycles, and boundaries. . . . . . . . . . . . . . . . . . . . . . . . . 11

3.2.1 Homology as a functor . . . . . . . . . . . . . . . . . . . . . . . . . 163.3 Homotopy Invariance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

3.3.1 Chain complexes and chain homotopy. . . . . . . . . . . . . . . . . 173.3.2 The prism operator . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

3.4 Relative Homology and the long exact homology sequence . . . . . . . . . 213.4.1 Reduced Homology . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

3.5 Excision . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253.5.1 Proof of Excision . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273.5.2 Mapping cylinders and cones. . . . . . . . . . . . . . . . . . . . . . 31

3.6 Applications to spheres: the degree of a map . . . . . . . . . . . . . . . . . 33

3.7 Cellular homology. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383.7.1 Cell complexes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383.7.2 Cell complex propaganda (not to be tested) . . . . . . . . . . . . . 403.7.3 Cellular Homology . . . . . . . . . . . . . . . . . . . . . . . . . . . 413.7.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

3.8 Mayer-Vietoris Sequence . . . . . . . . . . . . . . . . . . . . . . . . . . . . 463.9 Homology with coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

3.9.1 The Universal Coefficient Theorem for Homology . . . . . . . . . . 523.10 Covering spaces and the transfer. . . . . . . . . . . . . . . . . . . . . . . . 54

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4 Cohomology 564.1 The cup product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 604.2 The Kunneth formula. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

4.3 Manifolds and orientations . . . . . . . . . . . . . . . . . . . . . . . . . . . 684.4 The cap product and Poincare Duality . . . . . . . . . . . . . . . . . . . . 71

1 Introduction

Topology is the study of topological spaces ( e.g. subsets ofRn) and continuous mapsbetween them. The basic idea of algebraic topology is to study functors F fromtopological spaces to groups (or some other type of algebraic category). This means isthat for every topological space X, we assign a groupF(X), and to each continuous mapf :XY, we assign a group homomorphism F(f) :F(X) F(Y) such that

F(f g) =F(f) F(g)

for any pair of composable maps

X gY

fZ

and also that identity maps are sent to identity maps:

F(IdX) =I dF(X)

To see how this sort of thing may be useful, observe that if two spaces X and Yare isomorphic (i.e. homeomorphic), then F(X) andF(Y) must be isomorphic for every

functorF. It turns out that the most powerful way to prove that two spaces X, Y arenothomeomorphic is to find a functor such that F(X) and F(Y) are not isomorphic.

For another application, we begin with a definition. A subset A X is called aretract, if there exists a continuous map r : XA such that f(a) =a for all a A. 1.Inclusion of sets defines an injective map i: A X. IfA is a retract in X, then thereexistsr such that

r i= I dA.

For any functor, this means that

F(r) F(i) =F(r i) =F(IdA) =I dF(A).

In particular, this means that F(i) must be injective when A X is a retract (if not,F(r) F(i) = IdF(A) would not be injective, a contradiction). Using this idea, we willprove that the unit circle S1 is not a retract inside the unit disk D2.

The kinds of functors we will learn about in this course are the (singular) homologyand cohomology functors. These functors come in families labelled by non-negativeintegers called the degree (also called dimension): H0, H1, H2, ... for homology andH0, H1, H2, .... for cohomology. Both homology and cohomology take values inabeliangroups, though we will also study variations that take values in vector spaces.

1For example, the inclusion R R2 as the x-axis is a retract using the map r(x, y) = (x, 0)

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The historical motivation for homology theory came from vector calculus.2 Recall thatthere are various versions of the Fundamental Theorem of Calculus ( Stokes Theorem,Greens Theorem, the Divergence Theorem) that equate an integral over a manifold (curve,

surface, solid, etc.) with an integral over its boundary (set of points, a curve, surface,respectively). Homology emerged from efforts to understand how many independentsubmanifolds there are with respect to a given domain. Roughly speaking, the 0- homologyH0(X) is generated by points in X, the 1-homology H1(X) is generated by (oriented)closed curves inX, the 2-homology is generated by (oriented) closed surfaces, and so on.The homology class is trivial if the curve, surface, etc. is the boundary of a surface, solid,etc..

To see how this might work, consider the disconnected subset X R2 pictured inFigure1.

Figure 1: A space Xwith two path components

A point p in one component cannot be joined by a continuous path to a point q inanother component. It follows that p and qdetermine different elements [p] and [q] inH0(X) . We will show that there is an isomorphism H0(X)= Zn wheren is the numberof path-components ofX.

Consider now a annulus A in R2 (Figure2).The closed loop Crepresents an element in H1(A). It is intuitively clear thatS

1 isnot the boundary of a surface in A, so Crepresents a non-trivial element [C] in H1(A).Indeed, we will show that H1(A) = Z and that Crepresents one of the generators (theother generator is obtained by reversing the orientation on C). On the other hand, ifwe take a union of C with a curve D that winds around the annulus in the oppositedirection, we see that together they form the boundary of a surface (Figure 3). In termsof homology, this will mean that [C+D] = [C]+[D] = 0, or equivalently that [C] =[D].

2The development of homology theory is usually attributed to Poincare in the 1890s, though the sub-ject didnt really come into its own until the 1930s through the work of numerous other mathematicians.

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Figure 2: Loop in an annulus

Figure 3: Two loop in an annulus bound a surface

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There are many different kinds of homology that are defined in different ways. Theapproach we will take in this course is called singular homology. Singular homology hassome great theoretical advantages over others (such as simplicial homology and cellular

homology), but has the draw back of being difficult to calculate directly. Indeed, itwill take some time before we establish the fact that oriented submanifolds determinehomology classes which has been the basis of todays lecture.3 Instead, singular homologyis based on singular simplices which we will be introduced in ...

2 Review of Point-Set Topology

In this section we collect some basic facts from general topology that will be required inthis course. Proofs of these results can be found in any introductory textbook on generaltopology (e.g. Munkres Topology) or in the point-set topology notes I have posted on

D2L.Definition 1. Atopological space(or simplyspace) (X, ) is a setXand a collectionof subsets ofX, called the open sets, satisfying the following conditions:

i) and Xare open,ii) Any union of open sets is open,iii) Any finite intersection of open sets is open.

A set is calledclosedif its complement is open. Usually, we will denote the topologicalspace (X, ) simply byX.

Example 1 (Euclidean Topology). An open ball in Rn is a set of the form

B=B(p) :={x Rn| ||x p||< }

for some p Rn and > 0. A subset U Rn is called open if it is a union of openballs. Equivalently, U is open if for every p U, there exists an open ball B such that

p B U.

In the example above, we say that open balls form a basisfor the Euclidean topology.More generally, a collection of open sets B in a topological space X is called a basis ifevery other open set in Xis a union of sets in B.

Definition 2. A continuous map f : X Y between topological spaces is a map ofsets for which pre-images of open sets are open. I.e.

UY is open f1(U) :={x X|f(x) U} X is open

Definition 3. A homeomorphism is a continuous bijection f : X Y such that theinverse f1 is also continuous. This is the notion of isomorphism for topological spaces.

3It is possible to define a kind of homology theory using oriented manifolds directly, called bordism.One of the reasons that this approach is not standard in introductory courses is that the theory ofmanifolds gets complicated in dimensions larger than two.

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We will only rarely need use the abstract Definition2explicitly. More often, we willmake use of certain properties of continuous functions, including the following.

Proposition 2.1. LetX, Y andZbe topological spaces.

The identity map IdX :XXis continuous.

Iff :XY andg :Y Zare continuous, then the compositiong f :XZis continuous.

Any constant map f :XY is continuous.

The first two conditions above make topological spaces + continuous maps into acategory. We will speak more about categories later.

2.1 New spaces from old.Most of the topological spaces we encounter in this course are constructed fromRn usingthe operations below.

Definition 4. LetXbe a topological space and A Xa subset. Thesubspace topol-ogy on A is the topology for which V A is open if and only ifV = A U for someopen setU inX.

Example 2. Any subset ofRn acquires a subspace Euclidean topology. Unless otherwisestated, we will always assume subsets ofRn to have this topology.

Theinclusion mapi: A Xis continuous (with respect to the subspace topology).

In fact, we have the following special property: A map f : Y A from a topologicalspaceYis continuous if and only if the composition i f :Y X is continuous.

Definition 5. Theproduct spaceXYof two spacesXand Yis the Cartesian productof setsX Y, with abasisof open sets of the form U V whereUXandV Y areboth open.

The above definition iterates to define products of any finite number of spaces (infiniteproducts require a different definition).

Example 3. Then-fold product RR...R is homeomorphic to Rn with the Euclideantopology.

The key property of product spaces is that a map

F :ZX Y

is continuous if and only if the coordinate functions F = (F1, F2) are continuous as mapsfromZ toXand to Y respectively.

Definition 6. Let{X} be a (possibly infinite) collection of spaces indexed by . Thecoproduct space or disconnected union

X is the disjointunion of the sets X

withU

X is open if and only ifU X is open for all .

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The inclusions i0 : X0

X are all continuous. A map F :

X Y iscontinuous if and only if the composition F i : X Yare continuous for all .

Definition 7. An equivalence relation on a set X is a relation satisfying, for allx, y X

(i) x x

(ii) x y implies y x

(iii) x y and y z impliesx z.

Given any relation R on X, we can generate the smallest equivalence relation Rsuch that xRy implies xR y. Explicitly, we definexR y if and only if there exists afinite sequence{xiX}ni=0 forn 0 satisfying

x0= x,

xn= y and,

xiRxi1 orxi1Rxi for alli = 1,...,n.

Givenx X, the equivalence class ofx is

[x] :={y X|x y}

Notice that [x] = [y] if and only ifx y. The equivalence classes determine a partitionofX into disjoint sets. LetE := {[x]|x X} be the set of equivalence classes (we will

sometimes denote E=X/ ) . There is a canonical map

Q: XE, x[x]

called the quotient map.

Definition 8. LetXbe a topological space and let be an equivalence relation on theset underlying X. The quotient topology on E is the topology for which U E isopen if and only ifQ1(U) is open in X.

Observe that Q : X E is continuous and that a map f : E Y is continuous if

and only iff Q: XY is continuous.Example 4. SupposeXandYare topological spaces,A Xis a subspace, andf :AYis a continuous map. Define an equivalence relation on the coproduct X

Y generated

byf(a) a for all a A. We say that the quotient space (X

Y)/ is obtained byattachingX to Y along A using f.

Example 5. Suppose X is a topological space, and G is a group that acts on X viahomeomorphisms. Define two points in Xequivalent if they lie in the same orbit ofG.The quotient space in this case is called the orbit space and is denoted X/G.

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2.2 Connectedness and Path-Connectedness

LetIdenote the unit interval [0, 1] Rwith the Euclidean topology.

Definition 9. A space Xis calledpath-connectedif for any two points p, q Xthereexists a continuous map :IX such that(0) =p and (1) =q.

Definition 10. A space X is called connected if there is no proper subset A Xwhich is both openand closed. (proper means other thanXor, which are always bothopen and closed).

Observe that ifA Xis both open and closed, then the complement Ac is also bothopen and closed, and there is a natural isomorphism A

Ac =X. Thus spaces that are

not connected can be decomposed into a disconnected union of nonempty spaces.

Proposition 2.2. Path-connected spaces are connected.

The converse of Proposition 2.2 is not true in general. However all the connectedspaces we encounter in this course will also be path-connected.

Connectedness and path-connectedness are preserved under the following operations

A product of (path-)connected spaces is (path-)connected.

The continuous image of a (path-)connected space is (path-)connected.

Let{U} be a covering ofXsuch that each U is (path-)connected and the inter-sectionU is non-empty. Then X is (path-)connected.

2.3 Covers and Compactness

Definition 11. An open (closed) cover of a topological space Xis a collection of open(closed) sets{U}such that the union U = X.

Proposition 2.3. Let{U}be either an open cover or a finite closed cover ofX. A mapof sets

f :XY

between topological spaces is continuous if and only if the restrictionsf|U :U Y arecontinuous for all (whereU has the subspace topology).

The preceding proposition will be used in two ways: to test if a map fis continuous byconsidering the restrictions, and also to construct a map fby gluing together continuousmaps defined on the U that agree on overlaps.

Definition 12. A space X is called compact if every open cover {U} of X containsa finite subcover. I.e., there exists a finite collection {U1,...,Un} {U} such thatni=1Ui = X.

Proposition 2.4. A subspace ofRn is compact if and only if it is closed and bounded.

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Compactness is preserved under the following:

A closed subspace of a compact space is compact.

A finite union of compact spaces is compact.

A product of compact spaces is compact.

Iff :XYis continuous andXis compact, then the imagef(X) Y is compact.

2.4 Metric spaces and the Lebesgue number lemma

Definition 13. LetXbe a set. A metric on Xis a function

d: X X R0

called the distanceormetric function, satisfying

1. d(x, x) = 0 x = x (d separates points)

2. d(x, x) =d(x, x) (dis symmetric)

3. d(x, x) d(x, x) + d(x, x) (the triangle inequality)

A metric space (X, d) determines a metric topology on X, which is generated bythe basis of open balls B(p) ={x X|d(x, p)< }. IfA is a subset of a metric space XthenA becomes a metric space by restriction. The metric topology on A is the same as

the subspace topology onA.The following result will come up repeatedly.

Lemma 2.5 (Lebesgue number Lemma). LetA be an open covering of a compact metricspaceX. There exists >0, called theLebesgue number, such that for allp X, theopen ballB(p) is contained in someU A.

2.5 Hausdorff spaces

Definition 14. A space Xis called Hausdorff if for any pair of distinct points p, qX,there exist open sets U, V such thatp U, q V and U V =.

Proposition 2.6. Any metric space is Hausdorff. In particular, any subset of Rn isHausdorff.

The Hausdorff property is preserved under the following:

Products of Hausdorff spaces are Hausdorff.

Subspaces of Hausdorff spaces are Hausdorff.

Coproducts of Hausdorff spaces are Hausdorff.

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Definition 15. A space X is locally compact, if every point p Xis contained in anopen neighbourhood p Usuch that the closure U is compact.

For example, compact spaces are locally compact. Also, Rn is locally compact.

Proposition 2.7. Let Xbe a locally compact space and let Y be a Hausdorff space. If: XYis a continuous map that is also a bijection of sets, thenis a homeomorphism.

3 Singular Homology

3.1 Simplices

The standard q-simplexq is the simplex spanned by the zero vector e0 =0 and thestandard basis vectors e1,...,eq in R

q (Figure4). Thus,

q :={(t1,...,tn)| 0 for alli = 1,...,qand t1+ ...+ tq 1 }

Figure 4: The standard simplices

If X is a topological space, a singular q-simplex (or simply simplex)in X is a(continuous) map

: q X.

A singular 0-simplex in X is simply a point in X, a singular 1-simplex in X is acontinuous path in X, etc.. We can think of singular simplices as probes used to studythe space X.

Example 6. Letv0,...,vq be a set ofq+ 1-vectors in Rn for some n. Define

[v0,...,vq] : q Rn, (t1,...,tn)(1 t1 ... tq)v0+ t1v1+ ...+ tqvq.

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Example 7. Let1 and 2 be singular 2-simplices in X. Then21+ 32 S2(X) is a2-chain and

2(2

1+ 3

2) = 2

2(

1) + 3

2(

2)

= 2((0)1

(1)1 +

(2)1 ) + 3(

(0)2

(1)2 +

(2)2 )

= 2(0)1 + 2

(1)1 2

(2)1 + 3

(0)2 3

(1)2 + 3

(2)2

is a 1-chain in S1(X).

The boundary map can be understood schematically from Figure6, but be careful notto confuse singular simplices (which are maps) with their images (which are sets).

Figure 6: Boundary of simplices - intuition

Proposition 3.1. The compositionq1 q :Sq(X) Sq2(X) is the zero map. Drop-ping subscripts, we write this

2 = 0.

Proof. SinceSq(X) is generated by simplices, it suffices to check that q1 q() = 0 forall q-simplices .

It is an easy check that if 0 j < i q, the face maps satisfy

Fiq Fjq1 = F

jq F

i1q1.

Thus

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q1 q() = q1(

q

i=0

(1)i(i))

=

qi=0

(1)iq1( Fiq)

=

qi=0

(1)iq1j=0

(1)j( Fiq Fjq1)

=

qi=0

q1j=0

(1)i+j( Fiq Fjq1)

= 0ijq1

(1)i+j( Fiq Fjq1) +

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by Proposition3.1, Bq(X) is a subgroup ofZq(X). The qth degree singular homology ofXis the quotient group:

Hq(X) :=Zq(X)/Bq(X).

Example 8. The homology of a point. IfX = {pt} is a single point, then there isonly one singular simplex in each degree, which is the constant mapq : q {pt}. Thechain groups are

Cq({pt}) = Zq= Z.4

The boundary map satisfies (forq 1)

q(q) =

qi=0

(1)i(i)q

=

qi=0

(1)iq1

=

q1 ifq is even

0 ifqis odd

while0(0) = 0.It follows that forq 1

Zq({pt}) =Bq({pt}) =

0 ifqis even

Cq({pt}) ifqis odd

whileZ0({pt})= Z and B0({pt}) = 0. Thus

Hq({pt}) =Zq({pt})/Bq({pt}) =

0 ifq 1

Z ifq= 0

A space X for which Hq(X)=Hq({pt}) for all qis called acyclic meaning no cyclesthat are not also boundaries.

Proposition 3.2. Let{Xk} be the set of path components of a spaceX (indexed byk).Then

Hq(X) =kHq(Xk)for allq 0.

Proof. Because the standard q-simplex is path connected (indeed convex), the image ofa singular q-simplex : q X must be path connected and in particular must liewithin one of the path components of X. It follows that for all q we have a canonicaldecomposition,

Cq(X) =qCq(Xk).

4The case of a point is highly unusual in this respect. For most spaces Y, Cq(Y) has uncountablerank.

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Moreover, it is clear that the boundary map respects this decomposition, so that

Zq(X) =kZq(Xk)

andBq(X) =kBq(Xk)

and finally that

Hq(X) = Zq(X)/Bq(X)

= (kZq(Xk))/(kBq(Xk))

= k(Zq(Xk)/Bq(Xk))

= kHq(Xk)

We denote by 0(X) the set of path components of of a space X.

Proposition 3.3. There is a canonical isomorphism

H0(X) = Z0(X).

ThusH0(X)= Zn wheren is the number of path components ofX.

Proof. By Proposition3.2, it suffices to show that ifXis path connected, then there is acanonical isomorphism

H0(X) = Z.

Recall that a singular 0-simplex is the same thing as a point in X. ThusS0(X) =Z0(X) =

pX

Zp.

The standard one simplex 1 is equal to the unit interval [0, 1] R, so a singular 1-simplex is a continuous path in : [0, 1] X. Since Xis path-connected, for any twopoints p, q X, there exists a path such that (0) = p and (1) = q. Consequently,the boundary satisfies

() =(1) (0) =p q S0(X),

and thus

B0(X) =SpanZ{p q | p, q X} pZ

Zp.

Observe thatB0(X) is equal to the kernel of the homomorphism

:pZ

Zp Z

defined on generators by (p) = 1.It follows that descends to a homomorphism

H0(X) =Z0(X)/B0(X)= Z.

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3.2.1 Homology as a functor

Suppose that f : X Y is a continuous map. If is a q simplex for X, then thecomposition f is a q-simplex for Y. This defines a homomomorphism

Sq(f) :Sq(X) Sq(Y), Sq(f)(

a) =

af .

ClearlySq(IdX) =I dSq(X) andSq(f g) =Sq(f) Sq(g) for composable continuous mapsfandg. ThusSqis functor from topological spaces to abelian groups. It allows commuteswith the boundary map.

Lemma 3.4. q Sq(f) =Sq1(f) q.

Proof. It is enough to check for simplices.

Sq(f)() = (f )

=

qi=0

(1)i(f )(i)

=

qi=0

(1)if Fiq

=

qi=0

(1)if (i)

= Sq(f)(qi=0

(1)i(i))

= Sq(f)()

using associativity of composition.

It follows then thatSq(f) sendsZq(X) to Zq(Y) andBq(X) to Bq(Y) and thus inducesa homomorphism between the quotient groups

Hq(f) :Hq(X) Hq(Y).

It follows easily from the fact thatSqis a functor thatHq is a functor from topologicalspaces to abelian groups. It is common to use short hand

f = Hq(f)

though we will try to avoid doing so.

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3.3 Homotopy Invariance

Two continuous mapsf, g: XYare said to behomotopicif there exists a continuousmap

h: X IY

where I = [0, 1] is the unit inverval and both h(x, 0) = f(x) and h(x, 1) = g(x). Intu-itively, two maps are homotopic if one can be continuously deformed into the other.

The goal of this section is to prove the following theorem.

Theorem 3.5. Letf andg be homotopic maps fromX to Y. For allq 0, the inducedmaps on homology are equal: Hq(f) =Hq(g).

Two spacesXandYare calledhomotopy equivalentif there exist mapsf :XYand g : Y X such thatf g is homotopic toIdY andg f is homotopic I dX.

Corollary 3.6. IfXandYare homotopy equivalent, thenHq (X)=Hq (Y)for allq 0.

Proof. By Theorem3.5 and functoriality, we have

Hq(f) Hq(g) =Hq(f g) =Hq(IdY) =I dHq(Y)

and similarly Hq(g) Hq(f) = IdHq(X). The Hq(f) and Hq(g) are inverse isomorphismsbetween Hq(X) andHq(Y).

A space is called contractible if it is homotopy equivalent to a point. Examples ofcontractible spaces include all convex subspaces ofRn (exercise). By Corollary 3.6, a

contractible space X satisfies H

q

(X) = 0 for q 1 and H

0

(X) =Z

(i.e. contractiblespaces are acyclic).Before proving Theorem3.5,it will be helpful to introduce some abstract ideas about

chain complexes.

3.3.1 Chain complexes and chain homotopy

Achain complex (of abelian groups)

C := (Cq, q)qZ

is a sequence of abelian groups (Cq)qZ and homomorphisms q : Cq Cq1 such thatqq+1= 0 for all q Z.

q+3Cq+2

q+2Cq+1

q+1Cq

qCq1

q1Cq2

q2. . .

TypicallyCq = 0 for q


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We define Zq(C) = ker(q) and Bq(C) = im(q+1) and Hq(C) =Zq(C)/Bq(C), calledrespectively the q-chains,q-boundaries, andq-homology groups of the chain complex. IfzZq(C), denote by [z] Hq(C) the coset represented by z.

A morphism of chain complexes f : C C

is a sequence of homomorphisms(fq :Cq Cq)qZ that commutes with boundary maps: fq1q =

qfq for all q. In other

words, the following diagram commutes:

Cqq

fq

Cq1

fq1

CqqCq1

.

Example 10. Given a continuous map : X Y, the morphisms Sq() : Sq(X) Sq(Y) determine a chain morphism S() :S(X) S(Y).

A chain map f : C C induces a homomorphism in homology Hq(f) : Hq(C) Hq(C

) for all q Z by the same reasoning as in 3.2.1by the rule

Hq(f)([z]) = [fq(z)].

EachHq is a functor from chain complexes to abelian groups.Let f, g : C C be two chain maps. A chain homotopy between f and g is a

sequence of homomorphism (Pq :Cq Cq+1)qZ such that

q+1Pq+ Pq1q =fq gq.

Cq+1q+1

Cqq

fqgq

Pq

Cq1

Pq1

Cq+1 q+1Cq q

Cq1

.

The chain maps f and g are called chain homotopic if there exists a chain homotopybetween them.

Proposition 3.7. If chain maps f, g : C C

are chain homotopic, then the inducedmaps on homology are equal: Hq(f) =Hq(g) as maps fromHq(C) to Hq(C).

Proof. Let [z] Hq(C) be represented by z Zq(C). Then

fq(z) gq(z) =Pq1q(z) + q+1Pq(z) =q+1Pq(z)

is a boundary. Thus

Hq(f)([z]) Hq(g)([z]) = [fq(z)] [gq(z)] = [fq(z) gq(z)] = 0

so Hq(f)([z]) =Hq(g)([z]).

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3.3.2 The prism operator

Fort I= [0, 1], define

it: X X I, it(x) = (x, t).

Lemma 3.8. The two maps i0, i1 :XX I, determine chain homotopic chain mor-phismsS(i0) andS(i1).

Proof. The first step is to define a decomposition of q I Rn+1 into (q+ 1)-simplices.

Denote the vertices lying in q{0}byv0,...,vq and those lying in q {1}byw0,...,wq.For each i, the image of the affine simplex

[v0,...,vi, wi+1,....,wn] : q q I

can be thought of as the graph of a map from q toI, because composing the projectionq I q is the identity map. These graphs slice qI into the images of affine(q+ 1)-simplices

[v0,...,vi, wi,...,wq] : q+1 q I, i {0,...,q}.

Figure 8: Prism decomposition

For arbitraryX, define the prism operator Pq :Sq(X) Sq+1(X I) on simplicesby

Pq() :=

qi=0

(1)i( IdI) [v0,...,vi, wi,...,wq].

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This will be our chain homotopy.

q+1Pq() = q+1

qi=0

(1)i

( IdI) [v0,...,vi, wi,...,wq]

=

qi=0

ji

(1)i+j( IdI) [v0,.., vj ,...,vi, wi,...,wq]

+

qi=0

ji

(1)i+j+1( IdI) [v0,...,vi, wi, ...,wj...,wq]

On the other hand,

Pq1q() = Pq1(

qj=0

(1)j [e0,..., ej,...,eq])

=

qi=0

ji

(1)i+j( IdI) [v0,..,vi, wi,..., wj,...,wq].

Adding together, we get

q+1Pq() + Pq1q() =

qi=0

( IdI) [v0,..., vi, wi,...,wq]

qi=0

( IdI) [v0,...,vi, wi,...,wq]

= IdI [w0,...,wq] IdI [v0,...,vq]

= i1 i0 .

It follows thatq+1Pq+ Pq1q =Sq(i1) Sq(i0).

Proof of Theorem3.5. A homotopy between two mapsf, g: XYis a maph : XIYsuch thatf=hi0andg= hi1. By Lemma3.8,Sq(i0) andSq(i1) are chain homotopic,so Proposition3.7 implies Hq(i0) =Hq(i1). Finally we see that

Hq(f) =Hq(h) Hq(i0) =Hq(h) Hq(i0) =Hq(g).

In fact, it is not hard to show that S(f) and S(g) are chain homotopic via the chainhomotopySq+1(H) Pq.

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3.4 Relative Homology and the long exact homology sequence

A topological pair (X, A) consists of a space X and a subspace A X. A pair (X, A)gives rise to an inclusion of chain groups Sq(A) Sq(X) (technically the inclusion map

i: A Xdetermines injective homomorphismSq(i)). Define therelative chain groupof the pair to be the quotient group

Sq(X, A) :=Sq(X)/Sq(A).

The relative chain groups combine to form the relative chain complex

. . . S q+1(X, A) q+1Sq(X, A)

qSq1(X, A)

q1. . .

where the boundary map is defined by the following commutative diagram

Sq(X)

qSq1(X)

Sq(X, A)

qSq1(X, A)

where the vertical arrows are quotient maps. Note that qis well defined becauseqsends

Sq(A) to Sq1(A) and that 2

= 0 because 2 = 0. It follows that we can define relativecycles, relative boundaries, and relative homology as described in3.3.1, which are denoted

Zq(X, A), Bq(X, A), Hq(X, A)

respectively. Geometrically, a relative cycle inZq(X, A) is represented by a chain inZq(X)whose boundary lands in Sq1(A).

Remark 1. Observe that ifA= , then Sq(A) = 0 for all q. It follows thatSq(X, ) =Sq(X) and that Hq(X, ) = Hq(X). Thus it is possible to think of homology as just aspecial case of relative homology.

Amap of topological pairs

f : (X, A) (X, A)

is a continuous map f : X X such that f(A) A. Such a map determines a

morphism of chain complexesS(f) :S(X, A) S(X, A) and thus also a homomorphismon homology

Hq(f) :Hq(X, A) Hq(X, A).

The following properties are proven similarly to their counterparts for Hq(X).

Hqis functor from topological pairs to Abelian groups. I.e. Hq(f)Hq(g) =Hq(fg)andHq(Id(X,A)) =I dHq(X,A).

If {Xk} is the set of path components of X and Ak = A Xk, then there is acanonical isomorphismHq(X, A) = k

Hq(Xk, Ak).

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Leth: XIX be a homotopy between maps map such thatht(a) :=h(a, t) A

for all a A and t I. Then Hq(h0) =Hq(h1) as homomorphisms from Hq(X, A)toHq(X

, A).

The quotient morphisms Sq(X) Sq(X, A) fit together into a morphism of chaincomplexes j : S(X) S(X, A). Combined with inclusion chain morphism i : S(A) S(X) we get a commutative diagram

. . . Sq+1(A)

i

Sq(A)

i

Sq1(A)

i

. . .

. . . Sq+1(X)

j

Sq(X)

j

Sq1(X)

j

. . .

. . . Sq+1(X, A) Sq(X, A) Sq1(X, A) . . .

. (1)

By functoriality, these chain morphisms give rise to homology homomorphisms Hq(A)Hq(X) Hq(X, A) for all q0. The most important property of relative homology isthe existence of a connecting homomorphism

Hq(X, A) Hq1(A),

(denoted by using abuse of notation). Let Hq(X, A), be represented by a cyclej(z) Sq(X, A) which is the image of a chain z Sq(X). Then (z) must land inSq(A) Sq(X), because (j(z)) =j((z)) = 0 Sq(X, A) =Sq(X)/Sq(A). Define

([z]) = [(z)] Hq1(A).

(We leave it as an exercise prove that this homomorphism is well defined and does notdepend on the representatives z or z.). The connecting homomorphism permits us toextend to a long sequence of homomorphisms

Hq+1(X, A) Hq(A)

Hq(i) Hq(X)

Hq(j) Hq(X, A)

Hq1(A) . . . (2)

By convention, this sequence ends at H0(X) H0(X, A) 0.

Definition 16. A sequence of abelian groups and homomorphisms

A fB

gC

is calledexact at B if ker(g) = im(f).

Theorem 3.9. The sequence (3) is exact (meaning exact at all groups in the sequence).It is called the long exact homology sequence associated to the pair(X, A).

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Proof. We will only prove exactness at Hq(X, A) and leave the rest as an exercise. Con-sider

Hq(X)Hq(j)

Hq(X, A) Hq1(A)

If im(Hq(j)), that means we can choose a representative cycle z Cq(X) suchthat [j(z)] =. Since zis a cycle, it follows that (z) = 0, so

() = [(z)] = [0] = 0 Hq1(A).

Therefore im(Hq(j)) ker().Conversely, suppose that ker(). This means that we can choose a chainz Sq(X)

such that [j(z)] = and (z) =() for some Sq(A). Consequently,z is a cyclein Sq(X) and

Hq(j)([z ]) = [j(z )] = [j(z)] [j()] = [j(z)] =.

So ker() im(i).

Proofs like the one above are called a diagram chase, because they amount to chasingelements through a diagram like (1).

Denote by0(X, A) the set of path components ofXthat do not intersect A.

Proposition 3.10. There is a canonical isomorphism

H0(X, A)= Z0(X, A).

In particular, H0(X, A) = Zm wherem is the number of path components ofX that do

not intersectA.

Proof. Leti : A Xdenote the inclusion ofA intoX. Then we have an exact sequence

H0(A)H0(i)

H0(X)H0(j)

H0(X, A) 0.

Exactness implies thatH0(X, A)=H0(X)/im(H0(i)). We know (Prop3.3) thatH0(X) =Z0(X) and the image ofH0(i) is generated by those path components ofXthat containa path component ofA. The result follows.

One of the great merits of the long exact homology sequence is that it is functorial

with respect to maps of pairs.Proposition 3.11. Letf : (X, A) (Y, B) be a map of pairs. This induces homomor-phisms on homology such that the following diagram commutes.

. . . Hq+1(X, A)

f

Hq(A)

f

Hq(X)

f

Hq(X, A)

f

Hq1(A)

f

. . .

. . . Hq+1(Y, B) Hq(B) Hq(Y) Hq(Y, B)

Hq1(B) . . .

Proof. We leave verification up to the reader.

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The following lemma is helpful for calculations:

Lemma 3.12 (The Five Lemma). Consider a commutative diagram of abelian groups

A1

A2

A3

A4

A5

B1 B2 B3 B4 B5

where the rows are exact. If, , andare all isomorphisms, thenis also an isomor-phism.

Proof. Diagram chase.

Corollary 3.13. If f : (X, A) (Y, B) is a map of pairs such that two out of three

families of induced maps on homology{Hq(A) Hq(B)}qZ,

{Hq(X) Hq(Y)}qZ.

{Hq(X, A) Hq(Y, B)}qZ,

are isomorphisms in all degree, then the remaining family is isomorphisms in all degrees.

Proof. Simply apply the Five Lemma to the diagram:

. . . Hq+1(X, A)

f

Hq(A)

f

Hq(X)

f

Hq(X, A)

f

Hq1(A)

f

. . .

. . . Hq+1(Y, B) Hq(B) Hq(Y) Hq(Y, B)

Hq1(B) . . .

Example 11. If a map of pairs f : (X, A) (Y, B) restricts to homotopy equivalencesbetween X and Y and between A and B, then f : Hq(X, A) Hq(Y, B) is an isomor-phism in all degrees.

3.4.1 Reduced Homology

It is sometimes convenient to use a modified version of singular homology called reduced homologFor any space X, there exists a unique map to a point : X {pt}. Define the reducedhomology

Hq(X) := ker Hq().

It is an easy consequence of (8) that ifXhas n path components,

Hq(X)=

Hq(X) ifq 1

Zn1 ifq= 0

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More canonically, H0(X) is the kernel of the map Z0(X) Zthat sends each generatorto 1. For relative homology we define

Hq(X, A) =H

q(X, A)

ifA= is non-empty. Basically, reduced homology is designed so that Hq({pt}) = 0 forall degrees without exception and this sometimes makes calculations less clumsy.

Functoriality, homotopy invariance, and the long exact sequence all work for relativehomology. In particular, ifA is non-empty, then we have a long exact sequence

Hq+1(X, A) Hq(A)Hq(X)Hq(X, A)

Hq1(A) . . . (3)

Remark 2. If X is a path-connected space and p X, then the long exact sequencedefines a natural isomorphism

Hq(X, p) = Hq(X, p)= Hq(X)

in all degrees.

3.5 Excision

The last property we need before we can do calculations is called excision. Given anordered pair (X, A) we say a subspace ofB A can be excised if the inclusion (XB, A B)(X, A) induces isomorphisms

Hq(X B, A B)=Hq(X, A)

in all degrees q.

Theorem 3.14. If the closure of B is contained in the interior of A, then B can beexcised.

Corollary 3.15. IfV B A and

1. V can be excised, and

2. the inclusion(X B, A B)(X V, A V) determine homotopy equivalencesX B X V andA B A V,

thenB can be excised.

Proof. We want to prove that Hq(XB, AB) Hq(X, A) is an isomorphism. Byfunctoriality, it is enough to show that the homomorphismsHq(X B, A B) Hq(XV, AV) andHq(XV, AV) Hq(X, A) are isomorphisms. The first is an isomorphismby homotopy invariance (Example11) and the second is an isomorphism because V canbe excised.

We postpone the proof so that we can (finally!) do some actual calculations.

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Proposition 3.16. The homology groups of the unit sphereSn forn 1 satisfy

Hq(Sn) =

Z ifq= 0 orq= n

0 otherwiseProof. It will be more convenient to work with reduced homology, so our goal is to provethat

Hq(Sn) =

Z ifq= n

0 otherwise.

LetEn+and Endenote the upper and lower closed hemispheres ofS

n. Note that forn 1,En+ E

n

=Sn1. I claim that by an excision that

Hq(Sn, En)

= Hq(En+, S

n1), q Z, n 1.

Here I am excising the interior of the lower hemisphere En. This does not satisfy thehypotheses of Theorem3.14,but a slightly smaller open disk does and then I can applyCorollary3.15.

Now consider the long exact sequences (LES) associated to these pairs. Because En+is contractible, the LES of the pair (En+, S

n1) breaks into isomorphisms

0 Hq(En+, S

n1) Hq1(S

n1) 0

for all n 1 and all q Z. Likewise, the LES of (Sn, En) gives rise to isomorphisms

0 Hq(Sn)

Hq(S

n, En) 0

for all n 1 and all q Z. Combined, we obtain isomorphisms

Hq(Sn)= Hq1(S

n1)

for all n 1 and all q Z.SinceS0 is a disconnected union of two points, it follows that

Hq(S0) =

Z ifq= 0

0 otherwise.

The result now follows by induction.

Proposition 3.16hints at the special role that spheres play in homology. Later, we

will consider a special class of spaces built out of spheres called cell-complexes that areparticularly well suited to algebraic topology.

Theorem 3.17 (Brouwer Fixed Point Theorem). Letf :Dn Dn be a continuous mapfrom the closedn-diskDn to itself. There existsp Dn such thatf(p) =p.

Proof. For the sake of contradiction, suppose that no such p exists. Thenf(x) = x forall x Dn and we can define a continuous map r : Dn Sn1 as illustrated in Figure9

Notice that for points x Sn1, r(x) = x. This implies that r is retract. In par-ticular, this means r :Hn(D

n) Hn(Sn1) is surjective which contradicts the fact that

Hn1(Sn1)= ZandHn1(Dn) = 0.

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Figure 9: Brouwer retraction

3.5.1 Proof of Excision

We begin with some notation. LetV := {Vi} be an open cover of a space X. Definethe chain group SVq(X) to be the subgroup ofSq(X) generated by simplices whose imagelands entirely within someVi V. The proof of excision boils down to showing that everychain inSq(X) can be replaced - up to a boundary - by one in S

Vq(X), where the relevant

open cover is {X\ B,int(A)}. 5

The idea is to subdivide simplices into unions of simplices through a process called

barycentric subdivision. Iterating the process results in smaller and smaller simplicesuntil each one is contained in some set in the cover Vvia a Lebesgue number argument.

First we define barycentric subdivision for affine simplices. If [v0,...,vq] is an affinesimplex in Rn, the barycentre is the point

b= 1

q+ 1v0+ ...+

1

q+ 1vq.

The (geometric) barycentric subdivision of [v0,...,vq] is defined inductively to be the unionof simplices [b, w0,...,wq1] where [w0,...,wq1] arises in the barycentric subdivision of aface [v0,..., vi,...,vq].

Recall that the diameterof a set A Rn equalssup{|x y| | x, y A}.

Lemma 3.18. The diameter of (the image of) any simplex arising in the barycentricsubdivision of[v0,...,vq] is no greater than

qq+1 times the diameter of[v0,...,vq].

Proof. The distance between any two points v and

tivi in (the image of) the simplex[v0,...,vq] satisfies

|v

tivi|= |

ti(v vi)|

ti|v vi| max{|vi v|}.

5In fact, since the boundary operatorsendsSVq

(X) to SVq1(X), it actually forms a sub-chain complex

SV(X). Hatcher proves that the inclusionSV(X) S(X) is a chain homotopy equivalence and thusinduces an isomorphism on homology.

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Figure 10: Barycentric subdivision

becauseti0 andt0+...+tq = 1. Therefore the diameter of an affine simplex [v0,...,vq] isequal to the maximum distance between vertices max{|vi vj|}. Now let [b, w0,...,wq1]be a simplex in the subdivision of [v0,...,vq]. By induction on q, we may assume that foranyi, j 0,...,q 1, that

|wi wj| diam([w0,...,wq1])q 1

q diam[v0,..., vk,...,vq])

q

q+ 1diam([v0,...,vq]).

On the other hand, if one of the vertices is b we have a bound

|b wj| max{|b vi| | i = 0,...,q}.

Notice that

b= 1

q+ 1vi+

q

q+ 1bi

wherebi is the barycentre of the face [v0, ..., vi,...,vq]. Thus

|vi b|= q

q+ 1|vi bi| diam([v0,...,vn]).

completing the calculation.

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A consequence of Lemma 3.18 is that repeatedly iterating barycentric subdivisionallows us to decompose a given simplex into simplices of arbitrarily small diameter.

We now turn this geometric construction into an algebraic one. We will define a chain

map: B: S(X) S(X)

which is chain homotopic to the identity map. This meansB induces the identity mapon homology. For any chainc Sq(X) we will show that B

k(c) SVq(X) for sufficientlylarge k and the theorem will follow pretty easily from that.

To begin, assume X is a convex subset ofRn and consider only chains lying in thesubcomplex Saff(X) S(X) generated by affine chains.

Let [v0,...,vq] be an affine simplex in (a subset of) Rn. Givenb Rn, define

b[v0,...,vq] = [b, v0,....,vn]

and extend linearly to a map Saffq (X) Saffq+1(X).Observe that

b[v0,...,vq] = [v0,...,vq] +

qi=0

(1)i+1[b, ..., vi,...,vq] = [v0,...,vq] b[v0,...,vq]

or more succinctly

b = b (4)

Next we define the chain map B on Saff(X) Saff(X) by induction on degree. For

an affine simplex , denote b its barycentre. For zero simplex B0() = . Forq 1,defineBq() =bBq1().

A little thought confirms that the affine simplices arising in the chain Bq() are exactlythose arising in the geometric barycentric subdivision above.

We wish to show thatBq1= Bq to establish thatB is a chain map. By induction,suppose thatBq2= Bq1. For an affine q-simplex we have

Bq() = bBq1

= Bq1 bBq1

= Bq1 bBq22

= Bq1

using induction, (4), and 2 = 0.Next, we construct a chain homotopy

{Tq :Saffq (X) S

affq+1(X)}qZ.

such thatT + T =I d B.

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Lemma 3.19. Given an open coverV ofX, and a chainc S(X), there existsk 0such thatBk(c) SV(X).

Proof. A chain is a finite formal sum of simplices and B distributes over sums. Thus itis enough to prove that some k works for each simplex and then the largest of these willwork for the chain.

Let : q Xbe a singular simplex (a continuous map). The preimages {1(Vi) | Vi

V} is an open cover of q. Therefore, there exists a Lebesgue number d > 0 such thatsubset of q of diameter less than d is contained in some open set

1(Vi) in the cover.Applying Lemma3.18, by iterating barycentric subdivision a finite number k times, allsimplices in the decomposition are made to have diameter less than d. It follows thatBk() SV(X).

Proof of Theorem3.14. We are attempting to show that the map Hq(X\ B, A \ B) Hq(X, A) is an isomorphism.

To prove surjectivity we must show that every element ofHq(X, A) is represented bya chain Sq(X) contained in the subgroup Sq(X\ B). Let Hq(X, A) be representedbyz Sq(X). By Lemma3.19, we can choose k such that Bk(z) also represents , andevery simplex occurring in Bk(z) maps into either X\ B or A. Discard those simpliceslanding inAto get a chain c Sq(X\ B) representing.

To prove injectivity, suppose Hq(X\ B, A \ B) maps to 0 Hq(X, A). This meansthatis represented by z Sq(X\ B) and there exists Sq+1(X) such that

() =z+

where Sq(A). Choose k so that the simplices ofBk() are contained in one ofX\ B

orA. Decompose Bk() =1+ 2 where 1S(X\ B) and2 S(A). Then

1+ 2= Bk() =Bk=Bk(z) +Bk()

thus(1) B

k(z) =(2) +Bk()

lies in bothSq(X\B) andSq(A), so must lie inSq(A\B). It follows thatBk(z) Sk(X\B)represents and that Bk(z) = (1) + where = ((2) + Bk()) Sq(A \ B), soBk(z) is a relative boundary in Sq(X\ B, A \ B), and we conclude that = 0.

3.5.2 Mapping cylinders and cones

A subspaceA Xis called adeformation retractif there is a homotopyh: XIAsuch thath(x, 0) =x and h(x, 1) A for allx Xandh(a, t) =afor alla Aand t I.The inclusioni : A X is a homotopy equivalence (this was a homework problem).

Aclosed subspaceA Xis called aneighbourhood deformation retractif thereexists an open neighbourhood A UX such thatA is a deformation retract ofU.

Example 12. The inclusion ofS1 R2 is a neighbourhood deformation retract becauseit includes as a deformation retract into an open annulus.

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Proposition 3.20. Let A X be a neighbourhood deformation retract that intersectsevery path component ofX. Then there is a canonical isomorphism

Hq(X, A)= Hq(X/A)

whereX/A is quotient space ofXobtained by identifying all ofA to a point.

Proof. Since A U is a homotopy equivalence, we know by Example 11 that

Hq(X, A)=Hq(X, U)

for all q. By excision (Theorem3.14)

Hq(X A, U A)=Hq(X, U).

On the other hand, if we denote by A/Athe point inX/AthatA is collapsed to, it is nothard to see thatU /Adeformation retracts onto A/A. Thus

Hq(X/A)=Hq(X/A, A/A)=Hq((X/A) (A/A), (U/A) (A/A))=Hq(X A, U A),

where the first isomorphism follows from Remark2since X/A is path connected.

Example 13. The quotient spaceDn/Dn is homeomorphic to the sphereSn. It followsthat

Hq(Dn, Dn)= Hq(S

n)

which was basically what we used in the proof of Proposition 3.16.

Example 14. If X contains a contractible neighbourhood deformation retraction A,thenHq(X)=Hq(X/A) for all q. (Indeed, one may show that XX/Ais a homotopyequivalence.

The hypotheses of Proposition3.20hold in many situations, but not always, so it isconvenient to have a construction that works in general. Let f :Y Xbe a continuousmap. The mapping cylinder associated to fis the quotient space (or adjunction)

Cyl(f) := (Y I) fX= ((Y I) X)/

where the relation is generated by (y, 1) f(y) for ally Y. The inclusionX Cyl(f)is a homotopy equivalence with homotopy inverse C yl(f)X xx and (y, t)f(x).

Themapping coneoff :Y Xis the quotient space

Cone(f) :=Cyl(f)/(Y {0})

Proposition 3.21. Given any map of spaces f : Y X such that the image of fintersects every path component ofX, we can define a long exact sequence in homology

Hq+1(Cone(f)) Hq(Y) Hq(X) Hq(Cone(f)) Hq1(Y) ...

in casei : Y Xis a subspace inclusion, this is canonically isomorphic to the long exactsequence of the pair

Hq+1(X, Y) Hq(Y) Hq(X) Hq(X, Y) Hq1(Y) ...

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Proof. The subspace Y {0}is a closed subset ofC yl(f) and is a deformation retract ofthe open subsetY [0, 1), soY {0}is a neighbourhood deformation retract in C yl(f).Thus by ... we have canonical isomorphisms

Hq(Cone(f))= Hq(Cyl(f), Y {0}).

SinceY {0}is homotopy equivalent to Y andC yl(f) is homotopy equivalent to X, thelong exact sequence ... can be obtained from the LES of the pair (Cyl(f), Y {0}) byreplacing groups with isomorphic groups.

In casef :Y Xis subspace inclusion, then the homotopy equivalenceC yl(f) Xsending (y, t) Y I to f(y) restricts to a homeomorphism from Y {0} to Y. Theresulting morphism of long exact sequences

Hq(Y {0})

Hq(Cyl(f))

Hq(Cyl(f), Y {0})

Hq1(Y {0})

Hq1(Cyl(f))

Hq(Y) Hq(X) Hq(X, Y) Hq1(Y) Hq1(X)

which must be an isomorphism by the Five Lemma.

Example 15. The wedge sum Let Xk be a collection of spaces containing base pointspk Xk . The wedge product is the space

kXk = (kXk)/

where we identify basepoints pi pj for all i, j. If the base points neighbourhood defor-mation retracts then

H(kXk)=kH(Xk)

by Proposition3.20.

3.6 Applications to spheres: the degree of a map

Recall that our calculation ofHn(Sn) relied on the following sequence of isomorphisms

Hn(Sn)

= Hn(Sn, En) Hn(E

n+, S

n1)=

= Hn(Sn1)

We can use this to construct a cycle representing the generator ofH1(S1) by the Figure

12Indeed, the cycle we have constructed is the barycentric subdivision of a simplex

: 1 S1 that winds once around the circle. It is not hard to show that (exercise) that

any chain of 1-simplices that wraps once around the circle also represents the generatorofH1(S

1).Recall that Hn(S

n) = Z. Given a continuous map f : Sn Sn, the induced mapf : Hn(S

n) Hn(Sn) must be of the form f() = d for some integer d Z. We calld= deg(f) the degree of the map f, .

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Figure 12: Constructing a cycle generating H1(S1)

Figure 13: Cycles generating H1(S1)

Since Hn is a functor, we see immediately that deg(IdSn) = 1, that deg(fg) =deg(f)deg(g) for two maps f, g:Sn Sn, and that homotopic maps are have the samedegree.6

Proposition 3.22. A map f :Sn Sn that is not surjective has degree zero.

Proof. Suppose p Sn is not in the image off. Thenf factors through the inclusionmapSn Sn \ {p} Sn so by functorialityHq(f) factors throughHq(Sn \ {p})=0 andthus must be zero.

Given a space X, define the suspension S X:=X I/ to be the quotient ofX Iwhere collapsesX {0}and X {1}to distinct points. Iff :XY is a map, definethe suspension off

Sf :SXSY, Sf (x, t) = (f(x), t).

This defines the suspension functorfrom spaces to spaces.

Lemma 3.23. The suspension of a sphere satisfiesSSn =Sn+1. Given a map f :Sn

Sn, the suspensionSf :Sn+1 Sn+1 satisfiesdeg(f) = deg(Sf).

Proof. The homeomorphism SSn = Sn+1 is pretty clear; this is the picture where Sn

includes intoSn+1 as the equator. Because the long exact homology sequence is functorialwith respect to pairs and the excision isomorphism is canonical, we obtain a commutative

6It is also true that two maps from Sn to Sn are homotopic if and only if they have the same degree.The proof of this is beyond the scope of this course.

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diagram of with horizontal arrows isomorphisms

Hn(Sn)

(Sf)

Hn(Sn, En)

(Sf)

Hn(En+, S

n1)

(Sf)

Hn(Sn1)

f

Hn(Sn) Hn(S

n, En) Hn(E

n+, S

n1) Hn(Sn1)

so S(f) and fhave the same degree.

Proposition 3.24. Let rn : Sn Sn be a restriction of a reflection onRn+1 Rn+1.

Thendeg(f) =1.

Proof. Forn 1, we can identifyrn= Srn1, so by induction it suffices to prove the casen= 0. In this case, S0 ={N, S} is a pair of points and r0 transposes them. The pointsrepresent 0-simplices and H0(S

0) is generated by [N] [S]. We have

(r0)([N] [S]) = [r0(N)] [r0(S)] = [S] [N] =([N] [S])

so deg(r0) =1.

We define the antipodal map on Sn byx x.

Proposition 3.25. Iff : Sn Sn is a map with no fixed points (i.e. there is no pointp Sn such that f(p) = p), then f is homotopic to the antipodal map. In particular,deg(f) = (1)n+1.

Proof. Iffhas no fixed points, then the path tf(x) (1 t)xdoes not pass through theorigin. It follows that

h: Sn ISn, ht(x) = tf(x) (1 t)x

|tf(x) (1 t)x|

is a homotopy joining the antipodal map h0 to f =h1. Finally, note that the antipodalmap is equal to a composition of (n +1) reflections onSn Rn+1 so it has degree (1)n+1

by Proposition3.24.

Theorem 3.26 (Hairy Ball Theorem). Every continuous vector field on an even dimen-sional sphere has a zero.

Proof. A continuous vector field on a Sn is equivalent to a mapV :Sn Rn+1 such thatV(x) is orthogonal to x for all x Sn. If a non-vanishing vector field Vexists, then wecan define an associated map f : Sn Sn by f(x) = V(x)/|V(x)| which has no fixedpoints. By Proposition ..., this implies that deg(f) = (1)n+1

One the other hand, sincef(x) andx are always orthogonal, we can build a homotopy

h: Sn ISn, ht(x) =cos(t/2)x+ sin(t/2)f(x)

between the identity map and f, from with we conclude that deg(f) = 1. Ifn is even,this leads to a contradiction.

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Remark 4. In contrast with Theorem3.26, ifn is oddSn always admits a non-vanishingvector field. This is because S2m1 R2m = Cm and we can use complex scalar multica-tion to rotate each vector by 90 degrees. Explicitly,V(x1, y1,...,xn, yn) = (y1, x1,..., yn, xn).

Alternative approach:For n 0, the homology group Hn(Sn) is isomorphic to Z. There are two possible

isomorphismsHn(Sn)= Zdepending on a choice of generator. A choice of this generator

is called an (global) orientation ofSn.Given a point p Sn and an open neighbourhood p U Sn, we have canonical

isomorphisms

Hn(Sn)

=Hn(S

n, Sn {p}) =Hn(U, U {p}). (5)

composing the long exact sequence of the pair (Sn, Sn {p}) with excision. A choice oforientation for Hn(U, U {p}) = Z is called a local orientation ofSn at p. Becausethe isomorphism (5) is natural, an orientation ofSn determines local orientations at allpointsp Sn, and vice versa.

Now suppose that f : Sn Sn is a map and for some point p Sn the preimagef1(p) is a finite set of points {q1,...,qk} Sn.7 Suppose further that for some openneighbourhood p U the preimage f1(U) is a disjoint union of open sets V1 ... Vkfor which qi Vi. For each i, the restriction off induces homomorphism

Hn(Vi, Vi \ {qi}) Hn(U, U\ {p}).

Since both groups are isomorphic to Z, the homomorphism must be multiplication by aninteger di which we call the local degree.

Proposition 3.27. Under the conditions above, the degree of f is the sum of the localdegrees: deg(f) =

ki=1 di.

Proof. Fix an orientation Z = Hn(Sn) and use this to impose local orientations at all

points. We have a commutative diagram where natural and orientation isomorphisms areindicated by double lines.

Hn(Sn)

f

Hn(Sn)

Hn(Sn, Sn f1(p))

f Hn(Sn, Sn {p})

Z

A

ki=1 Hn(Vi, Vi {qi})

f Hn(U, U {p})

Zk B Z

7Such a point always exists iff is differentiable (Sards Theorem)

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In matrix notation, we have

A=

11

...1

B =

d1 d2 ... dk

So the composition is d1+ ...+ dk is the degree off.

In the simplest case,pandUcan be chosen so thatfrestricts to local homemorphismsVi U. In this case the local degrees are all1, so the degree is obtained by countingpointsq1,...,qk signs according to whetherfis locally orientation preserving or reversing.

Example 16. We can construct a map Sn Sn of degree d 2 as follows. Let A Sn

be the complement ofd disjoint open disksBi inSn. Let

q: Sn Sn =X/A=dSn

be the quotient map. The orientation on Sn induces local orientations and hence globalorientations on each sphere in the wedge sum. Let

p: dSn Sn

map each sphere by a degree 1 homeomorphism to Sn.The preimage (pq)1(y) of a generic point y Sn consists of a single point in each

disk Bi each with local degree is 1 because pq is a local homeomorphism. Therefore

deg(pq) =d. By precomposingpqwith a reflection, we can construct a map of degreed.

Consider the map given d Z

wd: S1 S1, wd(e

i) =eid

ford 1 we can see by Figure14that deg(wd) =d.

Figure 14: the winding mapw4

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Note thatwd is equal to the composition ofwd with a reflection, so deg(wd) =d.By suspension, we construct maps

Snwd: Sn+1 Sn+1

of degree d for any integer.

Theorem 3.28(Fundamental Theorem of Algebra). A complex polynomial functionf(z)of degreed 1 has a complex root.

Proof. The case d = 1 is obvious, so suppose that d 2. We assume f is monic forsimplicity so f(z) =zd +O(zd1). Assume thatf(z) has no complex roots. Then thereis a well-defined, continuous map

g: C S1, g(z) = f(z)

|f(z)| = z

d + O(zd1)

|zd

+ O(zd1

)|.

Define a homotopy h : S1 IS1 by

ht(ei) =g(

t

1 tei).

for t < 1 and extend by continuity for t = 1. We have h0(ei) = g(0) is a constant and

thus deg(h0) = 0. On the other hand, for large values ofz, g(z) becomes dominated bythe highest degree terms in the numerator and denominator, so in the limit t 1, wehave

h1(ei) =eid

so deg(h1) =d, which contradicts degree being a homotopy invariant.

3.7 Cellular homology

3.7.1 Cell complexes

LetDn :={x Rn||x| 1}

denote theunit diskor closed n-cellwith boundary

Sn1 =Dn :={x Rn||x|= 1}.

Given a topological space Xand a continuous map f : Sn1 X, we may construct anew space

Y := (X Dn)/

where we quotient by the equivalence relation generated by p f(p) for all p Sn1.We say that Y is obtained from X by attaching an n-cell; the map f is called theattaching map. More generally, if we have a collection of mapsf: S

n1 X, then weconstruct

Y = (X (

Dn))/

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wherep f(p) for all p Sn1 and .Acell complex(also calledCW-complex) is a space that is constructed inductively

by attaching cells. For instance,

A 0-dimensional complexX0 is a discrete set of points (i.e. a disconnected union of0-cells).

A 1-dimensional cell complexX1 is a space constructed by attaching a collection of1-cells toX0.

A 2-dimensional cell complex X2 is constructed by attaching 2-cells to X1.

and so on ...

In general, a cell complex Xmay have cells in arbitrarily high dimensions, in which case

it is called-dimensional. Eachn-cell determines acharacteristic map : Dn X.A subsetSXis open/closed if and only if1 (S) D

n is open/closed for all cells.

Example 17. A wedge ofn-spheres ISn is constructed by attachingImany n-cells ontoa pointX0= {p} by the only possible attaching map f :Sn1 {p}.

Example 18. The torus S1 S1 can be constructed by attaching a 2-cell onto a wedgeof two circles X=S1 S1. If we denote by aand b the loops defined by the two circlesin X, then the attaching map f :S1 Xis the loop a b a1 b1.

Example 19. More generally, the genus g surface g is constructed by gluing a 2-cell toa wedge of 2g circles. If the loops defined by the circles are calleda1, b1,...,ag, bg, thenthe attaching map sends S1 to the concatenation

gi=1[ai, bi], where [ai, bi] = aibia

1i b

1i

is the commutator.

A subspace A X is called a subcomplex if it is a closed union of cells (that is,of images of characteristic maps). Given a subcomplexA X, the quotient space X/Adefined by identifying all points in A with each other, is naturally a cell complex called aquotient complexofX.

Proposition 3.29. A subcomplexA of a cell complexXis a neighbourhood deformationretract. Thus, Hq(X, A)= Hq(X/A) we have a long exact sequence in homology

. . . Hq+1(X/A) Hq(A) Hq(X) Hq(X/A) Hq1(A) . . .

Proof. Skipped.

The subcomplex Xn Xconsisting of all cells of dimension n is called the n-skeleton ofX(by convention X1 = ). IfX is infinite dimensional, the topology onX satisfies S X is open (resp. closed) if and only ifS Xn is open (resp. closed) inXn for all n. In particular, a map f :XYis continuous if and only if the restrictionsfn: Xn Yare continuous for alln. We sayXhas the direct limit topology with respecttoXn.

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Figure 15: orientable surfaces

Lemma 3.30. Let X be a cell complex and C X a compact subspace. Then C iscontained within finitely many cells ofX.

Proof. Choose a sequence of points xi C lying in distinct cells. We will show that thesetS:= {xi} is finite. We begin by showing Sis closed.

First observe that

S Xis closed S Xn is closed in X, n

We use induction on n.ClearlySX0is closed inX0hence inX, because every subset ofX0is closed. Assume

by induction that S Xn1 is closed in X. Thus for any characteristic map

: Dk X

the pre-image 1 (S Xn1) is closed in Dk. Fork < n, the pre-image 1 (S Xn) =

1 )(SXn1) is closed in Dk. For k = n, the pre-image 1 (SXn) D

n equals1 (S Xn1) plus at most one point, thus it is a union of two closed sets, hence is closedin Dn. We deduce that S Xn is closed in Xn hence also in X. By induction, this holdsfor all n so Sis closed in X.

The same argument shows that every subset ofSis also closed, so Shas the discretetopology. ButS is a closed subset of the compact set C, so it is compact. We concludethatSis finite.

3.7.2 Cell complex propaganda (not to be tested)

We present some results showing that many interesting spaces are either homeomorphicor homotopy equivalent to cell complexes. The material in this section will not be tested.

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Definition 17. A real analytic function f : Rn R is a infinitely differentiablefunction such that at every point p Rn, fequals its Taylor series atp on some positiveradius. A real analytic set X Rn is the solution set finite collection of equations

f1(x) =... = fn(x) = 0, for fi real analytic.Example 20. Polynomial functions, exponential functions, trigonometric functions, etc.are real analytic.

Theorem 3.31 (Lojasiewicz 1964). Every real analytic setX Rn is homeomorphic toa cell complex.

LetX and Y be two topological spaces. LetCont(X, Y) be the set of all continuousmaps fromX toY with the compact-open topology.

Example 21. The space LY =Cont(S1, Y) is called the free loop space ofY.

Theorem 3.32 (Milnor 1959). If X and Y are cell complexes andX is compact, thenCont(X, Y) is homotopy equivalent to a cell complex.

Definition 18. A topological space X is called a (topological) n-manifold if it is Haus-dorff and if every point p is contained in an open neighbourhood p U X that ishomeomorphic to Rn.

Every open set in Rn is an n-manifold.

The sphere Sn is an n-manifold.

Surfaces of any genus are 2-manifolds.

The product of an m-manifold and ann-manifold is an m + n-manifold.

An example of a space that is locally Euclidean but is not a manifold is constructedby taking two copies of the real line R R = R {a, b} and forming the quotient by(t, a) (t, b) if t = 0. This space looks locally like R, but the points (0, a) and (0, b)cannot be separated by open sets.

Theorem 3.33. Every compactn-manifold is homotopy equivalent to a cell complex. Itremains an open question whether or not every compactn-manifold is homeomorphic to

a cell complex.

3.7.3 Cellular Homology

Lemma 3.34. IfXis a cell complex, then:

(a) Hq(Xn, Xn1) is zero ifq= n and is an free abelian group with generators corre-sponding to then-cells whenq= n.

(b) Hq(Xn) = 0 forq > n. ThusHq(X) = 0 forq >dim(X).

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(c) The inclusion i : Xn X induces an isomorphismHq(i) : Hq(Xn) Hq(X) forq < n.

Proof. By Proposition 3.29, we have isomorphism Hq(Xn, Xn1) = Hq(Xn/Xn1) andXn/Xn1 is a wedge of spheres indexed by the n-cells ofX. Property (a) follows.

Property (b) is proven by induction. Clearly true forn = 0. Now suppose it has beenproven forn 1. The long exact sequence of the pair contains

Hq(Xn1) Hq(Xn) Hq(Xn, Xn1)

where bothHq(Xn1) =Hq(Xn, Xn1) = 0 forq > n by induction and property (a). ThusHq(Xn) = 0 as well.

To prove property (c), consider the exact sequence

Hq+1(Xn+1, Xn) Hq(Xn) Hq(Xn+1) Hq(Xn+1, Xn).

By (a), the two groups on the end vanish ifq < nso Hq(Xn)=Hq(Xn+1). Repeating thisargument, we get

Hq(Xn)=Hq(Xn+1)=Hq(Xn+2)=...

which suffices ifX is finite dimensional. To take care of the infinite dimensional case,observe that Lemma 3.30 implies that every chain in Sq(X) must be in the image ofSq(Xn) for some n (since the union of images of simplices occurring in the chain is acompact subset ofX). Thus every cycle Zq(X) arises as the image of a cycle in Zq(Xn)for some n, and every boundary in Bq(Xn) arises as the image of a boundary in Bq(Xn)for some n. The result follows (details left to reader).

Define a homomorphismdn: Hn(Xn, Xn1) Hn1(Xn1, Xn2) by the commutativediagram

0

0

Hn(Xn+1)=Hn(X)

Hn

(Xn

)jn

Hn+1(Xn+1, Xn) dn+1

n+1

Hn(Xn, Xn1) dn

n

Hn1(Xn1, Xn2

Hn1(Xn1)

jn1

0

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where the diagonal maps occur in the long exact sequences of pairs. Notice that dndn+1=0 because it factors through n jn = 0. Thus (Hn(Xn, Xn1), dn)nZ forms a chaincomplex, called the cellular chain complex. The homology of the cellular chain complex

is called the cellular homology.Theorem 3.35. The cellular homology groups are naturally isomorphic to the singularhomology groups.

Proof. From diagram, we may identifyHn(X)=Hn(Xn)/im(n+1). Sincejn is injective,this is isomorphic to im(jn)/im(dn+1). By exactness, this is the same as ker(n)/im(dn+1).Finally, becausejn1 is injective, this is equal to ker(dn)/im(dn+1).

Theorem3.35is very useful for calculations, because it allows us to replace the usuallyuncountably infinite rank Sq(X) by the often finite rank Hn(Xn, Xn1). Before getting

started with examples, we want a more direct understanding of the boundary maps dn.Denote by {en} the set ofn-cells of a cell complex X, so that Hn(Xn, Xn1) is the

free abelian group generated by {en}.

Proposition 3.36. Forn >1, the cellular boundary map satisfies

dn(en) =

d,en1

whered,is the degree of the map

Sn1 Xn1 Sn1

defined by composing the attaching map of en with the quotient map Xn1 Sn1 =

Xn1/(Xn1 en).

Proof. The proof is based on the following commutative diagram

Hn(Dn, D

n)

=

Hn1(Dn)

f

(,) Hn1(Sn1 )

Hn(Xn, Xn1) n

dn

Hn1(Xn1)

q

jn1

Hn1(Xn1/Xn2)

=

q

Hn1(Xn1, Xn2) = Hn1(Xn1/Xn2, Xn2/Xn2)

where

is the characteristic map for en and f the attaching map.

q: Xn1 Xn1/Xn2 is the quotient map.

q : Xn1/Xn2 Sn1 is the quotient map obtained by collapsing everything be

thesphere to a point.

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Figure 16: non-orientable surfaces

which is identified as the wedge sum Sm Sn, which is a cell complex with cells indimension 0,m, and n (compare the case of a torus S1 S1).

Suppose thatn m and n >1. Then the (n+ 1)-skeleton is equal to Sm Sn so byLemma3.34, Hq(S

m Sn) =Hq(Sm Sn) for qn. It follows that the boundary mapin the cellular chain complex is trivial and that

Hq(Sm Sn) =

Z ifq= 0,m,n, or m + n

0 otherwise

Example 25. Real projective space RPn is the set of 1-dimensional vector subspaces

ofRn+1

. Topologically, RPn

is the quotient space ofRn

{0}by the relation v v for R a non-zero scalar. Equivalently, RPn is identified with quotient space Sn/ bythe relation v v.

By restricting to a closed hemisphere of Sn, we see that RPn can be obtained byattaching an n-cell to RPn1 using the quotient Sn1 RPn as the attaching map. Byinduction, RPn =e0 e1 ... en is a cell complex with one cell in each dimension n.Thus, the cellular chain complex looks like

0 Z dn Z

dn1

d1 Z 0

To understand the boundary maps, we must determine the degrees of the composition of

the attaching map and the quotient mapSq1

RPq1

q RPq1/RPq2 =Sq1.

Observe that q restricts to local homeomorphisms from each of the two open hemi-spheresSq1 Sq2 onto Sq1 {pt} so by ... the degree is either 0 or 2. Notice thatthese two restrictions are interchanged by precomposing with the antipodal map whichhas degree (1)q. Thus the local degrees have the same sign if q is even and differentsigns ifqis odd. Thus dq is multiplication by 0 ifqis odd and 2 ifqis even. The chaincomplex is

0 Z 0 Z

2 Z

0

2 Z

0 Z 0

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ifn is odd and0 Z

2 Z

0 Z

2

2 Z

0 Z 0

ifn is even. We obtain homology groups,

Hq(RPn) =

Z ifq= 0 orq= n and nis odd

Z2 if 0< q


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3.9 Homology with coefficients

So far we have developed singular homology theory for integer coefficients, meaning thatour chains are finite formal sumsa with coefficients a Z. More generally, it ispossible (and useful) to work with coefficients in any commutative ring R with identity1 R; in particular, this means we have a canonical ring homomorphism Z R. Themost interesting cases are when R = Zn is the ring of integers modulo n, or when R is afield, such as Q, R, Cor Zp for primep.

This generalization is pretty straight forward. DefineSq(X; R) to be the freeR-modulewhose chains are finite sums the form

a where the coefficients a are elements

of R. So far we have been studying the case Sq(X) = Sq(X;Z). The boundary mapq :Sq(X; R) Sq1(X; R) is defined as before

q(

a) =

(a) =

q

i=0

(1)ia(i)

and satisfies 2 = 0 giving rise to a chain complex S(X; R) = (Sq(X; R); q). Thisdetermines cycles Zq(X; R) := ker(q), boundariesBq(X; R) := im(q+1) and homology

Hq(X; R) :=Zq(X; R)/Bq(X; R),

all of which are now R-modules. We callHq(X; R)singular homology withR-coefficients.8

The proofs of all the main theorems carry over unchanged. In particular

Hq(; R) is a functor for topological spaces to the category ofR-modules. Thus forevery continuous map f :XY, there is a homomorphism ofR-modules

Hq(f; R) =f: Hq(X; R) Hq(Y; R)

such that (f g)= f g and (IdX)= I dHq(X;R).

Iff , g: XYare homotopic, then f = g.

A topological pair (X, A) gives rise to a long exact sequence ofR-modules

Hq(A; R) Hq(X; R) Hq(X, A; R) Hq1(A; R) ...

where the connecting homomorphism is defined using the same diagram chase asbefore.

The excision isomorphismHq(X\ B, A \ B: R)=Hq(X, A; R) hold under the samehypotheses as before.

H0(X; R) is a free R-module generated by the path components ofX.

8Note that an abelian group is the same thing as a Z-module. IfR is a field, an R-module is just anR-vector space.

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H0(pt; R) =R and Hq(pt; R) = 0 forq= 0. Define Hq(X; R) to be the kernel of thenatural mapHq(X; R) Hq(pt; R).

Hn(Sn; R) = R and Hq(Sn; R) = 0 for q= n. Iff : Sn Sn is a degree d map,thenf is simply multiplication byd R.

If X is a cell complex, then Hn(Xn, Xn1; R) is a free R-module with genera-tors corresponding to n-cells. We can form a cellular chain complex Cn(X; R) =Hn(Xn, Xn1; R) and boundary map defined by ... and cellular homology is natu-rally isomorphic to singular homology.

The Mayer-Vietoris Sequence is still exact and defined in the same way.

One of the big advantages in working with coefficients in a field F is that homologyare now vector spaces. This means for instance that short exact sequences always split

and this can simplify a lot of calculations.

Example 28. Recall that the cellular chain complex C(RP2) (Example25) is either of

0 Z 0 Z

2 Z

0

2 Z

0 Z 0

0 Z 2 Z

0 Z

2

2 Z

0 Z 0

depending on whethern is odd or even.Now consider the cellular chain complex C(RPn;Z2). The chain groups become free

Z2-modules and the boundary maps are all zero because 0 = 2 in Z2

0 Z2 0=2 Z2 0=2 Z2 0=2 0=2 Z2 0=2 Z20.

Thus

Hq(RPn;Z2) =

Z2 forq= 0,...,n

0 otherwise.

On the other hand, ifFis a field of characteristic other than 2, then 2 is a unit. Thecellular chain complexes becomes one of

0 F 0F

=F

0

=F

0F0

0 F

=F 0

F

=

=F 0

F0

and we get homology groups

Hq(RPn; F) =

F ifq= 0 or q= n andn is odd

0 otherwise.

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The only isomorphism invariant of a vector space is the dimension. Given a fieldF thedimensions dim(Hq(X; F)) are called Betti numbers ofX. These are basic invariantsof a space Xthat are often easier to compute than the homology groups Hq(X;Z).

Suppose thatXis a space for which dim(Hq(X); F) is finite in all degrees and is zeroin all but finitely many degrees. Define the Euler characteristic

(X) :=q=0

(1)q dim(Hq(X; F))

to be the alternating sum of the Betti numbers.

Proposition 3.40. IfXis finite cell complex, the Euler characteristic is well defined andsatisfies

(X) :=

n=0

(1)ncn

wherecn is the number of cells of dimensionn. In particular, (X)is a homotopy invari-ant ofXand is independent of the fieldF.

Proof. We will simplify notation to prove this result in a more abstract context. Supposewe have a finite length chain complex of vector spaces

0 CNn...

1C000

for which each Cq is finite dimension cq. For each q we have short exact sequences ofvector spaces

0 Zq Cq Bq10

and0 BqZq Hq 0.

Denote by zq, bq, hq the dimensions ofZq, Bq, Hq respectively. The rank nullity theoremgives us equations

cq =zq+ bq1, hq =zq bq

for all q Z. In case Cq =Cq(X; F) we get

(X) =q

(1)qcq

= q

(1)q(zq+ bq1)

=q

(1)qzq (1)q1bq1

=q

(1)q(zq bq)

=q

(1)qhq

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A surface of genus g has Euler characteristic 1 2g+ 1 = 2 2g.

CPn has Euler characteristic 1 + 0 + 1 + 0 + ...+ 1 =n+ 1.

A sphere has Euler characteristic 1 + 1 = 2 ifn is even and 1 1 = 0 ifn is odd.

A well known example of the Euler characteristic comes up for the boundaries of convexpolytopes, like a tetrahedron or a cube. The Euler characteristic is thenV E+ FwhereVis the number of vertices, Eis the number of edges, and Fis the number of faces. Sincethe boundary of a convex polytope must be homeomorphic to S2, it follows that

V E+ F = 2

Theorem 3.41 (Lefshetz Fixed Point Theorem). LetX be a finite cell complex and letf :XXbe a continuous map. The Lefshetz numberforfis the number

L(f) :=q=0

(1)qT r(Hq(f;Q))

whereT r(Hq(f;Q)) is the trace of the linear map Hq(f;Q) : Hq(X;Q) Hq(X;Q). IfL(f)= 0, thenfhas a fixed point.

Some examples

(Generalization of Brower Fixed Point Theorem) IfX is a finite cell complex suchthat H0(X;Q) = Q and Hq(X;Q) = 0 for q > 0, then L(f) = 1 for any map

f :XX. Consequently everyfmust have a fixed point.

Note that it is a necessary condition that X is finite, because the translation mapt: R R, t(x) =x+ 1 has L(t) = 1 but no fixed points.

A map of spheresf :Sn Sn with degreedhas Lefshetz numberL(f) = 1+(1)ndwhich vanishes if and only ifd= (1)n+1, recovering ...

Iff is homotopic to the identity map, then L(f) = (X). Thus if(X)= 0 thenanyfhomotopic to the identity must have a fixed point.

Proof. We will only briefly sketch the argument.Suppose that the map f : X Xrestricts to maps f : Xk Xk for each skeleton.

If this happens, it is not hard to show that fdetermines a morphism from the cellularchain complex ofX to itself:

. . . Cq+1(X)

Cq+1(f)

Cq(X)

Cq(f)

Cq1(X)

Cq1(f)

. . .

. . . Cq+1(X) Cq(X) Cq1(X) . . .

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Using an argument similar to the proof of Proposition ..., one shows that

L(f) =

q=0

(1)qT r(Cq(f;Q)).

In particular, this means that ifL(f)= 0, thenT r(Cq(f;Q))= 0 for someq. This impliesin particular that some cell ekXmust map onto itself, i.e. ekf(ek).

To go further, we must impose more assumptions on Xandf. First, we suppose thatX has the special property that the attaching maps of cells are all injective and hencethat the characteristic map Dk X are injective. Next, we suppose that fsends eachk-cell either in to the k 1-skeleton or homeomorphically onto a k-cell. These conditionsare satisfied ifXis a simplical complex and fis a simplicial map.

With these assumptions imposed, we deduce that L(f)= 0 then fsends some k-cellto itself. By the Brouwer Fixed Point Theorem, we deduce that there must be a fixed

point.To make this argument rigorous, we would have to show that any continuous map

f : X Xwithout fixed points can be replaced up to homotopy by a simplical mapf : X X without fixed points. This is true (see 2.C of Hatcher), but the proof israther long and technical.

3.9.1 The Universal Coefficient Theorem for Homology

The Universal Coefficient Theorem for Homology allows Hq(X; R) to be calculated fromHq(X;Z) and Hq1(X;Z). We need some definitions before stating the formula.

Every Abelian group G can be fit into a short exact sequence of the form

0 Zr A Zg G 0 (6)

where Zr and Zg are (possibly infinitely generated) free abelian groups on r and g gen-erators respectively. We call (6) a free resolution ofG. We think of the basis ofZg asthe generators ofG and the basis ofZr as relations. The homomorphism A is defined interms of ang r matrix.

Example 29. For example, every finitely generated abelian group is isomorphic to

G= Zf Zn1 ... Znt

for some choice f, n1,..., nt. The integer f is called the rank ofG. This fits into ashort exact sequence

0 Zt A Zf Zt G 0 (7)

whereA = (0, A) andA is the diagonal matrix with entries n1,...,nt.

Now let R be another abelian group. Given a free resolution (6) ofG, we obtain amap

Rr ARRg

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whereAR is given by the same integral matrix as A. We define

T or0Z

(G, R) := cok(AR) =Rg/im(AR).

T or1Z(G, R) := ker(AR).

Equivalently, these are the homology groups of the chain complex 0 Rr Rg 0.The group T or0

Z(G, R) is known as the tensor product ofG and R and is more com-

monly denotedG ZR := T orZ(G, R).

IfR is a ring, then both T or0Z

(G, R) and T or1Z

(G, R) are R-modules.

Theorem 3.42 (Universal Coefficient Theorem). For any space X and coefficient ringR, there is a non-canonical isomorphism

Hq(X; R) =T or0Z(Hq(X), R) T or1Z(Hq1(X), R).

Proof. Skipped. See Hatcher 3.A for a proof.

The moral of Theorem3.42is that homology groups with coefficents in Rcontain nomore information than homology groups with integer coefficients. Here are a couple ofeasy consequences.

Corollary 3.43. If Hq(X;Z) is free abelian in every degree, then Hq(X; R) is a freeR-module of the same rank in each degree.

Proof. For a free abelian group G = Zg, we form a resolution

0 0 Zg =G 0.

Thus T or0Z

(Zg, R) =Rg and T or1Z

(Zg, R) = 0 for any R for any R. It follows that

Hq(X; R) =T or0Z

(Hq(X), R) T or1Z

(Hq1(X), R) =Hq(X) ZR = Rn

wheren is the rank ofHq(X).

Corollary 3.44. Suppose that Hq(X; F) is finitely generated in all degrees. If F is afield, then

dimF(Hq(X; F)) rank(Hq(X;Z))

with equality ifFhas characteristic zero (e.g. F = Q, R, C).

Proof. For a finitely generated abelian group G, we have resolution (7) above. IfF is afield then T or0

Z(G, F) and T or1Z(G, F) are vector spaces of dimension f+t rank(A

F)

and t rank(AF). Consequently

Hq(X; F) =T or0Z

(Hq(X), F) T or1Z

(Hq1(X), F)

has dimension at least as large as the rank ofHq(X) with equality if and only ifrank(AF) =

t respectively. IfFhas characteristic zero, then equality holds.

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The deck transformations form a group under composition that acts on X. The coveringspace is called Galois if the deck transformation group acts freely and transitively onfibres. I.e. if for any two points x, x f1(y) there is a unique deck transformation

such that (x) =x

. In this case, the preimages of pointsf1

(y) are equal to the orbitsof a free group action on X.In the examples above, the deck transformation groups Zacting by translation on R,

Zd acting by rotation on S1, Z2 acting by the antipodal map on S

n, and Zp acting on S3.

All three examples are Galois.

Theorem 3.46. Suppose thatf :XY is an-sheeted covering map for finiten. Thenthere exists a natural homomorphism

: Hq(Y; R)Hq(X; R)

such thatHq(f) is multiplication byn onHq(Y; R).In particular, if R is a coefficient ring containing 1

n then must be injective. If

furthermore the covering is Galois, then the image of is

Hq(Y; R)=Hq(X; R)G

with the subgroup ofHq(X; R) of elements fixed by the deck transformation group G.

Proof. The transfer map is determined by a chain morphism :S(Y) S(X) definedon simplices : q Y to the sum

() =ni=1

i

where 1,..., nare the n lifts of (one for each sheet). It is not hard to see that =

so is a chain morphism. At the level of chains

f() =f(ni=1

i) =ni=1

f i= n.

ThusHq(f) () is multiplication by n.Now suppose thatn is invertible in R. Then 1nf is the identity onHq(Y; R) so

must be injective (in fact the inclusion splits!).Finally, suppose additionally that the covering is Galois with deck transformation

groupGof order n. Define a homomorphism

: Hq(X; R) Hq(X; R)G

by

() = 1

n

ni=1

g()

If Hq(X; R)G then() = and thus

im() =Hq(X; R)G.

Note also that n= f. Since f is surjective and n is a unit it follows that

im() = im(n) = im( f) = im()=Hq(Y; R).

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4 Cohomology

LetR be a commutative ring and letM be a R-module. The dual module

M :=H omR(M, R)

is the set ofR-module homomorphisms fromM toR. M is anR module under additionand scalar multiplication of functions. There is a natural isomorphism

(iMi)=

i

Mi ,

defined by the rule (1, 2,..., )(m1 +m2 + ...) =

i i(mi). In particular, for free moduleswe have

(iR)

=i

R

=i

R. (8)

More directly, (8) holds because a homomorphism out of a free module is specified bylisting where the free generators are sent.

Given a R-module homomorphism f :MN, define the transpose

f :N M

which sends N =H omR(N, R) tof() = f.

M f

f()=f

N

R

Dualization is a contravariant functorfromR-modules to R-modules. This meansthatI dM=I dM and (g f)

=f g. The first is obvious and the second follows fromassociativity of composition:

(g f)() = (g f) = ( g) f=f(g()) = (f g)()

M f

fg()

N

g P

R

Given a chain complex ofR-modules

... Cn+1n+1 Cn

nCn1n1 ...

we form the dual chain complex (or co-chain complex)

... Cn+1 n+1

Cn nCn1

n1 ...

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whereCn =Cn and n= n . Note that

n n+1= n

n+1= (n+1 n)

= 0 = 0

We can now define cochainsZn := ker(n+1), coboundariesBn := im(n) and cohomology

Hn :=Zn/Bn.

Definition 19.The singular cohomology of a pair of spaces (X, A), denoted Hq(X, A; R)q 0, is the cohomology of the singular cochain complex

Sq1(X, A; R) Sq(X, A; R) Sq+1(X, A; R) . . .

obtained by dualizing the singular chain complex ofX.

More geometrically, we can understand a singular co-chain Sq

(X; R) as a functionthat assigns a scalar to every singular simplex: q X. The pairing

Sq(X; R) Sq(X; R) R

is sometimes called integration, because it is an algebraic analogue of integrating adifferential form over parametrized manifold.

For example,

A 0-cochain is simply a function (of sets) f :XR, since 0-simplices corr

homology lecture notes

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