Algebraic Topo Chicago Notes

8/13/2019 Algebraic Topo Chicago Notes

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Math 317 - Algebraic Topology

Lectures by Benson Farb

Notes by Zev Chonoles

The University of Chicago, Fall 2012

Lecture 1 (2012-10-01) 1

Lecture 2 (2012-10-03) 3

Lecture 3 (2012-10-05) 7

Lecture 4 (2012-10-08) 9

Lecture 5 (2012-10-10) 12

Lecture 6 (2012-10-12) 14

Lecture 7 (2012-10-15) 16

Lecture 8 (2012-10-17) 19

Lecture 9 (2012-10-19) 22

Lecture 10 (2012-10-22) 24

Lecture 11 (2012-10-24) 26

Lecture 12 (2012-10-26) 28

Lecture 13 (2012-10-29) 31

Lecture 14 (2012-10-31) 34

Lecture 15 (2012-11-02) 37

Lecture 16 (2012-11-05) 38

Lecture 17 (2012-11-07) 40

Lecture 18 (2012-11-09) 42

Lecture 19 (2012-11-12) 44

Lecture 20 (2012-11-14) 49

Lecture 21 (2012-11-16) 51

Lecture 22 (2012-11-19) 54

Lecture 23 (2012-11-21) 57

Lecture 24 (2012-11-26) 59

Lecture 25 (2012-11-28) 64

Lecture 26 (2012-11-30) 67

Lecture 27 (2012-12-03) 71

Lecture 28 (2012-12-05) 73


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Introduction

Math 317 is one of the nine courses offered for first-year mathematics graduate students at theUniversity of Chicago. It is the first of three courses in the year-long geometry/topology sequence.

These notes were live-TeXed, though I edited for typos and added diagrams requiring the TikZ

package separately. I used the editor TeXstudio.

I am responsible for all faults in this document, mathematical or otherwise; any merits of thematerial here should be credited to the lecturer, not to me.

Please email any corrections or suggestions to [email protected].

Acknowledgments

Thank you to all of my fellow students who sent me corrections, and who lent me their own notesfrom days I was absent. My notes are much improved due to your help.

Copyright

Copyright c 2012 Zev Chonoles.

This work is licensed under a Creative Commons Attribution-ShareAlike 3.0 Unported License.This means you are welcome to do essentially anything with this work, including editing, adapting,

distributing, and making commercial use of it, as long as you

include an attribution of Benson Farb as the lecturer of the course these notes are based on,and Zev Chonoles as the person taking the notes,

do so in a way that does not suggest either of us endorses you or your use of this work, and

if you alter, transform, or build upon this work, you must apply to your work the same, orsimilar, license to this one.

More details are available at https://creativecommons.org/licenses/by-sa/3.0/deed.en US.
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Lecture 1 (2012-10-01)

Today Ill try to present a general idea of algebraic topology.

Definition. A categoryC is a set C(though not always really a set) of objects and a set

Mor(C) ={f :A B | A, B C}

of morphisms between the objects ofC. We can always compose morphisms (whose domains andcodomains are appropriately related), this composition is always associative, and for all objectsA C, there is a morphism idA: A A that acts as the identity for composition.

Examples. The following are some common examples of categories.

C= {topological spaces}, and Mor(C) ={continuous maps}

C= {topological spaces}, and Mor(C) ={homeomorphisms}

C= {vector spaces over a field K}, and Mor(C) ={K-linear maps}

C= {abelian groups}, and Mor(C) ={homomorphisms}

Definition. A (covariant) functor F : C D is a mapping that takes objects A Cto objectsF(A) D, and morphisms (f :A B) between objects ofCto morphisms (F(f) :F(A) F(B))between the corresponding objects ofD.

Well often refer to F(f) as f when the functor F is understood.

We can also define a contravariant functor to be a functor that reverses the direction of morphisms,

so thatF(f) :F(B) F(A). In this situation, the shorthand for F(f) will be f.

In either case, it must satisfy F(idA) =idF(A). It also must satisfy functoriality; that is, we musthaveF(f g) =F(f) F(g), or in shorthand (f g) =f g. This goes in the other order for

contravariant functors.Examples. The following are some common examples of functors.

C= {finite sets} with Mor(C) ={set maps}, D= {R-vector spaces} with Mor(D) ={linear maps},and F :C D defined on objects by

Sfree R-vector space on S

and for a morphism f :S T, F(f) is the linear map uniquely determined by sending thebasis elements to the basis element f(s).

C={R-vector spaces} with Mor(C) ={linear maps}, and F : C Cdefined on objects byV V, and on morphisms by (A: V W)(A :W V).

Broadly speaking,

algebraic topology construction of functors{spaces} {algebraic objects}.

The functors well look at in this class are Hi, Hi,1, and n.

Here are some demonstrations of the power of functors. Lets suppose we know there exists afunctorFi: {spaces} {abelian groups}such that Fi(D

n) = 0 for any n, Fi(Sn) = 0 forn =i, and

Fi(Si)= 0. We dont have to know what else it does.

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Example. From this assumption alone, we can prove that Sn= Sm n= m, and as a corollary,prove invariance of dimension, i.e. Rn= Rm n= m.

Proof. Suppose h : Sn Sm is a homeomorphism, with inverse g. Then applying the functor Fn,

g h= idSn = g h= idFn(Sn),

h g= idSm = h g= idFn(Sm).

This implies thath: Fn(Sn) Fn(S

m) is a bijection, and hence an isomorphism. But Fn(Sn)= 0

and Fn(Sm) = 0 unless n = m.

Example (Brouwer Fixed Point Theorem). Any continuousf : Dn Dn has a fixed point (n 2).

Proof. Suppose there are no fixed points. Then we can define r : Dn Dn = Sn1 by drawing aline from f(v) to v (this is possible only because we are assuming they are different) and extendingit until it hits the boundary ofDn, and setting that to be r(v).

v

f(v)

r(v)

You can prove that this is continuous on your own. Note that for any v Sn1, we will have thatr(v) =v. Thus, letting i : Sn1 Dn be the obvious inclusion map, we have that r i= idSn1 (we

say that r is a retraction ofDn onto Sn1). Because r i= idSn1, we have that

Fn1(r) Fn1(i) = idFn1(Sn1)

so that Fn1(r) must be surjective. But Fn1(r) : Fn1(Dn) Fn1(S

n1) cant be surjectivebecauseFn1(D

n) = 0 and Fn1(Sn1)= 0. This is a contradiction.

A final comment: a good method for turning invariants for spaces into invariants for maps is toassociate to a map f :XY its graph fX Y.

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Lecture 2 (2012-10-03)

There are some corrections for the homework: on problem 2a, it should read

H0(C) =

H0(C) Z

and on problem 3, you should show that the following diagram commutes:

n

[v0vn]

m Y

Xf

Simplices and -complexes

Definition. The standard (ordered) n-simplex [v0 vn] is the convex hull of the set {e1, . . . , en+1}whereei= (0, . . . , 1, . . . , 0) R

n+1. For example, when n = 2,

e0e1

e2

Equivalently, it is the set

ni=1 aiei Rn+1 |ai 0,ni=1 ai= 1 .We will often want to specify using coordinates. Let v0= 0, and vi= ei for i = 1, . . . , n + 1. Then

n = convex hull ({v1 v0, . . . , vn+1 v0}).

For example,0 = v0

1 = v0 v1

2 =

v0 v1

v2

3 =

v0

v1

v2

v3

Note that an ordered n-simplex hasn + 1 face maps; for each i = 0, . . . , n,

[v0 vi vn] = an ordered (n 1)-simplex (subsimplex of [v0 vn]).For example, for n = 2, we get canonical linear maps

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[v0v1v2]

[v1v2]

[v0v2]

[v0v1]

[v2]

[v1]

[v2]

[v0]

[v1]

[v0]

Definition. LetXbe a topological space. A -complex structure on X is a decomposition ofXinto simplices. Specifically, it is a finite collection S= {i}of simplices with continuous maps thatare injective on their interiors, that also satisfies

i i(i) =X, For allx X, there is a unique i such thatx Int(i(i)).

If: X is an element ofS, then all subsimplices of are also elements ofS. In otherwords,Sis closed under taking faces.

Example. The torus T2 can be given a -complex structure as follows:

v v

v v

a

a

b bc

Example. The Klein bottle can be given a -complex structure as follows:

v v

v v

a

a

b bc

Example. Let Xbe a topological space, and let {Ui} be an open cover ofX (assume indexed byan ordered set, induces order for simplices). The nerve of the cover is the -complex specified by

A vertex vi for each Ui

IfUi Uj = , the 1-simplex [vivj ]

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vi vj

UjUi

IfUi Uj Uk = , the 2-simplex [vivjvk]

vi vj

vk

UjUi

Uk

Simplicial Homology

We want, for any -complex X, and i 0, an abelian group Hi(X). We want this to be

1. Computable

2. Topologically invariant: ifX, Yare -complexes and X=Y, then Hi(X)=Hi(Y).

3. Non-triviality:Hn(Sn)= 0.

Step 1 (topology): We input a -complex structure on X. We output, for each n 0,

Cn(X) = free abelian group on {(ordered)n-simplices in X}.

Example. LetXbe a 2-simplex,

v0 v1

v2

Then we have

C0(X) = Z3 =[v0], [v1], [v2]

C1(X) = Z3 =[v0v1], [v0v2], [v1v2]

C2(X) = Z =[v0v1v2]

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Cn(X) = 0 for n 3

Example. LetXbe the torus,

v v

v v

a

a

b bc

T1

T2

Then we have

C0(X) = Z =[v]

C1(X) = Z3 =a,b,c

C2(X) = Z2 =T1, T2

Cn(X) = 0 for n 3

Note that we are abusing notation and identifying simplices with their maps to the space.

Step 2 (algebra): Because Cn(X) is the free abelian group on the simplices, a Cn(X) is uniquelyexpressed as

=r

i=1

aii, ai Z.

For all n 1, we define homomorphisms n : Cn(X) Cn1(X) by specifying them on thegenerators ofCn(X): for any n-simplex:

n X, we let

n=ni=0

(1)i|[v0 vivn].

The key property of these homomorphisms is that n n1 = 0. Note that we also needed theordering to define this.

We will compute some homology next class.

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Lecture 3 (2012-10-05)

LetXbe a -complex. This is essentially a simplicial complex with extra structure, namely, anordering of the vertices compatible with -maps.

For each n,

n(X) =Cn(X) = free abelian group on the set of allowable maps n X.

For each n-simplex in X(i.e. n eX), there are several -maps

n e X

n1

e

(these are the faces ofe)

Given e(v0, . . . , vn), we define

e=ni=0

(1)ie|(v0,..., vi,...,vn).

The key property of this construction is that = 0.

Now we do some pure algebra.

Definition. A chain complex is a collection of abelian groupsCiand homomorphismsi: Ci Ci1such thati1 i= 0. We write this as

Cn+2n+2Cn+1

n+1CnnCn1

n1

We say that Cn is a cycle ifn() = 0 and Cn is a boundary if there is some Cn+1such thatn+1() =. In other words,

n-cycles = ker(n), n-boundaries = im(n+1).

All boundaries are cycles, but not all cycles are boundaries. Letting Zn be the subgroup of cycles inCn, and Bn the subgroup of boundaries in Cn, we have Bn Zn, and we define

Hn= Zn/Bn.

Applying this to topology, we define

Hn(X) =nth homology group of (n(X), n).

Example. Consider the triangle

a b

c

x

yz

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We have that0= a,b,c= Z

3, 1= x,y ,z= Z3.

The map 1 sendsxb a, yc b, zc a

or expressed as a matrix,

1 =

1 0 11 1 00 1 1

.Thus,

ker(1) =x + y z= Z.

Of course, we have ker(0) = Z3, and im(1) is the set of things which sum to zero, soH0 = Z.

Example. Consider the 2-simplex

a b

c

x

yz

T

We have that0= a,b,c= Z

3, 1= x,y ,z= Z3, 2= T.

The map 1 is the same as before, and 2(T) =y z+ x. In this case we get

H0= Z, H1=0, H2=0.

Example. Consider the torus

v v

v v

y

y

x xz

A

B

We have that0= v, 1= x,y ,z, 2= A, B

with 0= 0, 1= 0, and2(A) =y + z x, 2(B) =y x + z

which implies that ker(2) =A B and im(2) =y+ z x. Thus,

H1= x,y,z/y+ z x= Z2, H2= A, B/A B= Z.

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Thus, weve shown that H0(X)= Z for Xconnected. Now we want to prove that

Hi(X) =d

j=1

Hi(Xj).

Let

n:d

i=1

Cn(Xi) Cn(X)

be defined by sending to (since XiX, a chain on Xi is literally a chain on X) and we canget an inverse map by noting that, for any simplex , its image (n) is connected, and thereforecan only lie in one connected component. Thus n is an isomorphism.

Note that commutes with , i.e.

+n =n

where+n

=Xin

(this fact is clear).

Now we will need a homological algebra fact. For any chain complexes C = {Cn, n} and C =

{Cn, n}, a chain map f :C C

is a collection of maps fn: Cn Cn such thatfn n=

n fn,

or as a diagram,

Cnn

fn

Cn1

fn1

Cn n

Cn1

The homological algebra fact is that any chain map f :C C functoriallyinduces a homomorphism

f : Hn(C) Hn(C). Here is a proof.

Given a Zn(C), or equivalently n= 0, we get that

n(fn ) =fn(n) =fn(0) = 0

sofn() Zn(C).

We map Zn(C) toHn(C) by sending to [], and likewise for C. To showf induces a compatible

map on homology (the dashed arrow)

Zn(C)

Zn(C)

Hn(C) =Zn(C)/Bn(C) Hn(C

) =Zn(C)/Bn(C

)

note that for any Bn(C), we have (f()) = f() =f(0) = 0, so that f() Bn(C

). Thusfis well-defined.

Now, to finish our proof, use the functoriality of the above map to see that, because is anisomorphism, it induces an isomorphism on homology.

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Lecture 5 (2012-10-10)

So far, weve seen how to start from a -complex Xand then form a complex of simplicial chains,and from there we use homological algebra to define the homology groups Hi(x). But wed like tobe able to start from a topological space without a specified -complex structure and be guaranteed

to get isomorphic homology groups. In other words, we want to show that

X=Y = Hi(X)=Hi(Y) for alli 0.

As is common in mathematics, this becomes easier if we enlarge our category, in this case to thehomotopy category.

Definition. Let X, Ybe topological spaces, and f, g : X Ymaps. A homotopy from f to g,denotedfF g, is a map F :X [0, 1] Y such thatF|X{0}= f andF|X{1}= g.

For t [0, 1], we write Ft : XY for the function Ft(x) =F(x, t), so that F0 = f and F1=g.

Definition. A map f :XY is a homotopy equivalence if there is some g :Y X such thatf g idY and g fidX. We say that XY.

Remark. fg is an equivalence relation, and XYis an equivalence relation.

Definition. X is contractible if there is a point x X such that i : {x} X is a homotopyequivalence.

Example. Homotopy equivalence cant tell you anything about contractible spaces, not even theirdimension, because Dn is contractible: the inclusion {0} Dn has as a homotopy inverse the mapsending everything to 0. The homotopy is Ft(v) =tv.

Example. Any homeomorphism f :XY is also a homotopy equivalence.

Proposition. SupposeHi is any functor from

topological spaces (or-complexes)with continuous maps

abelian groupswith homomorphisms

satisfying

1. Functoriality:(idX)= idHi(X), (f g)= f g.

2. Homotopy functor: iffg, thenf= g.

ThenH is a topological invariant; in fact, iff :X Y, thenf: Hi(X)

=Hi(Y).

Proof. Iff :X Y has g : Y Xas a homotopy inverse, then the fact that f g idY and

g fidXimplies that

f g= id = f surjective, g f8= id = f injective,

and hence f is an isomorphism.

How to Prove that Hi is a Homotopy Functor

Method 1: The Simplicial Approximation Theorem

The idea is that, given -complexes Xand Y with f :X=

Y,

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Lecture 6 (2012-10-12)

Remember, this is our goal:

Theorem. IfF :{ topological spacescontinuous maps

} { abelian groupshomomorphisms

} is a functor that is a homotopy functor, i.e.

fg = f= g,

thenh: X Y (i.e. X andYare homotopy equivalent) implies thath :F(X) F(Y) is an

isomorphism.

Given a spaceX, we defined its singular chain complex CSn (X) to be the free abelian group on theset{continuousn X}, and then defined the singular homology HSi (X) to be the homology ofthis chain complex. That this is functorial is trivial; the map f#: C

Sn (X) C

Sn (Y) induced by an

f :XY defined by f#(

aii) =

ai(f i) is a chain map.

Proposition. Givenf, g: XY withfg, thenf= g : HSi (X) H

Si (Y) for alli 0.

Corollary. A homotopy equivalenceh: X Y induces an isomorphismh: H

Si (X)

=HSi (Y).

Proof of Prop. Given a homotopy F :X [0, 1] Y with F0= f andF1 = g, we will construct achain map Pi: C

Si (X) C

Si+1(Y),

CSi+1(X) CSi (X)

Pi

CSi1(X)

Pi1

CSi+1(Y) CSi (Y)

CSi1(Y)

Morally,P is the (unique linear extension of) the prism operator which takes a simplex Ci(X)to a prism Pwhich is then sent to Yby the homotopy F; thus, the prism allows us to comparef with g . Of course, a prism is not a simplex, but we can view its image in Yas a chain in

Ci+1(Y) by appropriately subdividing it.

v0 v1

v2

w0 w1

w2

X [0, 1]

X {1}

X {0}

Y

f

g

F ( [0, 1])

Taking the boundary of the prism operator produces

P=

topg#

bottomf#

sidesP .

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A subdivision that works: given : [v0 vi] X, define P by

P [v0 vi+1] =i

j=0

(1)jF ( id)|[v0vjwj wi]

Homological algebra interlude

Definition. Let , : {Cn, n} {Cn,

n} be chain maps. A chain homotopy from to is a

homomorphismP :Cn Cn+1 such that

= P+ P .

Key Prop. If and are chain homotopic then = : Hi(C) Hi(C) for all i.

Apply this to our situation: letting = g#, = h# gives us what we want.

Proof of Key Prop. Given c Zi(C), we want to prove that ([c]) = ([c]). This is the case if

and only if(c) (c) Bi(C

). We have that c= 0, so

(c) (c) =P (c) + P(c) = P(c) .

Wed like the map Cn(X) CSn (X) to induce isomorphisms on homology, but were missing

something deep when were thinking that.

Developing Singular Homology

The setup of relative (singular) homology is a subspace A X. This induces an inclusionCn(A)Cn(X). We define Cn(X, A) by

0 Cn(A) Cn(X) Cn(X, A) 0

(well talk about exact sequences next time). This seems like a random algebraic thing to define,but well be able to relate Cn(X, A) toCn(X/A).

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Lecture 7 (2012-10-15)

There will be an in-class midterm the week after next.

Until further notice,Hi(X) will denote the singular homology groups; also could be simplicial, reallyjust pretend Im doing both at the same time.

Weve defined simplicial and singular homology, and proved that singular homology is a homotopy

functor. However, we cant compute singular homology. Thus, we want to show that Hi=HSi .

Relative Homology

Suppose thatXis a topological space andA Xa subspace, with i : A X the inclusion. Thisobviously induces an injective chain map i# : Cn(A)Cn(X); well identify Cn(A) with its imageunder i#. There is an induced map on homology i :Hn(A) Hn(X). Note that i is often notinjective, for example with X= D2 and A = S1, we getH1(S

1) H1(D2) is a map from Z to 0.

Definition. The relative chain group is defined to be

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

Becausen onCn(X) preservesCn(A), we get an induced map n: Cn(X, A) Cn1(X, A). Then

C(X, A) ={Cn(X, A), n}

is a chain complex, and we define the relative homology group to be Hi(X, A) =Hi(C(X, A)).

A cycle Zn(X, A) is just a chain Cn(X) such that Cn1(A).

A

Definition. Suppose that A Y. Then A is a deformation retract ofY , denoted Y

A, if thereis a homotopy F :Y [0, 1] Y such that F0 = idY, F1(Y) A, and for all t [0, 1], we haveFt|A= idA.

Definition. A pair (X, A) of spaces is reasonable if there is a neighborhoodY A such thatY

A.This is the case for all -complexes. This rules out crazy things like the topologists sine curve.

XY

A

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Theorem (Homology of quotient spaces). Let (X, A) be a reasonable pair. Then the quotientP : (X, A) (X/A, A/A) induces an isomorphism for alli 0,

P: Hi(X, A) Hi(X/A, A/A)=Hi(X/A).Fundamental Theorem of Homological Algebra

Let

An+1n+1An

nAn1

be a sequence of abelian groups, or chain complexes, or R-modules, . . . with n appropriatemorphisms. The sequence is exact at An if im(n+1) = ker(n).

Example. A short exact sequence (SES) is a sequence

0 A B

C0

which is exact at A, B , andC. Being exact at A is equivalent to being injective, being exact at Cis equivalent to being onto. Key example:

0

Cn(A)

Cn(X)

Cn(X, A) 0

This is a SES of chain complexes.

Theorem (Fundamental Theorem of Homological Algebra). LetA, B, andCbe chain complexes,and suppose that

0 A B C 0

is a SES of chain complexes. Then there exists a connecting homomorphism: Hn(C) Hn1(A)and a long exact sequence of abelian groups

Hn(A) Hn(B)

Hn(C) Hn1(A)

Hn1(B)

Start of proof. We need to define : Hn(C) Hn1(A). The natural candidate is

Zn(C)

Zn1(A) Hn1(A)

Hn(C) =Zn(C)/Bn(C)

A short exact sequence of chain complexes just means a diagram like this:

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...

...

...

0 An+1

n+1

n+1

Bn+1n+1

n+1

Cn+1

n+1

0

0 Ann

n

Bnn

n

Cn

n

0

0 An1n1

n1

Bn1n1

n1

Cn1

n1

0

... ...

...

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Lecture 8 (2012-10-17)

Theorem (Fundamental Theorem of Homological Algebra). LetA, B, andCbe chain complexes,and suppose that

0 A B C 0

is a SES of chain complexes. Then there exists a connecting homomorphism: Hn(C) Hn1(A)and a long exact sequence of abelian groups

Hn(A) Hn(B)

Hn(C) Hn1(A)

Hn1(B)

Proof. Our SES of chain complexes gives us the following diagram:

0 An+1n+1

n+1

Bn+1n+1

n+1

Cn+1

n+1

0

0 Ann

n

Bnn

n

Cn

n

0

0 An1n1 Bn1

n1 Cn1 0

We want to define : Hn(C)Hn1(A), mapping a homology class [c] to a homology class [a].Note that Hn(C) =Zn(C)/Bn(C) and Hn1(A) =Zn1(A)/Bn1(A).

There are three key rules to use in any diagram chase:

1. 2 = 0

2. exactness

3. chain maps

Given ann-cyclec Cn, so that c = 0, the fact that is onto implies that there is a b Bn suchthat (b) =c. Note that

0 =c= (b) =(b) = b ker() = im().

Thus, there is an a An1 such that (a) = b. Check thata Zn1(A):

(a) =(a) = b = 0.

Butis injective, so a = 0.

Thus, given [c] Hn(C), we can choose a representative c inZn(C) and produce an a Zn1(A).Now we need to check that regardless of the representative we choose, the cycle a represents the sameclass in homology. In other words, we want to check that the proposed map : Hn(C) Hn1(A)is well-defined.

Thus, suppose instead ofc wed chosenc + as a representative of [c], where Cn+1. We wantto show that will map it to the same [a].

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Because is surjective, there is someb such that (b + b) = c + . Thus, in particular(b) = ,and hence

(b) = (b) = = 0.

This implies that b ker() = im(), hence b = (a) for some a. Our goal is to show thatthere is an a such thata =a.

Note that (b + b) =b + b = b + = b, hence (b) = 0.

To finish, note that

((a + a) a) =(a + a) (a) =(a) = b = 0.

Now we need to prove the exactness of the long exact sequence (LES). Lets just prove exactness at

Hn(B) for example. Given [b] Hn(B) such that ([b]) = 0, we want to show that there is some[a] Hn(A) such that ([a]) = [b].

We know that b = 0, and because ([b]) = 0 Hn(C), we have that (b) =c for some c Cn+1.

Thus (b) = (b) =c = 0, so that b ker() =im(), and hence there is some a

such that(a) =b.

Finish as an exercise.

Now lets apply this theorem to a pair of spaces (X, A). The SES of chain complexes

0 Cn(A) Cn(X) Cn(X, A) 0

produces a LES

Hn+1(X, A) Hn(A) Hn(X) Hn(X, A) Hn1(A)

Well prove later using this tool that if (X, A) is reasonable, then Hn

(X, A)

=H

n(X/A).

Example. Let (X, A) = (Dn, Dn) for n 1. Note that Dn = Sn1.

Letk >1. The LES says that

Hk+1(Dn, Sn1) Hk(S

n1) Hk(Dn) Hk(D

n, Sn1) Hk1(Sn1) Hk1(D

n)

Any homology of the disk for k >1 is 0, because it is contractible. We can even show that directlyin singular homology, because homology is a homotopy functor and the disk is homotopy equivalent

to a point, whose singular homology we can compute. Thus, we get for any k >1

0 Hk(Sn) Hk1(S

n1) 0

and this just says that Hk(Sn)=Hk1(S

n1).

Lets look at what happens at the end.

H1(Dn, Sn1) H0(S

n1) H0(Dn) H0(D

n, Sn1) 0

0 ifn>1Z ifn=1

Z ifn>1Z2 ifn=1 Z 0

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Our inductive claim is that for all n 1 and k >1,

Hk(Sn) =

0 ifk= 0, n

Z ifk= 0, n.

Here is a table: H0 H1 H2

S0 Z2 0 0

S1 Z Z 0

S2 Z 0 Z

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Lecture 9 (2012-10-19)

Last time we computed thatHi(Sn) = Z ifi= n,0 ifi=n, , and thatHi(Dn) = 0 for all i. Thus, as acorollary we can now conclude the Brouwer fixed point theorem.

This theorem is a big deal. Thurston used the Brouwer fixed point theorem to classify homeomor-

phisms of hyperbolic 3-manifolds up to homotopy by taking the space of hyperbolic metrics, whichany homotopy class of homeomorphisms acts on, and looking at the various places a fixed pointcould be.

Now well finally prove that Hi (X)=HSi (X).

LetXbe a -complex andAa subcomplex ofX. There exists a chain map : Ci (X, A) CSn (X, A)

sending to itself. This induces a homomorphism : Hn(X, A) H

Sn (X, A).

Theorem. For all(X, A) andn 0, is an isomorphism.

Corollary (Amazing). Hn(X, A) doesnt depend on which-complex structure you choose!

Proof. Well do this for finite-dimensional X and A = . Read about other cases in Hatcher.

LetXk = k-skeleton ofX = {the k-simplices i: k X}.

We have a LES of the pair (Xk, Xk1) in each homology theory, and a commutative diagramcomparing them because is a chain map:

Hn+1(Xk, Xk1)

Hn(Xk1)

Hn(Xk)

Hn(Xk, Xk1)

Hn1(Xk1)

HSn+1(Xk, Xk1) HSn (X

k1) HSn (Xk) HSn (X

k, Xk1) HSn1(Xk1)

We want to induct on k to prove that Hn(Xk)

=HSn (X

k). If this is true for all k, then weredone, because (by our assumption that X is finite-dimensional), we have X=Xk for some k . Theclaim is true for k= 0 because we know how to look at points. By our inductive hypothesis, the2nd and 5th vertical arrows in our diagram are isomorphisms.

Now we will prove that the 1st and 4th vertical arrows are isomorphisms. Intuitively, taking thek-skeleton ofX(which is just a gluing ofk-simplices) and quotienting by the (k 1)-skeleton isjust crushing the boundaries of all of thek-simplices, turning the k-skeleton into a union and/orwedge of spheres.

We have that

Cn(Xk, Xk1) = 0 ifn =k,Zd ifn = k.whered = #(Xk \ Xk1). Thus, Hn(X

k, Xk1) is the same. Now we want to find HSi (Xk, Xk1).

Recall that Xk = {i : k Xk}. Define the map :

i(

k, k) i (Xk, Xk1). Then

induces a homeomorphism k/

k

=Xk/Xk1

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essentially by the definition ofXbeing a -complex. This is the same as our proof that

(Dn, Dn) (Dn/Dn, Dn/Dn) = (Dn/Dn, pt)

induces an isomorphism in homology. This is the same map that were using in our big diagram,so that map is the isomorphism (a priori, just knowing the domain and codomain are isomorphicdoesnt tell us that our map is an isomorphism).

Lemma (Five Lemma). Given a map between exact sequences

A

B

C

D

E

A B C D E

where, ,, are isomorphisms, thenis an isomorphism.

Now were done, as long as we can prove formally

Theorem (Excision). Given a topological spaceX andA, UX subspace withUint(A), theinclusioni: (X/U, A/U)(X, A) induces an isomorphism on homology.

Proof. Well do the case when -complex and each ofA, X U, A U is a subcomplex ofX.

We get chain mapsCn(X U) Cn(X) Cn(X, A)

and call the composition . The map is surjective, which is clear - any chain inXwithout anypart in A certainly is a chain in Xwithout any part in U. Then induces an isomorphism.

Cn(X U, A U) def

= Cn(X U)

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

Thus, we have proven that -complex homology and singular homology are isomorphic.

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Lecture 10 (2012-10-22)

The midterm will be on November 2nd in class; it will be closed book, and cover everything up tothe class before that.

Theorem (Homology of quotients). LetXbe a topological space, andA Xa reasonable subspace.

Then the quotient map p: (X, A) (X/A, A/A) induces for alli 0 an isomorphism

p: Hi(X, A) Hi(X/A, A/A) =from homework

Hi(X/A).This theorem is not as powerful as it may seem, because it only allows us to study smashing thesubspaceAto a point; we often want to glue two subspaces together without turning them into asingle point.

Proof. We will do the case thatAis open.

As always, we get the induced map p :Hn(X, A) Hn(X/A, A/A).

Hn(X, A)

p

=excision

Hn(X A, A A) =H n(X A)

(p|XA)

Hn(X/A, A/A) =Hn(X/A A/A)

The key is that (p|XA) is a homeomorphism onto its image.

We want to be able compute the homology of RPn, CPn, a knot S3 K, and other interestingspaces.

Two applications of what weve done so far include the Euler characteristic and the Lefschetz Fixed

Point Formula.Definition. Let X be a finite -complex. Let cn(X) denote the rank of C

n(X). The Euler

characteristic ofX is(X) =

n0

(1)ncn.

Lets look at some -complex structures on the disk:

0-simplices 1-simplices 2-simplices #0 #1 + #2

3 3 1 15 8 4 14 5 2 1

Theorem (Topological invariance of). LetXbe a finite-complex. Then

(X) =n0

(1)nbn(X)

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wherebn(X) is thenth Betti number ofX,

bn(X) = rank(Hn(X)/Tn(X)).

In particular, because XY = bn(X) =bn(Y) for all n 0, the right side, and hence the leftside, depends only on the homotopy type ofX.

The essential idea is that (letting lowercase letters denote the ranks of the groups typically denoted

by the corresponding uppercase letters):

hn= zn bn, cn= zn+ bn1

and using this in the alternating sum makes the proof pop out.

Hopf Trace Formula

LetC= {Cn, n} be a finite chain complex of finitely generated free abelian groups. Let f :C Cbe a chain map. We know that Cn= Z

d for some d, so picking a basis for Cn, we can think of eachfn Md(Z). Then tr(fn) is the trace of this matrix. Recall that

tr(ABA1) = tr(B),

so tr(fn) doesnt depend on choice of basis.

We also have that (fn) acts on Hn(C)/Tn(C), where Tn denotes the torsion subgroup ofHn(C).Thus, we can also consider tr((fn)).

Theorem (Hopf Trace Formula). We haven0

(1)ntr(fn) =n0

(1)ntr((fn))

Remark. LetCn= Cn(X), and f= (idX)#. Then we get that

n0

(1)n rank(Cn(X)) = n0

(1)nbn(X).

Proof. We have Bn Zn Cn, all free abelian groups. First, pick a basis 1, . . . , r for Bn;extend to (not a basis; a maximal Z-linearly independent set) z1, . . . , zs Zn; then extendto a basis ofCn. Thus, we have a basis for Cn that weve broken up into boundaries, plain oldchains, and cycles.

We want to compute, for each element of the basis above, the coefficient ofv in fn(v). Lets callthis (fn(v)).

Becausefn is a chain map,(fn(n+1j)) =(fn+1(j))

Then n0

(1)ntr(fn) =n0

(1)nn

k=0

(f(zk))

because all other terms cancel in pairs. Because the action off on [zi] Hi is just sending it to[f(zi)], the coefficents of the action off and off itself are the same. Thus, the left side is equal to

n0

(1)ntr((fn)).

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Lecture 11 (2012-10-24)

Neat fact: given closed A1, . . . , An+1 Sn such that

Ai = S

n, there exists a j such that Ajcontains a pair of antipodal points.

The Lefschetz Fixed Point TheoremLast time, we talked about the Hopf trace formula, where we had a complex C={Cn, n} and achain map : C C, which induces maps (n) : Hn(C)/Tn(C) , and deduced that

(1)ntr(n) =

(1)ntr((n)).

Lets do some setup first. Let Xbe a finite -complex, and let f :XXbe a continuous map.Consider (fn): Hn(X)/Tn(X) .

Definition. The Lefschetz number off is

(f) = n0

(1)n

tr((fn)).

Theorem (LFPT). If(f)= 0, thenfhas a fixed point.

Corollary. LetXbe a contractible-complex, e.g. X= Dn. Then anyf : XXhas a fixedpoint.

Proof of Corollary. By hypothesis, we have that Xis aspherical, i.e.

Hn(X) =

Z ifn = 0,

0 ifn >0

This implies that (fn) = 0 for all n >0, and hencetr((fn)) = 0 for all n >0. We also have (fromyour current homework)

tr((f0)) = tr(id) = tr([1]) = 1.

Thus,(f) = 1 0 + 0 = 1= 0.

Proof of Lefschetz. Suppose that f(x)=x for all x X.

Put a metric d on X. There is some > 0 such that d(f(x), x) > for all x Xbecause f iscontinuous, f(x)=x for all x, and because Xis compact. Now, by barycentrically subdividing, wecan put a -complex structure on Xsuch that diam()< d100 for all simplices .

By simplicial approximation (which we havent covered in detail, but you should know the statement

at least), we have that, possibly after further subdividing, there is a simplicial map h : XX suchthat h f andd(h(x), f(x)) 2 for allx X.

Thus, for allx X, we haved(h(x), x)> 10 > 100 , and becausehis simplicial, we haveh()=

for all simplices .

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We claim that this impliestr((hn)#) = 0 for all n 0. The matrix for (hn)# acting onCn(X)= Zd

looks like

1 2 d1 02 0...d 0

so, applying the Hopf trace formula,

(h) =

(1)ntr((hn)) =

(1)ntr((hn)#) = 0.

Butfh, so f = h, and so (f) = (h) = 0.

Corollary (of Brouwer). LetA be ann n matrix with real entriesaij >0. Then there exists areal eigenvalue with a real eigenvector(v1, . . . , vn) with allvi> 0.

Proof. This has applications to the adjacency matrix of a graph, and to probability.

Let X={rays through 0 in a positive orthant}, which is homeomorphic to n1. We have that AmapsX to Xby hypothesis. Then Brouwer implies that there exists an x Xsuch that Ax = x,i.e. there is a ray Rx such that Av = v Then we must have >0 and vi> 0 for all coordinates ofv = (v1, . . . , vn).

Remark(Weils Observation). Suppose thatf :XXis a homeomorphism of compact manifolds.

Assumption: Suppose that (f) = # of fixed points off, and that (fn) = # of periodic points offof period n, i.e. points with fn(x) =x.

Claim: There exist algebraic integers 1, . . . , r, 1, . . . , s such that

(f

n

) = # of periodic points offof period n= ni nj.The amazing thing is that the left side is obviously an integer, but the right side rarely is if youchoose arbitrary i andi.

The proof is just that

(fn) =i

(1)itr((fni ))

where (fni ) : Hn(X)/Tn(X) , and Hn(X)/Tn(X)= Zd, so that (fi) Mdd(Z) and hence its

eigenvalues1, . . . , s are algebraic integers, so that tr((fni )) =

ni.

Moreover, he observed that, ifXis a projective smooth variety defined over Z, then

X(Fpn) = ni njand he conjectured there was a relationship. Indeed, X(Fp) Frobp, and fix(Frob

np) =X(Fpn).

(1)i Frobnp

Hietal(X) =Hi(X(C))

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Lecture 12 (2012-10-26)

As a reminder, the midterm is Friday, November 2, 11am - 12:30pm.

Maps of Spheres

A mapf : Sn Sn induces a mapf: Hn(Sn) Hn(Sn). Recall thatHn(Sn)= Z, having [1 2]as a generator, where 1 and 2 aren-simplices glued along their boundaries to form the equatorof the sphere.

Definition. The degree offis the uniquedeg(f) Zsuch that f(z) =deg(f)z for allz Hn(Sn).

Note that

deg(f g)z= (f g)(z) =f(g(z)) =f(deg(g)z) = deg(g)f(z) = deg(f)deg(g)z,

and hencedeg(f g) =deg(f) deg(g). Also, iffg, thenf= g, and hence deg(f) =deg(g). Infact, the converse is true; that is, considering the map

{f : Sn Sn}/ Z

sending [f] to deg(f), this map is a bijection. This is an old theorem:

Theorem (Hopf). For anyf, g: Sn Sn, thenfg deg(f) = deg(g).

Here are some basic properties.

Iffis not surjective, then deg(f) = 0. This is because there is an x such that f(Sn) Sn x,and Sn x= Rn, so the fact that we have a commutative diagram

Hn(Sn) Hn(S

n)

Hn(Sn x)

=Hn(Rn) = 0

f

f

forcesf= 0.

We can consider the suspension of a map between spheres, f : Sn Sn. Note thatSn= Sn+1 (this is clear if you draw a picture). We get a commutative diagram

Hn(Sn) Hn+1(S

n+1)

Hn(Sn) Hn+1(S

n+1)

i

=

f f

implying that deg(f) = deg(f).

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Let d 1. Let d : S1 S1 be the map z zd. We give S1 two different -complex

structures,

X=

1

2

d

Y =

Then d(i) = for all i, so that the map

(d): H1(X) di=1 i

H1(Y) []satisfies

(d)([d

i=1 i]) = [(d)#d

i=1 i]

= [d

i=1 d(i)]

= [d

i=1 ]

=d[]

and hence deg(d) =d. Note that we implicitly used that this diagram commutes:

H1 (X) H1 (Y)

HS1(S1) HS1(S

1)

f

= =

fS

but this is just because (the isomorphism between simplicial and singular homology) isnatural, i.e. the composition of degrees doesnt depend on the chosen -complex structure.

Thus, for all n 1 and d Z there is a map f : Sn Sn of degree d, namely n1d.

In differential topology, while you can run into trouble with a few bad points, you cangenerically get the degree in a similar way. For example, consider the suspension d, which isthe map from S2 to S2 wrapping the sphere around itself so that a lune of angle 2

d is wrapped

over the entire sphere.

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Then the preimage of almost every point is a set ofd points, except the north and south poles,which have only one preimage (they are preserved).

Let Sn ={(x1, . . . , xn+1)|

x2i = 1} Rn+1. Let r : Sn Sn be the reflection map

r(x1, x2 . . . , xn+1) = (x1, x2, . . . , xn+1).

We claim that deg(r) =1. Here is the proof: consider the -complex X= n1 n2 where

n1 =n2 . Note thatX

= Sn. Then the induced map r#: Cn(X) Cn(X) just swaps thesimplices, 12 and 21. Also, note that Hn(X)= Z =[1 2]. Thus,

r([1 2]) = [r#(1 2)] = [2 1] =[1 2]

and hence deg(r) =1.

As a corollary, this implies that the antipodal map A(x1, . . . , xn+1) = (x1, . . . , xn+1) hasdegree (1)n+1, because

deg(A) = deg(r1

r2

rn+1

) = deg(r1

)deg(r2

) deg(rn+1

) = (1)n+1,

and hence AidSn when nis even, and A idSn when nis odd.

Which maps f : Sn Sn have no fixed points? Well, the antipodal map A, obviously; any others?

The Lefschetz number of any map f : Sn Sn is

(f) = tr(f: H0(Sn) )

= 1

+(1)n deg(f)

Note that iff has no fixed points, then we must have (f) = 0, hence deg(f) = (1)n+1, andthereforefA by Hopfs theorem. Thus, f : Sn Sn has no fixed points implies that fA.

Definition. Consider Sn1 as a subset ofRn. A continuous vector field on Sn1 is a continuousmap v: Sn1 Rn such thatv(z) z for all z Sn1.

Theorem. Then-sphere Sn, forn 1, admits a nowhere zero vector fieldv n is odd.

Proof. To see = , consider v(x1, . . . , xn+1) = (x2, x1, x4, x3, . . . , xn+1, xn). We needed nodd to be able to pair off the coordinates.

To see =, use the vector field v to prove idSn1 A. Ifv is our nowhere vanishing vector field,then consider the map vt: S

n1 I Sn1 defined by

vt(z) = cos(t)z+ sin(t)v(z).

As t goes from 0 to 1, this map slides a point z Sn along the great circle connecting z to zin the direction determined by v(z). This is a homotopy from idSn1 to A, sodeg(idSn1) = 1 =(1)n+1 = deg(A), hence n is odd.

Theorem (Adams, 1962). Let n 1, and write n+ 1 = 24a+b(2k+ 1) wherea,b, k Z, and0 b 3. Then the maximum number of linearly independent vector fields on Sn is precisely2b + 8a 1.

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Lecture 13 (2012-10-29)

CW-Complexes and Cellular Homology

Definition. CW-complexes are built up in an inductive process. The base of the inductive process

is X(0)

, the 0-skeleton, consisting of discrete points. The inductive step is as follows: assuming wehave constructed X(n1), we are given

a collection ofn-balls {Dn| I},

maps {: DnX

(n1)} attaching the boundaries of the balls to the (n 1)-skeleton.

Then we define

X(n) = X(n1)

ID

n

x (x) for all I , x Dn.

The spaceX=

n0 X(n) built in this manner is called a CW-complex. When X=X(N), we say

that X is N-dimensional.

Note that we give Xthe weak topology, i.e. we declare A Xto be open precisely when 1

(A) Dn

is open for all cells : Dn X.

Examples.

A 1-dimensional CW-complex is equivalent to a graph. Here is an example:

X0D1

D1

D1

X1

Then-sphere Sn can be given a CW-complex structure with one 0-cell p and one n-cell, byletting : Dn {p}be the constant map.

p p p

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The surface g of genusg 1 can be given a CW-complex structure with one 0-cell v, 2g1-cellsa1, b1, . . . , ag, bg, and one 2-cell : D

2 X(1) defined by the worda1b1a11 b

11 agbga

1g b

1g .

The 1-skeleton just looks like

a1

b1a2

b2

ag

bg

.

.

.

We attach D2 to the 1-skeleton X(1) as follows. We break up D2 into 4g arcs of angle 24g ,

D2

a1

b1

a1b1

and define by mapping an arc to the indicated edge in the 1-skeleton, with degree 1depending on the sign determined by the word (this is reflected in the orientations of the arcs

in the picture).

Real projective space RPn is defined to be

{lines through origin in Rn+1}= Rn+1 \ {0}uv i ffu=rv

for some rR

={v Rn+1 \ {0}: v= 1}

v v =

Sn

v v.

Denoting the upper hemisphere ofSn by Dn+,

Dn+

then we can also see that

RPn = Dn+

v v for all v Dn+.

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Note that Dn+ = Sn1. We have the quotient map : Sn1 RPn1 identifyingv and

v for all v Sn1; this is just the attaching map for the n-cell Dn+ when we give RPn a

CW-complex structure.

In general, RPn is a CW-complex with 1 i-cell fori= 0, 1, . . . , n. Thus, RP0 is just a point,RP1 is S1, and

RPn = RPn1 Dn+

v (v) for all v Dn.

How to Compute CW Homology

LetXbe a CW-complex. We define a chain complexCCW(X) ={CCWn (X), dn} by letting CCWn (X)

be the free abelian group on the n-cells, and the boundary maps dn :CCWn (X) C

CWn1(X) satisfy

dn1 dn= 0, but well skip their definition to get to some calculations.

Of course, lastly we define HCWi (X) =Hi(CCW(X)).

Examples.

Lets compute the CW homology ofSn, for n 2. The CW chain complex is just

CCWn+1(Sn) CCWn (S

n) CCWn1(Sn) CCW0 (S

n) 0

0 Z 0 Z 0

and therefore

HCWi (Sn)=CCWi (S

n) =

Z ifi= 0 or n,

0 otherwise.

This agrees with singular, and hence also simplicial, homology.

The CW chain complex for g is

0 Z Z2g Z 0

=

f=

a1, b1, . . . , ag, bg=

v

d2 d1

Everyai and bi is sent tov v= 0, so that d1= 0. Now note that d2 sends f to

a1+ b1 a1 b1+ a2+ + ag+ bg ag bg = 0

and therefore d2= 0. Thus,

HCWi (Sn)=CCWi (S

n) =Z ifi = 0 or 2,Z2g ifi = 1,

0 otherwise.

Morally, the boundary map in CW homology is measuring degree of the attaching maps ofthe cells.

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Remark. By the topological invariance ofHCWn and the Hopf trace formula, we can compute

(X) =n0

(1)n rank(CCWn (X)).

For example,

(Sn) = 1 + (1)n = 0 ifn odd,2 ifn even.

(RPn) = 1 1 + 1 + (1)n =

0 ifn odd,

1 ifn even.

(g) = 1 2g+ 1 = 2 2g.

To compute HCWn (X), we need to understand the boundary maps dn. We have

CCWn (X) = Z-span of{: Dn X

(n)}

CCW

n1(X) = Z-span of{ : Dn1 X

(n1)

}Consider the composition of the attaching map of an n-cell with the quotient map sendingX(n1) toX(n1)/X(n2):

Dn

1(Dn1)

2(Dn1)

=X(n1)/X(n2)

Then, for all , consider the composition of that with the quotient map from X(n1)/X(n2) to justone of the (n 1)-cells :

Sn1 =Dn X(n1) X(n1)/X(n2) (D

n1) = Sn1|Dn

quotient

morequotient

This is a map ,= Sn1 Sn1.

Lemma (Hatcher, p.141). For all, , we have

dn() =

d,

whered,= deg(,).

Lets work out the homology of the torus this way. Let X= T2. The CW chain complex is

C2 C1 C0

Z Z2 Z 0d2 d1

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We obviously have d1 = 0. Now lets think about a. It may be helpful to note that the mapsendingD2 toa(D

1) factors through the quotient collapsing the b arcs:

D2ab

a b

a

a

ba

We clearly have that deg(a) = 1 1 = 0, and similarly, deg(b) = 0. Therefore d2 = 0, and

HCWi (T2)=CCWi (T

2) =

Z ifi= 0, 2

Z2 ifi= 1

0 otherwise.

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Lecture 15 (2012-11-02)

Midterm in Ryerson 352 (the Barn) from 11am to 12:30pm.

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Lecture 16 (2012-11-05)

Today well discuss Mayer-Vietoris, which is one of the most powerful computational tools we have,

and well be moving to cohomology soon.

LetXbe a space and consider injective maps i: A X andj : B X(we may as well consider

Aand B as subsets ofX) with the property that A B=X, and (X, A) and (X, B) are good pairs.

Theorem (Mayer-Vietoris). There is a long exact sequence

Hn(A B) Hn(A) Hn(B) Hn(X) Hn1(A B)

Proof. There is a short exact sequence of chain complexes

0 Cn(A B) Cn(A) Cn(B) Cn(A + B) 0

z (i(z), j(z))

(x, y) x y

i# j#

where Cn(A + B) is the subgroup ofCn(X) consisting of chains of the form

aii+

cii wherei Cn(A) and iCn(B).

The fundamental theorem of homological algebra implies that we get a long exact sequence

Hn(A B) Hn(A) Hn(B) Hn(A + B) Hn1(A B)

whereHn(A + B) is the nth homology of the chain complex {Cn(A + B)}.

The key claim is that the inclusion h: Cn(A + B) Cn(X) induces an isomorphism on homology,

h : Hn(A + B)=

Hn(X) for all n 0. The idea of the proof of this key claim is as follows.

IfXis a -complex and we can triangulate Xsuch that the triangulation restricts to triangulationson A,B, and A B, then the key claim is easy since Cn(A + B)

=Cn(X). In general, we need

to subdivideX (use simplicial approximation), because for example, wed run into trouble with asimplex like this:

A B

Applications.

Let X = Sn, A = Dn+, B = Dn, so that A B = S

n1 (really, we mean a little regularneighborhood around these subsets).

A

B

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Lecture 17 (2012-11-07)

Cohomology

Fix an abelian group G, for example G= Z, R, Q, or Z/nZ.

For any abelian group A, G =Hom(A, G) is an abelian group under pointwise addition. For anyhomomorphism: A A, there is a dual homomorphism : (A) A defined by ,and for any f :A A andg: A A, we have (g f) =f g. Lastly, (idA)

= idA .

Thus, we have proved that Hom(, G) is a contravariant functor abelian groups

and homomorphisms

abelian groups

and homomorphisms

which acts on objects by A A, and acts on morphisms by (f :A B)(f :B A).

Given a chain complex C={Cn, n} of free abelian groups, its nth homology Hn(C) measures howfar the complex is from being exact at C

n.

We will now form a new chain complex C ={Cn, n}, where the n are just the duals of the n.Specifically, looking at n :CnCn1, we set n1 = (n)

:Cn1 Cn. Wed like to take the

homology ofC, but its going the wrong way:

C2 C1 C0 02 1 0

Note that ( ) = ( ) = 0, so we will call the s coboundary maps, and we call C a cochaincomplex. The group Cn is called the group ofn-cochains.

Definition. Given a chain complex C = {Cn, n} and abelian group G, the associated cochaincomplexC ={Cn, n}is defined as above, and the cohomology ofCwith coefficients inG is defined

to beHn(C; G) =

ker(n)

im(n1).

Given a chain map f : C C, we get an induced map on cohomology in the other direction,fn :H

n(C; G) Hn(C; G). This is because

f = f = f = f =f =f .

This also implies that iff is chain homotopic to g , then f =g.

Theorem (FTHA, cohomology version). Given a short exact sequence of chain complexes

0 A B C 0

there is a long exact sequence

Hn+1(C) Hn(A) Hn(B) Hn(C)

Next class we will prove the universal coefficient theorem, which will allow us to compare homology

with cohomology. Lets do an example first though. FixG= Z, and consider this chain complex:

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0 Z Z Z 0 2 0

The dual of the zero map is obviously the zero map. Note that Hom(Z, Z)= Z, with the identitymap idZ as the generator. It is easy to see that (2) : Z Z induces the map from Hom(Z, Z) toHom(Z, Z) which is also multiplication by 2.

Thus, corresponding cochain complex is

0 Z Z Z 0 2 0

Therefore, the homology and cohomology are

Hi(C) =

Z i= 0 Z

Z/2Z i= 1 00 i= 2 Z/2Z

=Hi(C, Z)So in general, how can we compare H(C; G) with H(C)?

We claim that there is a homomorphism : Hn(C; G) Hom(Hn(C); G). We construct it as follows.Given a Zn C

n =Hom(Cn, G), we have that = 0 C

n+1. Thus, for any Cn+1, we

have that ()() = ( )() =() = 0, so (Bn) = 0. Therefore, : Cn G factors througha map: Cn/Bn G, and then restricting to Zn, we get a map|Zn :Zn/Bn G, i.e. a mapfromHn(C) to G, which is exactly the kind of object we are claiming sends to.

You should check on your own that is well-defined, and that in general may have a kernel.

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Lecture 18 (2012-11-09)

The Universal Coefficient Theorem

Fix an abelian group G, and let C= {Cn, n}be a chain complex. As we discussed last time, this

gives us a cochain complex C

={Cn,

n

=n1}, and then the cohomology groups H

n

(C; G) withcoefficients in G.

Last time, we defined a map :Hn(C; G) Hom(Hn(C), G). It is easy to show that is onto, sowe get a short exact sequence

1 ker() Hn(C; G) Hom(Hn(C), G) 1

We want to compute ker(), since it represents the difference between homology and cohomology.

You should read in Hatcher about the functor Ext(, G), which goes from Ab to Ab, and also readabout free resolutions. The key step will be showing that ker() = Ext(Hn1(C), G). This thenimples the universal coefficient theorem, because we get a short exact sequence

1 Ext(Hn1(C), G) Hn(C; G) Hom(Hn(C), G) 1

The following proposition explains how to compute Ext(, G).

Proposition.

1. Ext(A B, G) = Ext(A, G) Ext(B, G).

2. Ext(A, G) = 0 ifA is a free abelian group.

3. Ext(Z/nZ, G) =G/nG, wherenG= {ng| g G}.

In fact, when G = Z, we have

Ext(Hn1(C, G))= Hn(C)

torsion(Hn(C)) torsion(Hn1(C)).

Remark. The functorExt(A, G) is called that because it has to do with extensions ofG by A, i.e.ways of picking a group so that

1 A G 1

Thus, cohomology is a key tool for the classification of groups.

Cohomology of Spaces

For any topological space X (respectively, any -complex, CW-complex), we define

Hn(X; G) :=Hn({Csingn (X)})

HnCW(X; G) :=Hn({CCWn (X)})

Hn(X; G) :=Hn({Cn(X)})

Then because the universal coefficient theorem is natural in C, and the isomorphisms between thevarious homology theories are natural, we can conclude that

Hn(X; G)=Hn(X; G)=HnCW(X; G)

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when they are all defined (i.e. when Xis a -complex, CW-complex). Moreover, naturality togetherwith the corresponding fact about homology implies that fg we have f =g (so that Hi is ahomotopy functor), and we even get the long exact sequence of a pair, excision, and Mayer-Vietoris

for cohomology, all from naturality.

Cup ProductFix a ring R (e.g. R = Z, Q, Z/dZ, R, Qp, . . .). Let X be a space. Considering the underlyingadditive abelian group ofR, we get the abelian groups Hn(X; R) for alln 0, and we let

H(X; R) =n0

Hn(X; R)

which is a (graded) abelian group. Our goal is to make H(X; R) into a ring. This additional ringstructure on cohomology will let us distinguish S1 S1 S2 and T2, even though

Hi(S1 S1 S2) =

Z i= 0Z2 i= 1Z i= 2 =Hi(T2).

Fora Ci(X; R) =Hom(Ci(X); R) and b Cj(X; R), for any (: [v0v1 vi+j ] X) Ci+j(X)

we define a b Ci+j(X; R) by

(a b)() =a(|[v0v1vi]) mult. in R

b(|[vivi+j ]).

We have the graded product rule for :

(a b) =a b + (1)ia b

This implies that if a= b= 0, then (a b) = 0 and a b= a b = 0, so that the cup productmap

:Ci(X; R) Cj(X; R) Ci+j(X; R)

induces a map in homology

:Hi(X; R) Hj(X; R) Hi+j(X; R),

defined by [a] [b] = [a b]. This is a bilinear map, so passing to the tensor product and puttingall degrees together, we get a homomorphism

:H(X; R) H(X; R) H(X; R),

In summary, H(X; R) is a graded commutative ring a b= (1)ijb a, where addition is justaddition of cocycles, and multiplication is . For any map f : X Y, we get an induced ringhomomorphismf(a b) =f(a) f(b).

The additive and multiplicative identity are the constant functions 0, 1 H0(X; R) =Hom(C0(X); R).

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Lecture 19 (2012-11-12)

Today well look at the cup product structure on the cohomology of a few spaces.

Lets start by considering S1 S1 S2 vs. T2.

Both of these spaces have CW-complex structures with one 0-cell, two 1-cells, and one 2-cell. Thedifference is that the 2-cell was attached via a homotopically trivial map in S1 S1 S2, whileit was attached via the the path aba1b1 in T2. While homotopically these are different, on thelevel of homology they cannot be distinguished because H1 is the abelianization of the fundamentalgroup. Surprisingly, the difference can be detected in the cup product structure.

The homology of both of these spaces is

H(S1 S1 S2)=H(T

2) =

Z in dim 0

Z2 in dim 1

Z in dim 2

0 in higher dim

Because all of these groups are free, the Ext groups in the universal coefficient theorem are trivial,and therefore the cohomology of both spaces is isomorphic to the homology.

What about the cup product structure on H? The only interesting case is in H1. We have thatH1 = Z2 =Hom(H1, Z) is generated by two 1-cocycles and, where (a) = 1 and (b) = 0, and(a) = 0 and (b) = 1.

By the anti-commutativity of the cup product in odd dimensions, for either space we have =( ) and =( ), and because H2 has no 2-torsion, this implies = = 0for either space.

What about ?

We map X= S1 S1 S2 down toY = S1 S1 by collapsing the 2-cell to a point:

b

a

ba

f

X

Y

We have that

H(Y) =

Z in dim 0

Z2 in dim 1

0 in higher dim

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Let , H1(Y) be generators ofH1(Y)= Z2. Clearly, = 0 because Yhas no cohomology indimension 2. Because we have

f()(a) =(f(a)) =(a), f()(b) =(f(b)) =(b)

and similarly with and , we can conclude that

0 =f(0) =f( ) =f() f() = .

Now lets look at T2. The lines a and b represent homology classes which will generate H1(T2), andwe will also draw lines a andb which will let us define the cocycles which are dual to [ a] and [b].

ab

b

a

We define cochains , by

() = (signed) # of intersections ofa ,

() = (signed) # of intersections ofb .

a

+ +

We claim that these cochains are actually cocycles. Looking at the intersections ofa with the facesof a 2-simplex,

0 2

1

+

+

+

++

+

and adding up all the signs on [01] + [12] [02], by the classification of 1-manifolds we know thereare an even number of intersections and the sum must cancel.

Thus, and represent classes [], [] H1(T2). We have

[]([a]) = 1, []([b]) = 0, []([a]) = 0, []([b]) = 1.

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Triangulating the torus, and drawing a and b ,

b

a

Recall that, by definition, we compute

by having eat the front face of a triangle, and eatthe back face, i.e. (T01)(T12). We will compute ( )[T

2] by adding up the contribution from

all of the triangles in our triangulation.

However, in all but two triangles, there is no contribution because either a or b does not passthrough at all, so that or vanish. The only interesting piece is

00 1

1 22

b

a

1

+1

We see that([] [])([T2]) = (1) [(+1)(1)] + (+1) [(0)(0)] = 1,

which isnt 0, so the cup product structures on S1 S1 S2 and T2 are different.

Now lets do a similar calculation for Sg, the closed oriented surface of genus g .

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We know that

H(Sg) =

Z in dim 0

Z2g in dim 1

Z in dim 2

0 in higher dim

Taking inspiration from what we did with the torus, we define analogous curves a1, . . . , ag and

b1, . . . , bgon Sg, and then we define cocyclesiandiwhich count the signed number of intersections

with a i or bi, respectively. Thus

i(ai) = 1 i(bi) = 1

i(bj) = 0 for all j i(aj) = 0 for all j

i(aj) = 0 for all j =i i(bj) = 0 for all j=i

To compute cup products, we choose a triangulation, so that the fundamental class [Sg] can berepresented by a sum of triangles. Then

([i]

[i])([Sg]) = 1, [i]

[j] = 0, [i]

[j] = 0, [i]

[j ] = 0 for i =j.This is a non-degenerate, alternating (i.e. antisymmetric) pairing on H1 (such a form is calledsymplectic). This is a general phenomenon - in even-dimensional manifolds, the cup product onmiddle-dimensional cohomology gives a nice pairing to the top-dimensional cohomology (which is

just Z when it is orientable). If the dimension of the manifold is 0 mod 4, then the middle dimensionis even, and we get a symmetric form; if the dimension of the manifold is 2 mod 4, then the middle

dimension is odd, and we get an anti-symmetric form.

Lets finish with RPn. So far, weve been working with 2-dimensional manifolds, so looking forthings that 1-dimensional submanifolds intersect with gave us other 1-dimensional submanifolds. In

general, we need to look at 1-codimensional submanifolds.

Because RPn is non-orientable, we cant talk about the signed number of intersections, but we canwork in Z/2Z.

ViewingRPn as Dn/(1on Dn), there aren different copies ofRPn1 in RPn, each perpendicularto one of the coordinate axes. Choose one, which we will call a, and define by

() = # of intersections of with a (mod 2)

In particular, () = 1= 0 (mod 2), where is the coordinate axis perpendicular to the chosencopy ofRPn1.

a

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Lecture 20 (2012-11-14)

Kunneth Theorem

Our goal is to computeH(X Y; R) in terms ofH(X; R) andH(Y; R). Recall that the product

X Ycomes with natural projections

X Y

X Y

P2P1

Definition. The product on cohomology, : H(X; R) H(Y; R) H(X Y; R), is definedby sendinga b to P1 (a) P

2 (b).

The map is bilinear, and induces a homomorphism ofR-modules

: H(X; R) RH(Y; R) H(X Y; R).

In fact, we can make this into a homomorphism of graded rings by defining

(a b) (c d) := (1)|b||c|(ac bd)

where|b| and|c| are the degrees ofb and c.

Theorem (Kunneth). IfX andY are CW-complexes, and ifH(Y; R) is a freeR-module, thenis an isomorphism.

Example. The cohomology ring ofTn over Z is

H(Tn, Z)=[x1, . . . , xn] =ni=0

i[x1, . . . , xn]

where i[x1, . . . , xn] is the free abelian group on

{xi1 xik |i1< < ik},

so that rank(i[x1, . . . , xn]) =ni

.

Proof. We induct using Kunneth:

H(Tn, Z)=H(Tn1 S1; Z) =H(Tn1) H(S1).

For example,

H3(T4) =H3,0 H2,1 H1,2 H0,3

= [H3(T3) H0(S1)] [H2(T3) H1(S1)] [H1(T3) H2(S1)] [H0(T3) H3(S1)]

= [Z Z] [2Z3 Z] 0 0= Z 2Z3.

When is an isomorphism, we have

Hn(X Y; Z) =

p+q=n

[Hp(X) Hq(Y)].

In Hodge theory, we see that there is a decomposition like this even when were not looking at aproduct manifold.

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Prologue to Poincare duality

Definition. A second countable, Hausdorff space M is an n-dimensional manifold (possibly wthboundary) if for anym M, there is a neighborhood Um such that either Um= R

n=Bn (the openball in Rn), or Um= R

n+= {(x1, . . . , xn)| xi 0} (the closed upper half space). Key fact:

{m Mfor which the latter holds}= an (n 1)-dimensional manifold M.

Definition. A compact manifold Mis called triangulable if there is a simplicial complex K=M.Most of the manifolds which occur in real life are triangulable, and we might as well assume that

all manifolds well use in class will be triangulable.

Definition. An orientation on a simplex [v0 vn] is a choice of one of the two equivalence classesof (n + 1)-tuples (vi0 vin), where two (n + 1)-tuples are equivalent when the permutation sendingone to the other is an even permutation.

Definition. Mis orientable if one of the following equivalent statements holds.

1. There is a triangulation on Mwhich can be oriented compatibly.

2. All triangulations onMcan be oriented compatibly.

We know that under any reasonable definition, a disk ought to be orientable; giving the disk atriangulation and assigning the same orientation to each simplex, we get

This motivates the definition of compatible orientation to be a choice of orientation on each n-simplexi such that ifi j is a (codimension 1) face of each, theni andj induce opposite orientations

on that face.

Exercise.

1. Prove that g is orientable for all g 1.

2. Prove that the Mobius strip and Klein bottle are not.

3. How about RPn?

Theorem (Poincare duality). Let M be a closed (i.e. compact, with no boundary), orientablemanifold of dimensionn. Then

Hi(M; R)=Hni(M; R).

Corollary. IfM is a closed manifold of odd dimension, then(M) = 0.

Proof of corollary. Letbi= dim(Hi(M; R)). Then we know bi= dim(Hi(M; R)) by the universal

coefficient theorem, but by Poincare duality bi is also equal to dim(Hni(M; R)). Thus

(M) =b0 b1+ b2 b2k+1

cancels out, becauseb0= b2k+1, b1= b2k, etc.

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Lecture 21 (2012-11-16)

Another Corollary of Poincare Duality

Here is a useful algebraic fact: any non-degenerate skew-symmetric bilinear form on Zn is, after

changing basis, of the form J . . .J

whereJ=

0 11 0

. In particular, n must be even. This fact, together with Poincare duality, imply

Corollary. LetMbe a closed, oriented, (4k+ 2)-dimensional manifold. Then because we have anon-degenerate pairingHni(M) Hi(M) Z, rank(Hi(M, Z)) =(M) must be even.

Continued Prologue to Poincare duality

IfMbe a closed, orientable n-manifold, then Hi(M;Q

)=H

ni

(M;Q

). Poincares idea about thiswas as follows: in a -complex, we compute Hi using Ci (M). We can build a dual CW-complex

(it wont be a -complex in general, we may get a polygonal complex) by putting a 0-cell in thecenter of every n-simplex, adding a 1-cell connecting two n-simplices if they meet on a codimension1 subface, etc.

Thus,is an isomorphismCi (X)

=CCWni(X).

Thus,{Ci (X), }computes Hi(X)=Hi(M), and {C

n1(X), } computes H

ni(X)=Hni(M).

Classification of Surfaces

Theorem 0.1. LetMbe a closed 2-manifold. ThenM is homeomorphic to one of:

g whereg 0 (whenM orientable), or

g# RP2

(whenM non-orientable).Theorem 0.2 (2-dimensional Poincare conjecture). The following are equivalent:

1. (M) = 2

2. M= S2

3. Every loop M (i.e. embedding: S1 M) separatesM into 2 components,

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Statement 2 implies statement 3 by the Jordan separation theorem, and we already know thatstatement 2 implies statement 1.

We will be assuming the following classical result:

Theorem (Rado, 1920s). Every surface is triangulable.

Proof of Theorem 2. We will show that statement 1 implies statement 2. Let K be a triangulationofM. First, pick a maximal tree T in K(1). Note thatTK(0).

Now build a connected graph K as follows: V() consists of the 2-simplices ofK, and E()consists ofK(1) T(0).

Here is an example on a tetrahedron:

We have

(M) =V(K) E(K) + F(K)

=V(T) (E(T) + E()) + V()

= [V(T) E(T)] + [V() E()]

=(T) 1

+ () 1

2,

where we have used the following lemma.

Lemma. Let be a finite connected graph. Then() 1 with equality if and only if is a tree.

Proof of lemma. Trivial application of induction (check for yourself).

Thus, we have that(M) 2, with equality if and only if(T) =() = 1, which is the case if andonly if is a tree. But ifT Mis a tree, there is a neighborhood ofT homeomorphic to D2. Beingvery careful, we can thicken and Tto neighborhoods each homeomorphic to D2, and such that

nbhd(T) nbhd()= S1 =D21= D22.

This implies thatM= S2

.Now, note that statement 3 implies that is a tree: if it werent, it would have a loop , and would have to separate M; but then it would separate two vertices ofK(0), which would contradictthat T= .

Thus, statement 3 implies statement 1.

Proof of Theorem 1. We are given M, which has (M) 2 by the above.

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Does(M) =

2 ?

YesM= S2

No a loop in M thatdoes not separate M

A neighborhood of is homeomorphic to either S1 [0, 1] or the Mobius strip (the latter is onlypossible ifMis not orientable).

cut

2

1

M1

M2

LetM be M1 with 1 and2 filled in by disks. Thus, M is a new closed, connected surface, and

(M) =(M1) + (M2) (S1 S1).

It is easy to check that (M)> (M); in fact, (M) =(M) + 2.

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Lecture 22 (2012-11-19)

The Mayer-Vietoris Argument

Definition. An open cover {Ui} of an n-manifoldM is a good cover if each Ui=Rn and for all

i1, . . . , ir,Ui1 Uir= R

n

or is empty.Theorem. Every smooth manifoldMhas a good cover. Of course, ifM compact, then we canchoose the good cover to be finite.

Proof. First, we will prove that any smooth manifoldM admits a Riemannian metric. Take a coverofM by coordinate patches Ui, and on each coordinate patch, let , i be the pullback of thestandard Riemannian metric on Rn. Let Pi be a partition of unity subordinate to the cover. Then , =

Pi , i is a Riemannian metric on M.

Now we appeal to a theorem of Whitehead (or Alexander): ifMis any Riemannian manifold, theneverym M has a convex neighborhood Um, where convex means that there is a unique geodesicinUm between any two points. Because the intersection of convex sets is convex, and convex = R

n,

we are done.

Remark. Good covers are cofinal in the set of covers ordered by refinement.

Theorem. If a manifoldM has a finite good cover, then the ith homologyHi(M) ofM is finitelygenerated for alli 0.

Proof. Mayer-Vietoris implies that for open U, V M, we have an exact sequence

Hi(U) Hi(V) Hi(U V) Hi1(U V)

Thus, by the rank-nullity theorem,

dim(Hi(U V)) dim(Hi1(U V)) + dim(Hi(U)) + dim(Hi(V)),

where dim here means the minimial number of generators.

IfU1, . . . , U r is a good cover, then U=U1 Ur1 is an open submanifold ofM, and we canlet V =Ur. By induction on the size of a minimal open cover, we have that Hi(U) and Hi(V) arefinitely generated, and thus we are done by our observation above.

The base case is just M= Rn which is trivial.

Kunneth theorem via a Mayer-Vietoris argument

Theorem. LetH(X) =H(X; Q). Then for any manifoldsM1, M2 with finite good covers (in

fact we only need one of them to have this), we have

H(M1 M2)=H(M1) H

(M2).

Proof. We will show the map : H(M1) H(M2) H(M1 M2) defined by sending a b toa b is an isomorphism.

LetU1, . . . , U r be a good cover ofM. We induct on r.

For r = 1, since Rn M2 M2, we have H(M2)=H

(Rn M2)=H(R) H(M2).

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Now suppose that {U, V} is a cover of M1. Mayer-Vietoris implies that we have a long exactsequence

Hi(U V) Hi(U) Hi(V) Hi(U V)

Tensoring withHni

(M2) throughout (?), we get

Hi(U V) Hni(M2) (Hi(U) Hi(V)) Hni(M2) Hi(U V) Hni(M2)

which is still exact because tensoring with a vector space preserves exactness.

Summing the above exact sequence over all i, we get an exact sequence

ni=1

[Hi(U V) Hni(M 2)]

nth degree ofH(UV)H(M2)

which maps to

Hn((U V) M2)

Check that forms a chain map between these two exact sequences, and use the 5 lemma. (?)

Orientation Revisited

Letx Rn, and let B be a neighborhood ofx. We have

Hi(Rn, Rn {x})=

excision

Hi(Rn, Rn B)

=Hi(Rn/(Rn B))=Hi(B/B)=Hi(Sn)=

Z ifi = n,

0 otherwise.

Now let Mbe a manifold. For any x M, we can find a neighborhood B ofx lying in a coordinatechart. By the same reasoning,

Hi(M, M {xi})=Hi(B, B {x})

=

Z ifi= n,

0 otherwise.

A choice of generator ofHn(M, M {x})= Z is called a local orientation at x.

Now let Mbe a triangulated, closed, and connected n-manifold for which there exists a compatibleorientation on each simplex of the triangulation. Let {i} be the set of n-simplices, and let=

i Cn(M); this sum makes sense because Mis compact, so there are finitely many terms.

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Because M is a manifold without boundary, and the orientations on i j cancel, we have that =

i = 0. Thus, Zn(M). But Cn+1(M) = 0 because M is n-dimensional; therefore,

[] Hn(M) is of infinite order. This homology class is called the fundamental class ofM, and isdenoted [M].

Corollary. The Klein bottleKis not orientable sinceH2(K) = 0.

Check for yourself that for any x M, the inclusion map i : (M, )(M, M x) induces a mapi: Hn(M) Hn(M, M x) that takes a generator to a generator.

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Lecture 23 (2012-11-21)

Poincare Duality

Definition. Let Xbe a space. The cap product is a pairing between certain homology groups and

cohomology groups ofX. Fork , we define

:Ck(X) C

(X) Ck(X) by taking Ck(X),a singular k-chain : [v0 vk] X, and C(X), a singular -cochain, and mapping them to

= (|[v0v])|[v+1vk].

It is easy to check the following properties:

is bilinear

( ) = (1)( )

(Zk Z) Zk, i.e. cycle cycle = cycle

(Bk Z) Bk, i.e. boundary cycle = boundary

These facts imply that the cap product descends to a bilinear map :Hk(X) H(X) Hk(X).

Theorem (Poincare Duality). LetMbe a closed, orientedn-manifold. Then for any0 i n,the map DM :H

i(M) Hni(M) is an isomorphism, whereDM is defined by

DM([]) = [M] [].

Corollary. The top homology group Hn(M) is isomorphic to Z, and [M] is a generator.

Proof idea. We want to use a Mayer-Vietoris argument, but theres an immediate problem: thebase case is false!

Hn(Rn) = 0= Z =H0(R

n).

To overcome this, we define Hnc(M), cohomology with compact support, which will satisfy

Hnc(M)=Hn(M) for M compact

Hic(Rn)=

Z ifi = n,

0 otherwise.

Then we prove, using a Mayer-Vietoris argument, for all (not-necessarily-compact) connected,oriented manifolds Ywithout boundary that the map

DY :Hic(Y) Hni(Y)

is an isomorphism, and then finally, extend to the non-compact case.

Cohomology with Compact Support

LetXbe a locally finite -complex. Define Cic(X; R), the i-cochains with compact support, to be

Cic(X; R) :={| = 0 outside a finite # of simplices} Ci(X; R) := Hom(Ci(X), R).

Clearly,(Cic) Ci+1c , so that={C

ic(X), }is a cochain complex. We then define the cohomology

ofXwith compact support to be Hic(X) :=Hi().

Note that Hic is only a (contravariant) functor when considering proper maps between spaces.

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Example. Lets consider X= R. We give it the following triangulation:

ThenC0(R; Z) just consists of the functions on R, and given C0(R; Z), we have = 0 only if(v) =(w) for all w, v R, i.e. is constant. Therefore, if C0c (R), we must have = 0. Thus

Z0c (R) = 0, and thus H

0c (R) = 0.

Now we claim that H1c (R; Z)= Z. Let : C1c (R) Z be the map sending to

eX(1)(e). Then

for any C0c (R), we have[i, i + 1] =(i + 1) (i),

so that () = 0. Therefore, induces a homomorphism : H1c (R) Z. It is easy to check that is a bijection, and therefore H1c (R; Z)

= Z.

Note that{KX |K compact}is a poset under inclusion. We obviously have thatKL impliesX KX L, so for any inclusion KL, we get a map Hi(X, X K) Hi(X, X L), andthis gives us directed system of abelian groups.

Theorem. For any spaceX,Hic(X)=lim

K

Hi(X, X K).

Proof. Do on your own.

It turns out that there is a relative fundamental class [(M, M K)] Hn(M, M K), whereM isann-manifold. We can extend to Hk(X, A) H

(X, A) Hk(X, A), and then use the abovetheroem to define

:Hk(X) Hc(X) Hk(X).

Proposition. The cohomology with compact support of Rn is

Hi

c(Rn

; Z)= Z ifi= n,0 otherwise.

Proof. Let Kr =B0(r). We get an increasing sequence of compact sets, K1 K2 , and hencea decreasing sequence X K1 X K2 . This is a cofinal sequence in the poset mentionedabove. Because

Hi(Rn, Rn B0(r))=

Z ifi = n,

0 otherwise

and all of the maps in the directed system are the identity, we will get the same thing when wetake the algebraic limit.

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Lecture 24 (2012-11-26)

Today well be talking about the fundamental group. It may seem like it would be similar tohomology, but in homology, we only deal with abelian / linear things, whereas the fundamentalgroup is usually non-abelian.

The Fundamental Group

We start by considering maps from [0, 1] to X. If, : [0, 1] X satisfy (1) =(0), then we candefine their composition (a b) : [0, 1] X by

(a b)(t) =

(2t) if 0 t 12 ,

(2t 1) if 12 t 1.

We define a path to be a map from [0, 1] to X, with the endpoints treated specially. We definetwo paths to be homotopic when they are homotopic rel their endpoints.

Thus, if and are paths with (0) =(0) and(1) =(1), then we say when there is somehomotopyF : [0, 1]2 Xsuch that F0= , F1= , Ft(0) =(0) =(0), and Ft(1) =(1) =(1).

Now well prove this is an equivalence relation.

For any patha, we havea a because we can define a homotopy as follows (imagine time progressesas one goes bottom to top; this homotopy is constant on the gray lines):

(0) (1)

Suppose and are paths with (0) =(0) and (1) =(1), such that (say, via a pathhomotopyF : [0, 1]2 X from F0 = to F1 =). Then we have , because we can definea path homotopy G: [0, 1]2 X from G0 = to G1 = simply as Gt(s) = F1t(s). Pictorally,(again, imagine time progresses as one goes from bottom to top):

F

(0) (1)

= =

(0) (1)

F G

(0) (1)

= =

(0) (1)F

Lastly, if, , are paths with (0) =(0) =(0) and(1) =(1) =(1), such that and via homotopies F : [0, 1]2 X fromF0 = to F1= and G : [0, 1]

2 X from G0= toG1 = , we have because we can define a homotopy H : [0, 1]

2 X from H0= to H1= by

Ht(s) =

F2t(s) if 0 t

12 ,

G2t1(s) if 1

2 t 1.

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Pictorally,

F

(0) (1)

= =

(0) (1)

F G

(0) (1)

= =

(0) (1)

G

H

(0) (1)

= =

(0) (1)

F

G

Note also that if and , then we also have ( ) ( ):

(0) (1)

= =

(0) (1)

F G

If [] denotes the equivalence class of with respect to , then this demonstrates that descendsto an operation on equivalence classes, i.e. [ ] = [] [].

Moreover, is associative (when is defined) because there is a homotopy

(0) (1)

( )

( )

There are also identities for (be careful, because is only defined for paths whose endpoints arethe same). We define 1p to be the constant map from [0, 1] to p X. Then1(0) = becausethere is a homotopy

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1(0)

(0) (1)

and similarly 1(1) = .

Lastly, there are inverses. Ifis a path, we define 1 to be the same path but with the oppositeorientation, i.e. 1(t) =(1 t). Then 1 =1(0), because there is a homotopy

1

1(0)

(0) (0)

(it is constant on the gray lines), and similarly with 1 = 1(1).

A structure satisfying these properties is called a groupoid. It is the same as a group, except thereare some elements which may not be able to be composed.

Thus, given any space X gives rise to a groupoid, called the fundamental groupoid ofX. Theconstruction of this groupoid didnt favor any point in Xover any other, but if we break the

symmetry a bit and choose some point x Xto be our basepoint, then we can define a group asfollows: let 1(X, x) be the set of equivalence classes of paths : [0, 1] Xwhich both start andend atx, i.e. (0) =(1) =x. Then1(X, x) forms a group under the operation , and the identityof1(X, x) is just1x. (Note that we usually want to assume Xis connected.) We say that 1(X, x)is the fundamental group ofXwith basepoint x.

How does 1(X, x) depend on our choice ofx?

Given two pointsx and y, choosing a path fromx to y determines a homomorphism from 1(X, x)to 1(X, y) by sending [] to [

1 ]. In fact, it is easy to see that it must be an isomorphism(take1 as a path fromy tox). Ifx = y, then this isomorphism is just the inner automorphismthat is conjugation by [], because [1 ] = [1] [] [].

Thus, we often say somewhat loosely that a group is the fundamental group of a space X, eventhough th

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