# 0708-1300/Class notes for Tuesday, February 26

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## A Homology Theory is a Monster

Page 183 of Bredon's book

Bredon's Plan of Attack: State all, apply all, prove all.

Our Route: Axiom by axiom - state, apply, prove. Thus everything we will do will be, or should be, labeled either "State" or "Prove" or "Apply".

## Typed Notes

The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.

### First Hour

Recall we had defined for a chain complex the associated homology groups: ${\displaystyle H_{p}(C_{*}):=\ ker\partial _{p}/im\partial _{p+1}}$

From this we get the pth homology for a topological space ${\displaystyle H_{p}(X)}$

We have previously shown that

1) ${\displaystyle H_{p}(\cup X)=\oplus H_{p}(X_{i})}$ for disjoint unions of spaces ${\displaystyle X_{i}}$

2) ${\displaystyle H_{p}(pt)=\mathbb {Z} \delta _{p0}}$

3) ${\displaystyle H_{0}(connected)=\mathbb {Z} }$

4) ${\displaystyle H_{1}(connected)\cong \pi _{1}^{ab}(X)}$ via the map

${\displaystyle \phi :[\gamma ]_{\pi _{1}^{ab}}\mapsto [\gamma ]_{H_{1}}}$

${\displaystyle \psi :\sigma \in C_{1}\mapsto [\gamma _{\sigma (0)}\sigma {\bar {\gamma _{\sigma (1)}}}]}$ where ${\displaystyle \sigma _{x}}$ is a path connecting ${\displaystyle x_{0}}$ to x.

We need to check the maps are in fact inverses of each other.

Lets consider ${\displaystyle \psi \circ \phi }$. We start with a closed path starting at ${\displaystyle x_{0}}$ (thought of as in the fundamental group). ${\displaystyle \phi }$ means we now think of it as a simplex in X with a point at ${\displaystyle x_{0}}$. ${\displaystyle \Psi }$ now takes this to the path that parks at ${\displaystyle x_{0}}$ for a third of the time, goes around the loop and then parks for the remaining third of the time. Clearly this is homotopic this composition is homotopic to the identity.

We now consider ${\displaystyle \phi \circ \psi }$. Start with just a path ${\displaystyle \sigma }$. Then ${\displaystyle \psi }$ makes a loop adding two paths from the ${\displaystyle x_{0}}$ to the start and finish of ${\displaystyle \sigma }$ forming a triangular like closed loop. We think of this loop as ${\displaystyle \sigma '\in C_{1}}$

Now, we start from c being ${\displaystyle c=\sum a_{i}\sigma _{i}}$ with ${\displaystyle \partial c=0}$. So get ${\displaystyle \sum a_{i}(\partial \sigma _{1})=\sum a_{i}(\sigma _{i}(1)-\sigma _{i}(0))}$

So ${\displaystyle \Psi (c)=[\gamma _{\sigma _{i}(0)}\sigma _{i}{\bar {\gamma _{\sigma _{i}(1)}}}]_{\pi _{1}}}$ which maps to, under ${\displaystyle \phi }$, ${\displaystyle \sum a_{i}(\gamma _{\sigma _{i}(0)}+\sigma _{i}-\gamma _{\sigma _{i}(1)})=\sum a_{i}\sigma _{i}=c}$ ( in the homology gamma's cancel as ${\displaystyle \partial c=0}$)

Axiomized Homology

We now will move to an approach where we prove that our defined homology satisfies a series of established homology axioms that will allow us to apply the machinery of general homology to our specific "singular" homology defined via simplexes.

Axiom 0) Homology if a functor

Definition The "category of chain complexes" is a category whose objects are chain complexes (of abelian groups) and morphisms which is a homomorphism between each abelian group in one chain and the corresponding group in the other chain such that the resulting diagram commutes. I.e, ${\displaystyle Mor((C_{p})_{p=0}^{\infty },(D_{p})_{p=0}^{\infty })=\{(f_{p}:C_{p}\rightarrow D_{p})_{p=0}^{\infty }\ |\ f_{p-1}\partial _{p}^{C}=\partial _{p}^{D}f_{p}\}}$

Now, in our case, the chain complexes do in fact commute because ${\displaystyle \partial }$ is defined by pre-composition but f is defined by post-composition. Hence, associativity of composition yields commutativity.

Claim

Homology of chain complexes is a functor in the natural way. That is, if ${\displaystyle f_{p}:C_{p}\rightarrow D_{p}}$ for each p induces the functor ${\displaystyle f_{*}:H_{p}(C_{*})\rightarrow H_{p}(D_{*})}$

The proof is by "diagram chasing". Well, let ${\displaystyle c\in C_{p}}$, ${\displaystyle \partial c=0}$.

Let ${\displaystyle f*[c]=[fc]}$. Now ${\displaystyle \partial fc=f\partial c=0}$. Furthermore, suppose ${\displaystyle c=\partial b}$. Then, ${\displaystyle f_{*}c=fc=\partial fb}$ so therefore ${\displaystyle fc=\partial \beta }$ some ${\displaystyle \beta }$. This shows ${\displaystyle f_{*}}$ is well defined.

Thus, for ${\displaystyle c\in H_{p}(C_{*})}$ get ${\displaystyle fc\in H_{p}(D_{*})}$ via the well defined functor ${\displaystyle f_{*}}$.

### Second Hour

1) Homotopy Axioms

If ${\displaystyle f,g:X\rightarrow Y}$ are homotopic then ${\displaystyle f_{*}=g_{*}:H_{p}(X)\rightarrow H_{p}(Y)}$

Applications: If X and Y are homotopy equivalent then ${\displaystyle H_{*}(X)\cong H_{*}(Y)}$

Proof:

let ${\displaystyle f:X\rightarrow Y}$, ${\displaystyle g:Y\rightarrow X}$ such that ${\displaystyle f\circ g\sim id_{y}}$ and ${\displaystyle g\circ f\sim id_{x}}$. Well ${\displaystyle f_{*}\circ g_{*}=id_{H(Y)}}$ and ${\displaystyle g_{*}\circ f_{*}=id_{H(X)}}$

Hence, ${\displaystyle f_{*}}$ and ${\displaystyle g_{*}}$ are invertible maps of each other. Q.E.D

Definition

Two morphisms ${\displaystyle f,g:C_{*}\rightarrow D_{*}}$ between chain complexes are homotopic if you can find maps ${\displaystyle h_{p}:C_{p}\rightarrow D_{p+1}}$ such that ${\displaystyle f_{p}-g_{p}=\partial _{p+1}h_{p}+h_{p-1}\partial _{p}}$

Claim 1

Given H a homotopy connecting f${\displaystyle ,g:X\rightarrow }$ Y we can construct a chain homotopy between ${\displaystyle f_{*},g_{*}:C_{*}(X)\rightarrow C_{*}(}$Y)

Claim 2

If ${\displaystyle f,g:C_{*}\rightarrow C_{*}}$ are chain homotopic then they induce equal maps on homology.

Proof of 2

Assume ${\displaystyle [c]\in H_{p}(C_{*})}$, that is, ${\displaystyle \partial c=0}$

${\displaystyle [f_{*}c]-[g_{*}c]=[(f_{*}-g_{*})c]=[(\partial h+h\partial )c]=0}$ (as ${\displaystyle \partial c=0}$ and homology ignores exact forms)

Hence, at the level of homology they are the same.

Proof of 1

Consider a simplex in X. Now consider its image, a simplex, in Y under g and f respectively. Because of the homotopy we can construct a triangular based cylinder in Y with the image under f at the top and the image under y at the bottom.

Define ${\displaystyle h\sigma }$ = the above prism formed by ${\displaystyle \sigma }$ and the homotopy H.

${\displaystyle (f_{*}-g_{*})\sigma =h\partial \sigma +\partial h\sigma }$

This, pictorially is correct but we need to be able to break up the prism, ${\displaystyle \Delta _{p}\times I}$ into a union of images of simplexes.

Suppose p=0, i.e. a point. Hence ${\displaystyle \Delta _{0}\times I}$ is a line, which is a simplex.

Suppose p=1 which yields a square. Adding a diagonal divides the square into two triangles, so is clearly a union of simplexes.

Suppose p=2. We get a prism which has a triangle for a base and a top. Raise each vertex on the bottom to the top in turn. This makes the prism a union of three simplexes.

In general for ${\displaystyle \Delta _{p}\times I}$ let ${\displaystyle f_{i}=(l_{i},0)}$ and ${\displaystyle g_{i}=(l_{i},1)}$ for vertexes ${\displaystyle l_{i}}$

Then, ${\displaystyle h\sigma =\sum _{i=0}^{p}(-1)^{i}H\circ (\sigma \times I)\circ [f_{0}\cdots f_{i}g_{i}g_{i+1}\cdots g_{p}]}$

which is in ${\displaystyle C_{p+1}(Y)}$

So have maps ${\displaystyle Y\leftarrow _{H}X\times I\leftarrow \Delta _{p}\times I\leftarrow \Delta _{p+1}}$

Claim:

${\displaystyle \partial h+h\partial =f-g}$

Loosely, ${\displaystyle \partial h}$ cuts each ${\displaystyle [f_{0}\cdots f_{i}g_{i}g_{i+1}\cdots g_{p}]}$ between the f_i and g_i and then deletes an entry. h\partial however does these in reverse order. Hence all that we are left with is ${\displaystyle [f_{0}\cdots f_{p}]-[g_{0}\cdots g_{p}]}$