Homology I

Recall definitions of simplicial homology and singular homology.

Remark. The only reason we choose to use simplices is that its boundary is very easy to define and calculate; when dealing with geometric intuition, it's perfectly fine to replace the simplex with anything that is easy to define an oriented boundary.

Homotopy Invariance

Theorem. Homotopic maps induces same homomorphisms between homology groups.

This theorem agrees with our intuition that 1. homology groups count holes, and 2. homotopy doesn't change holes.

We first give a intuitive "proof" for this. We draw the image of $f$, and continuously modulate it until it become $g$. While doing this, we draw the "trace" of this image. Then every chain will be strectched out, becoming a prism; Therefore, corresponding chains will cancel out due to this prism.

Sketch of proof. Suppose $f$ and $g:X\to Y$ are homotopic through $F:X\times I\to Y$, and consider a singular simplex $\sigma :{\mathrm{\Delta }}^n\to X$. Consider the prism $F\circ (\sigma \times \mathrm{id}):{\mathrm{\Delta }}^n\times I\to X\times I\to Y$. Divide this prism into disjoint simplices $[v_0,\cdots ,v_i,w_i,\cdots ,w_n]$, giving the prism operators

Geometrically the boundary of the prism is made from the top face, the bottom face, and the sides. Algebraically this is expressed through the equation

Using this relation, we have, for every cycle $\alpha$, $g_{\mathrm{♯}}(\alpha )-f_{\mathrm{♯}}(\alpha )=\partial P(\alpha )$ is a boundary, so $g_{\mathrm{♯}}$ and $f_{\mathrm{♯}}$ determine the same homology class.

A map $P$ between chain maps $f_{\mathrm{♯}}$ and $g_{\mathrm{♯}}$ satisfying $\partial P+P\partial =g_{\mathrm{♯}}-f_{\mathrm{♯}}$ is called a chain homotopy.

Proposition. Homotopy equivalences induce isomorphisms of homology groups.

Exact Sequences and Excision

Relative Homology Groups

When defining the chain groups, we can ignore chains in $A$, replacing $C_n(X)$ by $C_n(X,A)=C_n(X)/C_n(A)$. These are called the relative chain groups, and we can define the relative boundary map and relative homology groups on them.

Notice, however, that $H_n(X,A)$ is generally not isomorphic to $H_n(X/A)$. A good way to think about this is: considering relative homology only allow us to "move around freely inside $A$", but the region $A$ itself cannot be ignored. To give an example, take $X$ to be a disk, and $A$ the same disk without its center. Then $X/A$ is contractible ($X/A$ is the two-element space called the Sierpiński space), giving a trivial $H_2(X/A)$, but (intuitively) there is a non-trivial chain in $H_2(X,A)$: any 2-dimensional chain (imagine a surface) which covers the center (the reader should verify that it is actually a cycle). Intuition gives that $H_2(X,A)\cong \mathbb{Z}$, by counting the number of times the chain covers the center. Actually this is true.

However, for some nice pairs $(X,A)$, $H(X/A)$ and $H(X,A)$ do coincide, as we will show later.

Now we try to extend the chain complex $\cdots \to H_{n+1}(X)\to H_n(X)\to H_{n-1}(X)\to \cdots$. We take a subspace $A$ of $X$. Apparently there is an inclusion $H(A)\stackrel{i_{\ast }}{\to }H(X)$. The quotient map $j:C_n(X)\to C_n(X,A)$ induces a quotient map $j_{\ast }:H_n(X)\to H_n(X,A)$, by replacing every element of $C_n$ by its homology class. Intuitively, $i_{\ast }$ and $j_{\ast }$ has the extra property that $\mathrm{im}{i}_{\ast }=\ker j_{\ast }$.

The boundary map $H_n(X,A)\to H_{n-1}(A)$ is more complicated. For a relative $n$-cycle $\gamma$, there is an $n$-chain $\beta \in C_n(X)$ such that $j\beta =\gamma$. Then $\partial \beta$ should be trivial relative to $A$, as formally shown in $j\partial \beta =\partial j\beta =\partial \gamma =0$. Therefore $\partial \beta$ is a chain in $A$; i.e. $\partial \beta =i\alpha$ for some $\alpha \in C_{n-1}(A)$. Also, $\partial \alpha =0$ since $0=\partial \partial \beta =i\partial \alpha$. To sum up: loosely speaking, $\gamma$ is the relative version of $\beta$, and $\alpha$ is the boundary of $\beta$; both $\alpha$ and $\gamma$ are cycles (but of different spaces). It should intuitively make sense to define $\partial [\gamma ]=[\alpha ]$, where $[\cdot ]$ denotes homology class; after all, $\alpha$ is just "the boundary of $\gamma$ adjusted to the correct space". (The $\beta$ is required only because we want to define "the boundary of a relative cycle" properly.)

(We still need to verify the boundary map is well-defined; i.e. independent of the choice of $\beta$. This should be intuitively obvious, and we omit the proof here.)

As $\alpha$ is a boundary, it should be in the trivial homology class of $H_n(X)$; this is expressed rigorously by $i_{\ast }\partial [\gamma ]=[\partial \beta ]=0$, or equivalently $\mathrm{im}\partial \subset \ker i_{\ast }$. Conversely, every cycle $\alpha \in C_{n-1}(A)$ corresponds to some $\beta \in C_n(X)$ such that $j\beta =i\alpha$; this implies $\partial [j\beta ]=[\alpha ]$, so $\mathrm{im}\partial =\ker i_{\ast }$.

$\partial \beta =0$, therefore $\partial j_{\ast }[\beta ]=0$, which means $\mathrm{im}{j}_{\ast }\subset \ker \partial$. Conversely, if $\partial [\gamma ]=0$, then $\gamma =j\beta$; we want some chain ${\beta }^{\prime }$ making $j{\beta }^{\prime }=0$ and $\beta -{\beta }^{\prime }$ a cycle, then $j(\beta -{\beta }^{\prime })=c$. Taking ${\beta }^{\prime }=i{\alpha }^{\prime }$ where $\partial {\alpha }^{\prime }=\alpha$ clearly does this. Therefore, $\mathrm{im}{j}_{\ast }=\ker \partial$.

All these discussions gives us an exact sequence:

Excision

The core of excision is the idea that, if we remove a set in $A$ (but with enough space left), then the relative homology $H(X,A)$ should not be affected.

The Excision Theorem. If $Z\subseteq A\subseteq X$ such that $\mathrm{cl}Z\subseteq \mathrm{int}A$, then the inclusion $(X-Z,A-Z)\to (X,A)$ induces isomorphisms $H_n(X-Z,A-Z)\to H_n(X,A)$.

Setting $B=X-Z$, we get a more convenient version:

The Excision Theorem, second version. For $A,B\subseteq X$ such that $\mathrm{int}A\cup \mathrm{int}B=X$, the inclusion $(B,A\cap B)\to (X,A)$ induces isomorphisms $H_n(B,A\cap B)\to H_n(X,A)$.

Proof. We want a map $C_n(B)/C_n(A\cap B)\to C_n(X)/C_n(A)$, which induces isomorphisms on homology groups. Notice the quotient $C_n(B)/C_n(A\cap B)$ is free with basis the singular $n$-simplices in $B$ that do not lie in $A$. We now make the first step forward: we replace $C_n(A\cap B)$ by $C_n(A)$. To make this operation available, we must modify $C_n(B)$ to include chains in $A$; a rational choice is $C_n(B)+C_n(A)=:C_n(A+B)$. We've managed to get closer to the target: $C_n(B)/C_n(A\cap B)\cong C_n(A+B)/C_n(A)$. Now we need to prove the inclusion $\bar{\iota }:C_n(A+B)/C_n(A)\to C_n(X)/C_n(A)$ induces isomorphisms on homology groups. We first show that the inclusion $\iota :C_n(A+B)\to C_n(X)$ is actually a chain homotopy equivalence.

Lemma. The inclusion $\iota :C_n(A+B)\to C_n(X)$ is a chain homotopy equivalence.

Sketch of proof. The key is that we can do iterated barycentric division to a chain in $X$, making it small enough so that every divided region is either in $A$ or in $B$ (by Lebesgue's theorem). Therefore, we need to 1. define "barycentric division" for singular simplices; and 2. establish a chain homotopy equivalence between the original chain and the divided one.

We first consider barycentric division of linear simplices $\lambda :{\mathrm{\Delta }}^n\to {\mathbb{R}}^m$, and establish the required notions. For a point $b\in {\mathbb{R}}^m$, we can form a simplex by adding $b$ to $\lambda$, namely, the cone with apex $b$ and bottom $\lambda$. We denote this cone by ${\mathrm{cone}}_b\lambda$. Notice that $\partial {\mathrm{cone}}_b\lambda =\lambda -{\mathrm{cone}}_b\partial$ (the boundary of a cone is its sides and its bottom); therefore $\partial {\mathrm{cone}}_b+{\mathrm{cone}}_b\partial =\mathrm{id}$, so ${\mathrm{cone}}_b$ is a chain homotopy between $\mathrm{id}$ and $0$.

We now define the barycentric division $S$. For a linear simplex $\lambda$, denote its barycenter by $b_{\lambda }$. We then define recursively as: $S\lambda ={\mathrm{cone}}_{b_{\lambda }}S\partial \lambda$. (Convince yourself that this is the barycentric division.) Intuition and calculation both gives $\partial S=S\partial$, meaning $S$ is a chain map. To show $S$ is chain homotopic to $\mathrm{id}$, we again need to divide a prism: we connect all simplices on the bottom and the sides to the barycenter of the top, as indicated in the image.

This division induces a prism operator $T$, satisfying (again, intuitively and rigorously through calculation) $\partial T+T\partial =\mathrm{id}-S$.

We can easily extend this set of notions to singular simplices, by compositing $\lambda$ with a singular simplex $\sigma :{\mathrm{\Delta }}^n\to X$. Now we're ready for iterated division. A chain homotopy between $\mathrm{id}$ and $S^m$ is given by the tower of prisms: $D_m=\sum _{i=0}^{m-1}T{S}^i$. (Imagine extending the prism above by building an identical prism upon each divided region on the top; iterate $m$ times.) For each singular simplex $\sigma :{\mathrm{\Delta }}^n\to X$, there is a least integer $m(\sigma )$ such that after $m$ divisions every region of $\sigma$ is either in $A$ or in $B$; i.e. $S^m\sigma \in C_n(A+B)$. We now define $D:C_n(X)\to C_{n+1}(X)$ by $D\sigma =D_{m(\sigma )}\sigma$ (i.e. a prism tower between $\mathrm{id}$ and a sufficiently small division). We then claim that $\rho =\mathrm{id}-\partial D-D\partial$ is a chain homotopy inverse for the inclusion $\iota$. The proof is pure calculation, and is omitted here.

We can now return to the proof of the excision theorem. Knowing that $\partial D+D\partial =\mathrm{id}-\iota \rho$ and $\rho \iota =\mathrm{id}$, we try to make quotient maps to actually get to $C_n(A+B)/C_n(A)$ and $C_n(X)/C_n(A)$. Remarkably, all maps in these equations ($\partial$, $D$, $\iota$, $\rho$, $\mathrm{id}$) are $C_n(A)$-stable. Therefore, we can factor out all chains in $A$, and the quotient map $\bar{\iota }$ thus induces an isomorphism on homology groups.

With the help of excision, we can finally relate $H(X,A)$ and $\tilde{H}(X/A)$ for some nice $X$ and $A$.

Corollary. If $A$ is the deformation retract of some of its neighbourhood, then the quotient map $q:(X,A)\to (X/A,A/A)$ induces isomorphisms $q_{\ast }:H_n(X,A)\to H_n(X/A,A/A)\cong {\tilde{H}}_n(X/A)$.

Proof. Suppose $A$ is the deformation retract of $V$, one of its neighbourhoods. We have a commutative diagram

The two unlabelled horizontal maps are isomorphisms by excision. $\phi$ is an isomorphism by the long exact sequence and the fact that $H_n(V,A)=0$ (the pairs $(V,A)$ and $(A,A)$ are homotopy equivalent). The same argument shows that $\psi$ is also an isomorphism. The rightmost $q_{\ast }$ is an isomorphism since $q$ restricts to a homeomorphism on the complement of $A$. From commutativity, the leftmost $q_{\ast }$ is also an isomorphism.

There's another way of converting relative homology to reduced homology using cones, which works for all pairs $(X,A)$; suppose $CA$ is the cone from point $p$ to $A$, then

the first isomorphism is from the long exact sequence, second from excision, third from the deformation retraction of $CA-p$ onto $A$.