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I've been trying to find an example of a not too obscure space for which one needs the excision theorem to compute the homology groups:

Excision: If $Z \subset A \subset X$ where $A, U$ are subspaces of $X$ and $U$ is a subspace of $A$ then if $\bar{Z} \subset int(A)$ the following map is an isomorphism:

$i_\ast : H(X,A) \rightarrow H(X-Z, A-Z)$.

Example: For example if $X=D^2$ and $A=D^2 - \partial D^2$ and $Z = \{ \ast \}$ then this tells me that H(D^2, A) = H(S^1, \{ \ast \}) = \tilde{H}(S^1) which is $\tilde{H_1}(S^1) = \mathbb{Z}$ and $\tilde{H_n}(S^1) = 0$ for $n \neq 1$.

But I can also compute this using exactness:

$H_n(D^2, S^1) = 0$ for $n \neq 2$ and

$H_2(D^2, S^1) = \mathbb{Z}$.

I have two questions about this: What am I doing wrong? They should be the same.

And do you have an example where I actually need excision? It seems to me there is always a different way to get the homology groups and I don't actually need excision at all.

Many thanks for your help.

  • 0
    Yes, inclusion $i:(X-Z,A-Z)\to (X,A)$ induces an isomorphism $i_{*}:H_p(X-Z,A-Z)\to H_p(X,A)$ on homology (by the excision theorem).2011-12-11

3 Answers 3

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Here's an example of how I've seen excision used.

Proposition: Let $M$ be a surface. Then $H_2(M,M\setminus\{*\})\cong\mathbb Z$.

Proof: The point $*$ is contained in some closed disk $D\subset M$ with boundary $\partial D\cong S^1$. Now apply excision with $Z=M\setminus D$. Then you get $H_2(M,M\setminus\{*\})\cong H_2(D,D\setminus\{*\})\cong H_2(D,\partial D)$ and from the long exact sequence of the pair $(D^2,S^1)$, you show that $H_2(D,\partial D)\cong\mathbb Z$. (As you mentioned.) $\Box$

The analogous result for $n$-manifolds is very useful for defining what an orientation of a topological manifold is.

  • 0
    does a similar result hold for stratified spaces?2016-12-24
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Here are two examples of where excision is a useful tool.

1) Local homology groups (Jim gave a specific example of this). For $x\in X$, the local homology at $X$ is the relative homology $H_*(X,X\setminus\{x\})$. Using excision, it's straightforward to show that these groups depend only on a neighbourhood of $x$. That is, if $U$ is an open neighbourhood of $x$, then $H_*(X,X\setminus\{x\})=H_*(U,U\setminus\{x\})$.

2) Excision is used in showing that the relation $H_*(X,A)=\tilde H_*(X/A)$. As the definition of excision you gave is from Hatcher's book, I'll refer you to proposition 2.22 in the book for a proof of this fact, wherein you can see how excision is crucial to the proof.

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‎J.J. Rotman ,Algebraic Topology, exercise:5.19 Page ‎104‎

If ‎$‎‎A\subset‎ X$ ,‎ ‎then ‎there ‎is ‎an ‎exact ‎sequence‎ \begin{equation*} ‎\cdots ‎‎\rightarrow ‎‎\tilde{H}_n(A)‎‎\rightarrow‎ \tilde{H}_n(X)‎\rightarrow ‎H_n(X,A)‎\rightarrow‎‎\tilde{H}_{n-1}(A)‎\rightarrow ‎‎\cdots ‎,‎ \end{equation*}‎ which ends‎ \begin{align*} &‎\cdots‎\rightarrow‎ ‎‎\tilde{H}_0‎‎(A)‎\rightarrow ‎‎\tilde{H}_0(X)‎\rightarrow‎ H_0(X,A)‎\rightarrow ‎0‎\\‎ (Hint:‎\tilde{S}_{*}(X)&/‎\tilde{S}_{*}(A)=S_{*}(x)/S_{*}(A)‎‎). \end{align*}‎ Now exercise:5.20 Page ‎104‎

‎ ‎‎‎Show that ‎$‎‎H_1(D^2,S^1)=0$ ?

Solution:

$‎‎S^1\subset D^2~‎ ‎\Longrightarrow‎$‎ ‎with ‎Exe ‎2.19‎ ‎‎\begin{align*}‎‎ &‎‎\cdots ‎‎\rightarrow ‎‎\tilde{H}_n(S^1)‎\rightarrow ‎‎\underbrace{‎\tilde{H}_n(D^2)‎}_{0}‎\rightarrow ‎H_n(D^2,S^1)‎\rightarrow ‎‎‎ ‎\underbrace{‎\tilde{H}_{n-1}(S^1)‎}_{0}‎\rightarrow ‎‎\cdots \\‎ & ‎\cdots‎‎‎\rightarrow ‎‎\tilde{H}_2(S^1)‎\rightarrow ‎‎\underbrace{‎\tilde{H}_0(D^2)‎}_0‎‎ ‎\rightarrow ‎H_2(D^2,S^1)‎\rightarrow‎‎\underbrace{‎\tilde{H}_1(S^1)‎}_{‎\mathbb{Z}‎}‎\rightarrow‎‎\underbrace{‎\tilde{H}_1(D^2)‎}_{0}‎\\‎ &\rightarrow ‎H_1(D^2,S^1)‎\rightarrow ‎‎\underbrace{‎\tilde{H}_0(S^1)‎}_{0}‎\rightarrow‎‎\underbrace{‎\tilde{H}_0(D^2)‎}_{0}‎\rightarrow‎ H_0(D^2,S^1)‎\rightarrow ‎0‎‎‎‎‎ ‎‎‎\end{align*}‎ ‎because ‎$‎‎D^2$ is ,convex, ‎contractible ‎and ‎path ‎connected ‎then‎ ‎\begin{align*}‎‎ & ‎‎\tilde{H}_n(D^2)=H_n(D^2)=0 ‎\qquad‎‎‎\forall ‎n‎\geqslant ‎1\\‎ & ‎H_0(D^2)‎\cong ‎‎\mathbb{Z}‎‎‎\Longrightarrow‎ ‎‎\tilde{H}_0(D^2)=0‎‎‎ ‎\end{align*}‎‎ ‎‎‎‎‎‎‎‎‎‎and ‎‎‎\begin{eqnarray*}‎ ‎\tilde{H}_q(S^n) ‎&\cong ‎& \left\{%‎ ‎\begin{array}{ll}‎ ‎‎\mathbb{Z}‎‎\qquad &‎‎ ‎‎q=n‎ \\‎‎ ‎0 & ‎\text{‎other}‎ ‎\end{array}‎ ‎\right. ‎\qquad‎ ‎‎\forall ‎n‎\geqslant ‎0‎ ‎\end{eqnarray*}‎ ‎then‎ ‎\begin{align*}‎‎ ‎0‎\rightarrow ‎H_0(D^2,S^1)‎\rightarrow ‎0‎‎\qquad ‎‎&\Longrightarrow ‎H_0(D^2,S^1)\cong 0\\‎ ‎0‎\rightarrow ‎H_1(D^2,S^1)‎\rightarrow ‎0‎‎\qquad &‎‎\Longrightarrow ‎H_1(D^2,S^1)\cong 0\\‎‎ ‎0‎\rightarrow ‎H_2(D^2,S^1)‎\rightarrow‎‎‎\mathbb{Z}‎\rightarrow‎ ‎‎0‎‎\qquad ‎‎&\Longrightarrow ‎H_2(D^2,S^1)\cong ‎\mathbb{Z}‎\\‎‎ ‎0‎\rightarrow ‎H_n(D^2,S^1)‎\rightarrow ‎0‎‎\qquad &‎‎\Longrightarrow ‎H_n(D^2,S^1)\cong 0‎\qquad ‎n\neq 2‎\\‎‎‎ ‎\end{align*}‎