# Excision in homology: $H(D^2, S^1)$

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.

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You want to compute $H_*(D^2,S^1)$ using excision. So don't you want $A=\partial D^2$? But then $A$ has empty interior so you are stuck. –  Grumpy Parsnip Aug 20 '11 at 14:05
oh noes, of course! Thanks! –  Rudy the Reindeer Aug 20 '11 at 14:59
If you're computing homology, the map $i_*$ should go the other way around. –  Olivier Bégassat Aug 21 '11 at 2:26
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). –  Amitesh Datta Dec 11 '11 at 23:33

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.

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What is the analogous result for $n$-manifolds? –  Exterior Jan 27 at 21:02
@Exterior: $H_n(M,M\setminus \{*\})\cong \mathbb Z$ –  Grumpy Parsnip Jan 27 at 22:46
But this more general results cannot be obtained via excision and the long exact sequence alone? –  Exterior Feb 13 at 13:02
@Exterior: yes, you can do this. The proof is the same! –  Grumpy Parsnip Feb 13 at 16:07
Great, thank you! –  Exterior Feb 13 at 22:15

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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