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.
Best Answer
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.