Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
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

2 Answers 2

up vote 5 down vote accepted

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.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.