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.
 A: 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.
A: ‎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*}‎
A: 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. 
