Isomorphism between $S^n$ and $S^n - \{x\}$ I wanted to understand the local degree of a map $f: S^n \to S^n$ and by doing this I faced a problem I couldn't solve by myself, although it might be very easy to see: I wanted to show, that for $n>0$, $x \in S^n$ there is an isomorphism $H_n(S^n) \simeq H_n(S^n, S^n-\{x\})$. 
I used the long exact sequence for the pair $(S^n, S^n - \{x\})$ and for $n>1$ the statement is clear:
$...\to \underbrace{H_n(S^n-\{x\})}_{= 0} \to H_n(S^n) \to H_n(S^n, S^n-\{x\}) \to \underbrace{H_{n-1}(S^n-\{x\})}_{= 0} \to ...$ 
that is by exactness $H_n(S^n) \simeq H_n(S^n, S^n-\{x\})$.
Now for n=1: Again i tried it with the exact sequence:
$...\to \underbrace{H_1(S^1-\{x\})}_{= 0} \to \underbrace{H_1(S^1)}_{= \mathbb{Z}} \to H_1(S^1, S^1-\{x\}) \to \underbrace{H_0(S^1-\{x\})}_{= \mathbb{Z}} \to 0$.
but now I can't see why the middle part is an isom. Can someone help me out? :) 
 A: You forgot the $H_0(S^1) = \mathbb{Z}$ (you wrote $0$ instead). The long exact sequence is actually:
$$\dots \to \underbrace{H_1(S^1 \setminus \{x\})}_{= 0} \to \underbrace{H_1(S^1)}_{= \mathbb{Z}} \to H_1(S^1, S^1 \setminus \{x\}) \to \underbrace{H_0(S^1 \setminus \{x\})}_{= \mathbb{Z}} \to \underbrace{H_0(S^1)}_{= \mathbb{Z}} \to \underbrace{H_0(S^1, S^1 \setminus \{x\})}_{= 0}$$
And now the induced map $H_0(S^1 \setminus \{x\}) \to H_0(S^1)$ is an isomorphism (because both spaces only have one path-component). So the image of $H_1(S^1, S^1 \setminus \{x\}) \to H_0(S^1 \setminus \{x\})$ is zero (= the kernel of the next morphism), and thus by exactness $H_1(S^1) \to H_1(S^1, S^1 \setminus \{x\})$ is surjective. It is also injective (again by exactness), so it's an isomorphism.

The easiest method is probably to use reduced homology, though. The long exact sequence becomes:
$$\underbrace{\tilde{H}_1(S^1 \setminus \{x\})}_{= 0} \to \tilde{H}_1(S^1) \to \tilde{H}_1(S^1, S^1 \setminus \{x\}) \to \underbrace{\tilde{H}_0(S^1 \setminus \{x\})}_{= 0}$$
and so the claim is immediately proven (because $\tilde{H}_1 = H_1$).
