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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? :)

Donut
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1 Answers1

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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 $\tilde{H}_1 = H_1$).

Najib Idrissi
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  • First of all. Thank's a lot for helping me!!! I'm a very beginner, so i have to comment on some things here to learn sth.: First of all the second part using reduced homology is clear! Now I want to understand the first version, though. You state, that because $S^1$ and $S^1-{x}$ are path connected the map between the homologygroups are isomorphic. I just can see that both are isomorphic to $\mathbb{Z}$, but not that the map between them, let's call it $i$, is surjective...I think it is injective, because it is induced by the (injective) inclusion map $i: S^1-{x} \to S^1$. – Donut Feb 13 '16 at 14:32
  • Ok, it is surjective by exactness, because $H_0(S^1, S^1-{x})=0$. But why is this again 0 ? – Donut Feb 13 '16 at 14:37
  • @Aceras An injective continuous map doesn't necessarily induce an injective map in homology: think about $S^1 \subset \mathbb{R}^2$. Do you know that $H_0(X)$ is the free abelian group generated by the path components of $X$? – Najib Idrissi Feb 13 '16 at 15:10
  • I think, i have read this in Hatcher's Algebraic topology. This is because each simplex is path connected and so a continous map must map each simplex to exactly one of the path components. So the homology group of the top. space X splits into the direct sum of the homology groups of the path components. – Donut Feb 13 '16 at 15:21