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Finding the homology group of $H_n (X,A)$ when $A$ is a finite set of points

I want to work out the homology of a sphere $S^2$ quotient a set of finite points(say p points) in the sphere called $A$.

So need to work out $H_{i}(S^2,A)$.

So I know that I got a long exact sequence

$0 \rightarrow H_{2}(A) \rightarrow H_{2}(S^{2}) \rightarrow H_{2}(S^{2},A) \rightarrow H_{1}(A) \rightarrow H_1 (S^{2}) \rightarrow H_{1}(S^{2},A) \rightarrow H_{0}(A) \rightarrow H_{0}(S^{2}) \rightarrow H_{0}(S^2,A) \rightarrow 0$

Reasoning is that the higher groups are zero as we can't have something of higher than 2 simplices be mapped onto $S^2$.

So we have this awful group $0 \rightarrow \mathbb{Z} \rightarrow H_2 (S^{2},A) \rightarrow 0 \rightarrow 0 \rightarrow H_{1}(S^2,A) \rightarrow \mathbb{Z}^{p} \rightarrow \mathbb{Z} \rightarrow H_{0}(S^2,A) \rightarrow 0$

This proves $H_2(S^{2},A) \cong \mathbb{Z}$. But, I'm unsure about the rest. I know you can say for $H_{0}(S^{2},A)$ that since A is sitting inside $S^2$ we can homotopy the points together through $S^2$ and sort of break the chain when you factor it. Well, I think it's something like that but don't understand the reasoning.

How do you calculate $H_{1}(S^2, A)$?

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marked as duplicate by Rudy the Reindeer, Jonas Teuwen, robjohn, t.b., Zev Chonoles May 19 '12 at 21:43

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up vote 2 down vote accepted

$H_1(S^2,A)$ injects into $H_0(A)$ (since the term before it vanishes), so it's isomorphic to the kernel of the map $f:H_0(A) \to H_0(S^2)$. But $f$ takes each generator of $\mathbb{Z}^p$ onto the same generator of $\mathbb{Z}$, so $H_1(S^2,A)$ can be identified with the subgroup $\mathbb{Z}^p$ consisting of the elements whose coefficients sum to zero.

Similarly, $H_0(S^2,A)$ vanishes (since $f$ is surjective and the next term in the sequence is 0).

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Isn't $H_{1}(S^2,A) \cong \mathbb{Z}^{p}$. I can see the subgroup, but can't you get more? – simplicity May 19 '12 at 18:50
Why do you think $H_1(S^2,A) \cong \mathbb{Z}^p$? The fact that the previous term in the sequence is zero is precisely why you can't get more. – Micah May 19 '12 at 19:04
I removed my previous "intuitively", which isn't actually a good way of thinking about it. A better intuition would be to think of $H_1(S^2,A)$ as the free abelian group on the edges of some arbitrary spanning tree whose vertices are the elements of $A$, which clearly has rank $p-1$. – Micah May 19 '12 at 19:11
Isn't there an injection $\mathbb{Z}^p \rightarrow \mathbb{Z}^{p}$ – simplicity May 19 '12 at 19:12
The point is, knowing that it's induced by inclusion means you should know exactly what the map $\mathbb{Z}^p \to \mathbb{Z}$ is; it's not "just something". $H_0(A)$ has one generator for each point in $A$; the inclusion map sends all of those generators to the generator of $H_0(S^2)$. The kernel of that map has rank $p-1$. – Micah May 19 '12 at 20:49

You can think of $S^2/A$ as being $S\cup CA$, where $CA$ is the cone over $A$. As you mention, we can move all the points of $A$ into a single point of $S^2$, so this is homotopic to $S^2\vee SA$, where $SA$ is the suspension over $A$. Thus our homology satisfies $$ H_\ast(S^2,A)\cong H_\ast(S^2)\oplus \tilde{H}_\ast(SA).$$

And with $SA$, we can contract one of the "edges" to get a wedge of $p-1$ circles. You can finish from here.

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