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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 $H_{1}(S^2, A)$?