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It is one of the problems in Hatcher's book.

I need to find the homology group of $H_n (X,A)$ when $A$ is a finite set of points and $X$ is $S^2$ or $T^2$.

I figured out that for $n>1$, I could use the long exact sequence and make $H_n (X,A)$ isomorphic to $H_n (X)$.

However, I am stuck with $H_1 (X,A)$ and $H_0 (X,A)$.

Can anyone give me an idea how I can find these?

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For $H_0 (X,A)$ (in both cases, $S^2$ and $T^2$) use that $H_n(X,A) \cong \tilde{H_n}(X/A)$ (Hatcher, page 124). You know that $X/A$ is path-connected so you get $H_n (X/A) = \mathbb Z$ and hence $H_n(X,A) = 0$. – Rudy the Reindeer Apr 9 '12 at 17:12
thanks! that gives me half of the answer! :-) – Emily Apr 10 '12 at 15:27
Welcome : ) Glad I could help. – Rudy the Reindeer Apr 10 '12 at 15:54
up vote 7 down vote accepted

Suppose $A=\{x_1,\dots,x_k\}$. Then $H_0(A)=\mathbb{Z}^k$ and $H_i(A)=0$ for $i>0$.

Whether $X=S^2$ or $T^2$ we have $H_0(X)\cong\mathbb{Z}$, and like Matt N said in his comment in either case $H_0(X,A)\cong\tilde{H}_0(X/A)=0$.

If $X=S^2$ then $H_1(X)=0$ so the l.e.s. has a portion like $$0\rightarrow H_1(X,A)\rightarrow \mathbb{Z}^k\rightarrow \mathbb{Z}\rightarrow 0 $$ and so $H_1(S^2,A)\cong\mathbb{Z}^{k-1}$.

If $X=T^2$ then $H_1(X)=\mathbb{Z}^2$, so we have $$ 0\rightarrow \mathbb{Z}^2\rightarrow H_1(X,A)\stackrel{\partial}{\rightarrow} \mathbb{Z}^k\rightarrow \mathbb{Z}\rightarrow 0$$ Then $\ker\partial\cong\mathbb{Z}^2$ and its image is $\cong\mathbb{Z}^{k-1}$. I believe this is enough to conclude $H_1(T^2,A)\cong \mathbb{Z}^{k+1}$

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thank you for a generous explanation! :-) But I am still confused about that last sentence. I guess you have used the splitness of the exact sequence. However, how do we know that the sequence splits? If we have a short exact sequence then, $\mathbb Z$ being free abelian would be enough, but can we extend this idea to the long exact sequence as above? – Emily Apr 10 '12 at 15:55
My reasoning is like this: $H_1(X,A)$ is a finitely generated abelian group, so which one is it? We know it has no torsion, since the image and kernel of $\partial$ are both free, so it is completely determined by the rank. But then the rank has to be $(k-1)+2$ – you Apr 10 '12 at 17:15
(hopefully that reasoning is valid...) – you Apr 10 '12 at 17:25
That explains to me! But can't we say that im(round) is enough for $H_1 (X,A)$ being free? I am asking this to make sure I am understanding right. – Emily Apr 10 '12 at 17:45
Well, we know that the image of $\partial$ is in a free abelian group, so if $H_1(X,A)$ has torsion it will be in the kernel of this homomorphism. But exactness tells us that $\ker\partial\cong\mathbb{Z}^2$, which has no torsion. – you Apr 10 '12 at 22:35

This uses Proposition 2.22 in Hatcher(and you must prove $(X,A)$ is a good pair).

Without saying too much, although I guess "you" did, I solved this by finding a homotopy equivalence between the space $S^2/A$ or $T^2/A$ and a CW complex. In the former case, you can see the homotopy between $S^2/A$ and a CW complex given by $k+1$ 0-cells(where $A$ is a collection of $k$ points), $2k$ 1-cells, and 2 2-cells. Graphically we arrange the first $k$ 0-cells into a lovely regular $k$-gon, with an outlier $x$ in the back. We use the first $k$ 1-cells to add edges to our $k$-gon, and the second $k$ connecting the vertices of the $k$-gon to the outlier. Then the two cells are each attached with their boundaries glued to the $k$-gon. By contracting the 1-cells attached to $x$, we see this is homotopic to $S^2/A$, but this has a CW complex structure, so it is easier to compute.

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This is an interesting point of view! It became geometrically clear, thank you! :-) – Emily Apr 10 '12 at 16:09
Glad you enjoyed it! – John Stalfos Apr 10 '12 at 16:39

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