Question on calculating $H_n(S^1 \times S^1,A)$ where $A$ is a finite set of points

Question

So, around $H_1(S^1 \times S^1,A)$, our long exact sequence looks like:

$0 \rightarrow H_1(S^1 \times S^1) \rightarrow^h H_1(S^1 \times S^1,A) \rightarrow^g H_0(A) \rightarrow^f H_0(S^1 \times S^1) \rightarrow 0$

Using exactness, I calculate that $ker(g) = \mathbb{Z}^2$ and $im(g) = \mathbb{Z}^{n-1}$.

And the correct answer is that $H_1(S^1 \times S^1,A) = \mathbb{Z}^{n+1}$.

So my question is how do I know that this is a situation where I know that $H_1(S^1 \times S^1,A) = ker(g) \oplus im(g)$?

Answer 1

Okay, to be fair, I gave you more of an alternative than I answered your question in my previous answer. I'll not delete it though, as I believe it's an efficient alternative (in particular, you don't have to worry at all about computing the image of $g$).

Well, considering what you have done, it's clear that you can write the short exact sequence:

$$0\to \ker(g)\xrightarrow{\imath}H_1(S^1\times S^1,A)\xrightarrow{g}\text{im}(g)\to0$$

where $\imath$ is the inclusion. The argument is now similar to what I used in my previous answer: you proved that $\text{im}(g)\cong\mathbb{Z}^{n-1}$, which is a free abelian group. This implies that the sequence splits which, as we're dealing with abelian groups, is equivalent to having your required isomorphism: $$H_1(S^1\times S^1,A)\cong\ker(g)\oplus\text{im}(g)$$

Answer 2

You should consider the long exact sequence for $\textit{reduced}$ homology here. So that exact sequence becomes:

$$0\to \tilde{H}_1(S^1\times S^1)\xrightarrow{h}\tilde{H}_1(S^1\times S^1,A)\xrightarrow{g}\tilde{H}_0(A)\xrightarrow{f}\tilde{H}_0(S^1\times S^1)\to0$$

$S^1\times S^1$ is path connected so $\tilde{H}_0(S^1\times S^1)$ is trivial and thus you have the short exact sequence:

$$0\to \tilde{H}_1(S^1\times S^1)\xrightarrow{h}\tilde{H}_1(S^1\times S^1,A)\xrightarrow{g}\tilde{H}_0(A)\to0$$

Because $\tilde{H}_0(A)$ is free, this sequence splits and, since you're dealing with abelian groups, this implies that $$\tilde{H}_1(S^1\times S^1,A)\cong\tilde{H}_0(A)\oplus\tilde{H}_1(S^1\times S^1)$$

The result follows from $\tilde{H}_1(S^1\times S^1)\cong \mathbb{Z}^2$ and the fact that if $A$ is a finite set of $n$ points, we have $\tilde{H}_0(A)\cong\mathbb{Z}^{n-1}$.