Computing $H_1(S^1, S^0)$

Question

I am trying to compute $H_1(S^1, S^0)$. This is what I've done so far

Since $S^0 \subseteq S^1$ we get an exact sequence $$\widetilde{H_1}(S^0) \to \widetilde{H_1}(S^1) \to H_1(S^1, S^0) \to \widetilde{H_0}(S^0) \to \widetilde{H_0}(S^1)$$

where $\widetilde{H_n}(X)$ denotes the reduced $n$-th homology group of $X$. It turns out that $\widetilde{H_1}(S^0) \cong 0 \cong \widetilde{H_0}(S^0)$ and $\widetilde{H_1}(S^1) \cong \mathbb{Z} \cong \widetilde{H_0}(S^0)$ so the above exact sequence becomes the following exact sequence

$$0 \to \mathbb{Z} \to H_1(S^1, S^0) \to \mathbb{Z} \to 0$$

Letting $f$ be the homomorphism from $\mathbb{Z}$ to $H_1(S^1, S^0)$ in the above sequence we see that $H_1(S^1, S^0)/f[\mathbb{Z}] \cong \mathbb{Z}$ by the first isomorphism theorem.

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Now I'm kind of stuck at this point, I'm not sure how to proceed further to calculate $H_1(S^1, S_0)$. Firstly am I on the right track and if so, how can I proceed?

Answer 1

You have that

$H_1(S^1,S^0)\sim H_1(S^1,S^0)/ f(\mathbb{Z})\times \mathbb{Z}= \mathbb{Z}\times \mathbb{Z}$

by nullity-rank theorem

Answer 2

(Credit must go to @Mees de Vries for walking me through this in the comments above)

One can check using exactness that $g : H_1(S^1, S^0) \to \mathbb{Z}$ (from the exact sequence in the question) is surjective. Since $g$ is surjective, there exists a $\alpha \in H_1(S^1, S^0)$ such that $g(\alpha) = 1$. Now since $\mathbb{Z}$ is a cyclic group (under addition), we can define a homomorphism $u : \mathbb{Z} \to H_1(S^1, S^0)$ by just specifying $u(1)$. Define $u(1) = g(\alpha)$. One can then check that $g \circ u = 1_{\mathbb{Z}}$ and hence by the splitting lemma there exists an isomorphism $h : H_1(S^1, S^0) \to \mathbb{Z} \oplus \mathbb{Z}$ and thus $H_1(S^1, S^0) \cong \mathbb{Z} \oplus \mathbb{Z}$.