Elementary way to show the exact sequence $0 \to M \to \mathbf Z^2 \to \mathbf Z \to 0$ implies $M = \mathbf Z$ I am computing the singular homology of spheres by induction.
In the process, I have come across the following short exact sequence
$$0 \to H_1(S^1) \to \mathbf Z^2 \to \mathbf Z \to 0.$$
I wonder if there is a more direct way to conclude that $H_1(S^1) = \mathbf Z$ than my argument:
it is a split short exact sequence of modules over a ring (for example, $\mathbf Z$ is a free module, so the third map has a section) ; therefore, $H_1(S^1) \oplus \mathbf Z$ is isomorphic to $\mathbf Z^2$. But $\mathbf Z^2$ is a free module over a principal domain, hence, so is $H_1(S^1)$. Because of the direct sum, I conclude that $H_1(S^1)$ has rank $1$.
I have the impression that it is overkill.
Thanks.
 A: $H_1(S^1)$ is the kernel of the map $\mathbb{Z}^2\to\mathbb{Z}$. Every such homomorphism is of the form $(x,y)\mapsto ax+by$ for some $a,b\in\mathbb{Z}$. In this case the map is surjective, so $\gcd(a,b)=1$, and the kernel is the infinite cyclic group generated by $(b,-a)$.
A: You could tensor it with $\mathbb{Q}$ (or any field of characteristic 0) to show that $H_1(S^1)$ is isomorphic to $\mathbb{Z} \oplus F$ for some finite abelian group $F$.
The reason for this is that you end up with the short exact sequence of vector spaces
$$
0 \to H_1(S^1) \otimes \mathbb{Q} \to \mathbb{Q}^2 \to \mathbb{Q} \to 0
$$
and so dimension counting tells you that $H_1(S^1) \otimes \mathbb{Q} = \mathbb{Q}$. Now, by the fundamental theorem of finitely generated abelian groups (ok, this is perhaps a little bit of overkill---there may be a simpler reason for this), we know that $H_1(S^1) \cong \mathbb{Z}^{\oplus n} \oplus F$ for some finite abelian group $F$. The above argument tells us that $n = 1$.
Why is the fundamental theorem of finitely generated abelian groups relevant? Well, I suppose it depends on how you define $S^n$. If you are doing so as a CW complex, or anything like that, you know that (due to compactness) there are finitely many cells in each dimension, which lets you know that this theorem is applicable.
It is not too hard to show that $F$ must be zero, since the exact sequence above gives an injection $F \to \mathbb{Z} \oplus \mathbb{Z}$.
