# $H^1$ is $H_1$ mod torsion

I'm reading Hatcher's survey on 3-manifolds (https://www.math.cornell.edu/~hatcher/Papers/3Msurvey.pdf) and noting that the fundamental group determines all homology groups of a closed orientable 3-manifold $M$.

$\pi_1(M)$ gives us $H_1(M)$ by Hurewicz. Duality says $H_2(M) \cong H^1(M)$. I don't quite understand why $H^1(M)$ is $H_1(M)$ mod torsion. Hatcher states that this is due to the universal coefficient theorem but I can't connect the theorem to the claim (I'm probably missing something obvious).

By the universal coefficient theorem, there is a short exact sequence $$0\to\operatorname{Ext}(H_0(M),\mathbb{Z})\to H^1(M)\to \operatorname{Hom}(H_1(M),\mathbb{Z})\to 0.$$
Since $H_0(M)$ is free, $\operatorname{Ext}(H_0(M),\mathbb{Z})=0$ and $H^1(M)\cong\operatorname{Hom}(H_1(M),\mathbb{Z})$. Now if $H_1(M)$ is finitely generated (as it is for any compact manifold), it is isomorphic to $\mathbb{Z}^n\oplus T$ for some $n$ and some torsion group $T$. Since $\operatorname{Hom}(T,\mathbb{Z})=0$, we have $$H^1(M)\cong \operatorname{Hom}(H_1(M),\mathbb{Z})\cong \operatorname{Hom}(\mathbb{Z}^n\oplus T,\mathbb{Z})\cong \operatorname{Hom}(\mathbb{Z}^n,\mathbb{Z})\cong \mathbb{Z}^n\cong H_1(M)/T.$$
That is, $H^1(M)$ is isomorphic to $H_1(M)$ mod its torsion subgroup.
(Note that this isomorphism is not natural--what is true is that $H^1(M)$ is naturally isomorphic to the dual of $H_1(M)$ mod torsion, whenever $M$ is such that $H_1(M)$ is finitely generated.)
• (The dual of $H_1(M)$ and the dual of $H_1(M)$ mod torsion are the same thing, in your last sentence) – Najib Idrissi Nov 2 '16 at 12:10
The universal coefficient theorem gives you a short exact sequence $$0\to \operatorname{Ext}^1_\mathbb{Z} (H_{n-1} (X), \mathbb{Z}) \to H^n (X) \to \operatorname{Hom}_\mathbb{Z} (H_n (X), \mathbb{Z}) \to 0$$ For $n = 1$ we have $H_0 (X) = \mathbb{Z}$, the Ext vanishes, and we obtain an isomorphism $$H^1 (X)\cong \operatorname{Hom}_\mathbb{Z} (H_1 (X),\mathbb{Z}).$$ Now Hatcher is trying to say that if $H_1 (X)$ is a finitely generated abelian group (!), then $\operatorname{Hom}_\mathbb{Z} (H_1 (X),\mathbb{Z})$ is isomorphic to the torsion-free part of $H_1 (X)$. (The isomorphism is not canonical, it depends on the choice of generators.)