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If $X$ is a very nice topological space, for example a finite simplicial complex, then is it true that the cohomology with compact supports $H^1_c(X,\mathbf{Z})$ is torsion-free? I have seen an assertion in a paper that seems to be tantamount to this statement (unless I've made a slip and misread between the lines) but my topology is weak :-( and my hopelessly paging through Hatcher has not yet come up trumps...

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    Hmm. I have made my spaces too nice. In the application $X$ is a Shimura variety :-/ so I need to replace "finite simplicial complex" by something like "smooth real manifold which is homotopic to a finite simplicial complex" or some such thing...2011-04-04

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As Joe Johnson says, in the compact case $H^1_c(X)=H^1(X)$. Now you can use the fact that $H^1(X)=Hom(\pi_1(X),\mathbb Z)$, which is an abelian group under the operation $(\phi_1+\phi_2)(g)=\phi_1(g)+\phi_2(g)$. Since $\mathbb Z$ is torsion-free, so is this group of homomorphisms. Thus $H^1(X)$ is indeed torsion-free.

This is not true for homology though, since $H_1(\mathbb{RP}^2)=\mathbb Z_2$.

Also, here's another way to think about it for connected spaces. The reduced cohomology has a cochain complex $0\to C^1(X)\to C^2(X)\to \cdots$. So $H^1(X)=\ker\delta^1\subset C^1(X)$. Since $C^1(X)$ is a free abelian group, so is any subgroup. Hence $H^1(X)$ is torsion-free.

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    @Kevin: That's true, I only addressed the compact case to keep things simple. It's also true in the noncompact case. In fact, at least for nice spaces, $H^1_c(X)=H^1(\hat X)$, where $\hat X$ is the one point compactification of $X$.2011-04-04
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I just noticed this!

Why not just argue sheaf-theoretically? We have the exact sequence of sheaves $0 \to \mathbb Z \to \mathbb Z \to \mathbb Z/p \to 0,$ which induces a corresponding short exact sequence of $H^0_c$s, and hence we get a long exact sequence beginning with $H^1_c$: $0 \to H^1_c(X,\mathbb Z) \to H^1_c(X,\mathbb Z) \to H^1(X,\mathbb Z/p) \to \ldots.$ The fact that the first arrow is injective is the torsion-freeness statement that you want.

Summary: thinking sheaf-theoretically, the same argument that works for $H^1$ works for $H^1_c$ as well.