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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  • $\begingroup$ 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... $\endgroup$ – Kevin Buzzard Apr 4 '11 at 18:01

As Joe Johnson says in his now deleted post, 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 $H_1(\mathbb{RP}^2)=\mathbb Z_2$.

[Edited to remove an alternative false proof.]

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  • $\begingroup$ Am I understanding you correctly here: you are answering the question only in the case that $X$ is compact? $\endgroup$ – Kevin Buzzard Apr 4 '11 at 17:59
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    $\begingroup$ @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 $X$. $\endgroup$ – Cheerful Parsnip Apr 4 '11 at 18:32
  • $\begingroup$ @cheerfulparnship what, why for reduced cohomology you have this cochain complex? Usually it is the dual of the augmented complex and I don't see why that's equivalent... $\endgroup$ – Nick A. Mar 6 at 18:56
  • $\begingroup$ @NickA. for connected spaces, the (dual of the) augmentation map is an isomorphism, implying that the map $C^0\to C^1$ is the zero map. $\endgroup$ – Cheerful Parsnip Mar 6 at 19:45
  • $\begingroup$ @cheerfulparsnip I don't see why this is true. For example the domain of the augmentation map is countable but the target is (usually) uncountable. Also this would imply that the kernel of the zeroth coboundary map is the whole zeroth cochain group which I also believe to be false. I am through my phone so I cannot tex to give a full argument but I will once I go home :) $\endgroup$ – Nick A. Mar 10 at 15:14

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.

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