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Let $\Gamma$ be a group and $A$ be an abelian group and let's take group zero-homology and zero-cohomology, $H_0(\Gamma,A)$, $H^0(\Gamma,A)$. Is there any relation between $H^0(\Gamma,A)$ and $Hom(H_0(\Gamma,A),\mathbb{Z})$? And changing $\mathbb{Z}$ by other group?

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Did you mean for $A$ to be a non-trivial $\Gamma$-module? –  Hurkyl Aug 21 '12 at 1:35
    
Yes, of course. –  iago Aug 21 '12 at 12:09

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up vote 4 down vote accepted

For any module $M$ the natural quotient map $A \to H_0(\Gamma,A)$ induces an isomorphism $Hom(H_0(\Gamma,A),M) \cong H^0(\Gamma,Hom(A,M)).$ (Here we regard $Hom(A,M)$ as a $\Gamma$-module by the contragredient action.)

Thus, in order to get a relationship between $Hom(H_0(\Gamma,A),\mathbb Z)$ and $H^0(\Gamma,\mathbb Z)$, you would need an isomorphism of $\Gamma$-modules $A \cong Hom(A,\mathbb Z)$. For this, $A$ should be a finitely generated free $\mathbb Z$-action, with a self-dual $\Gamma$-action.

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