Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.