# Cohomology of hom

Im having troubles in proving the following result: Let $C^{\bullet}$ be a complex of $R$-modules ($R$ noetherian ring) with non-zero modules in positive degree, and let $M$ be an $R$-module. Assume that $H^i(C^{\bullet})=0$ for $i<n$. Then $H^n( Hom_R(M,C^{\bullet}))=Hom_R(M,H^n(C^{\bullet}))$ Any hint on how to do it?

-

This may not be true, because $Hom(M, -)$ is not in general exact.

Consider the following example. Let $R = \mathbb{Z}$ be the ring of integers and consider the exact complex $C ^{\bullet}$ given by

$\ldots \rightarrow 0 \rightarrow \mathbb{Z} \rightarrow \mathbb{Z} \rightarrow \mathbb{Z} _{2} \rightarrow 0 \rightarrow \ldots$,

where the first non-trivial map is multiplication by $2$. Let $M = \mathbb{Z} _{2}$. Since the cohomology of $C^{\bullet}$ is trivial, we also have $Hom (\mathbb{Z} _{2}, H^{i}(C^{\bullet})) = 0$ for all $i$. But $Hom(\mathbb{Z} _{2}, C^{\bullet})$ looks like this

$\ldots \rightarrow 0 \rightarrow 0 \rightarrow 0 \rightarrow \mathbb{Z} _{2} \rightarrow 0 \rightarrow \ldots$,

so it has non-trivial cohomology.

-
Thanks a lot! you are right, it doesn't happen in general, thanks again!! –  Miguel Dec 28 '12 at 10:54