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Axel
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Kunneth formula for cohomology

Is there an algebraic Kunneth formula for cohomology?

More precisely assume $A_{*}, B_{*}$ are chain complexes of free $R$-modules ($R$ is a $PID$) and $M, N$ are $R$-modules. Then the map $\sum H^n(A_{*},M)\otimes H^m(B_{*},N)\rightarrow H^{n+m}(A_{*}\otimes B_{*}, M\otimes N)$ is defined as usually.

Is there an exact sequence of $R$-modules involving the map above analogous to the corresponding well-known Kunneth formulas for homology and universal coefficients theorems for homology and cohomology?

The problem here which confuses me is that in general for two free $R$-modules $A$ and $B$, $Hom(A,M)\otimes Hom(B,N)\neq Hom(A\otimes B, M\otimes N)$ so one can not just take the cochain complexes $Hom(A_{*}, M), Hom(B_{*}, N)$ then consider them as chain complexes with the "reversed" order and apply a usual Kunneth formula for homology as was suggested for example in J.P.May "Coincise course of algebraic topology".

This strategy would work say for cellular cohomology of finite $CW$-complexes but not in general.

Axel
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