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.
Best Answer
Yes, this is Theorem 5.5.11 in Spanier's "Algebraic Topology" text.
The conditions are that the torsion product $\operatorname{Tor}_R(M,N)=0$ and either $H(A;M)$ and $H(B;N)$ are of finite type, or $H(B;N)$ is of finite type and $N$ is finitely generated.
Then there is a natural short exact sequence $$ 0 \to H(A;M)\otimes H(B;N)\to H(A\otimes B; M\otimes N)\to Tor_R(H(A;M),H(B;N))\to 0 $$ (where the second map raises degree by one) and this sequence splits.