$\DeclareMathOperator\Hom{Hom}$Is there an algebraic Künneth 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 Künneth 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 Künneth formula for homology as was suggested for example in J.P. May "Concise course of algebraic topology".


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



Künneth formula for cohomology
**Edit:** The answer below assumes that $A$ and $B$ are bounded below (or above). This is not explicitly mentioned in the statement in Spanier, but seems to be necessary.




---


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.


