Künneth formula for cohomology

Question

$\DeclareMathOperator\Hom{Hom}$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.

Answer 1

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.

Answer 2

Axel, I have no idea what you are suggesting I suggested, but in fact the natural map of cochain complexes $$\omega\colon Hom(X,M)\otimes Hom(X',M') \to Hom (X\otimes X', M\otimes M')$$ for chain complexes $X$ and $X'$ and abelian groups $M$ and $M'$ is defined explicitly on page 134 of Concise. It is an isomorphism when $X$ and $X'$ are free and of finite type. When just their homologies are of finite type (and they are bounded below, like the chains of a space), they are chain homotopy equivalent to chain complexes $Y$ and $Y'$ that are free and of finite type, and it follows that $\omega$ then induces an isomorphism on (co)homology. Then we can apply the chain level K\"unneth theorem to deduce the result quoted from Spanier. That is all there is to its proof. While I didn't include this in my book either, I disagree with Allen about it being at all unnatural or unuseful. In applications, we constantly use that, with field coefficients and spaces of finite type, the cohomology of a product is the tensor product of the cohomologies. This is in fact an isomorphism of cohomology algebras over the field of coefficients. For obvious examples, consider the torus $T = (S^1)^n$ or products of real or complex projective spaces. The latter examples are especially important in the study of characteristic classes.

Answer 3

Prof. May, thank you very much for your comment. Your answer is clarifying the situation.

The source of my confusion was that on the page 136( in my version of Concise) where you define the map $\omega$ in the Chapter "Relations between $\otimes$ and Hom", you do say that under some finiteness assumptions we may apply the usual Kunneth formula for the reversed complexes.

However, as you did not mention neither any direct topological applications nor extension to the case when only the homologies are of finite type ( which you did here) on that page it made me think that the finiteness assumptions $\textbf{directly}$ only work in the quite restricted case of finite $CW$-complexes and can not even be directly applied to the singular homology ( as singular complex is almost never finitely generated).

Now of course the situation is much more clear to me.