# 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.

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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.

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Thank you Mark! This is quite what I wanted to see. It seems to me that the finite type conditions you've mentioned somehow reflect the fact that $Hom(A, M)\otimes Hom(B, N)\neq Hom(A\otimes B, M\otimes N)$ so one has to impose some finiteness assumptions? –  Axel Sep 21 '12 at 7:37
Yes, that's right. The finiteness assumptions are there exactly to make your unequality an equality. I wonder why the more modern textbooks (such as Hatcher and May) don't state this result? Dold's book has it also, as Proposition VI.12.16. –  Mark Grant Sep 21 '12 at 8:21
This was omitted from my book for two reasons. First, it is rarely needed for applications, where it usually suffices just to combine the homology version and the universal coefficient theorem. And second, the cohomology version has always seemed to me a little unnatural, as evidenced by the extra finiteness assumption and the presence of the Tor functor, which in the universal coefficient theorems appears in the homology version rather than the cohomology version. A lighter toolkit is always preferable, and specialized tools can always be added for specialized applications. –  Allen Hatcher Sep 21 '12 at 12:29
@Allen Hatcher: Thank you for taking the time to respond to my comment, which I hope you won't take as a slight against your wonderful book! –  Mark Grant Sep 21 '12 at 12:45
@Sigur: If you wanted to compute the cohomology of $A\otimes B\otimes C$, one way would be to apply the above Kunneth sequence twice: once to compute the cohomology of either $A\otimes B$ or $B\otimes C$, then again to compute the cohomology of $A\otimes B\otimes C$. –  Mark Grant May 26 at 9:17

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.

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I agree completely that the case of field coefficients and finite type is very important and is used all the time. My comment about unnatural and unuseful was directed just at the general version where the Tor term comes into play, which I took to be the version under discussion. Of course the "unnatural" appellation is subjective so opinions on that may differ. –  Allen Hatcher Sep 21 '12 at 21:45
Agreed, but it is always fun to disagree with you :) –  Peter May Sep 21 '12 at 23:22
Prof. May, as my comment to your post was too long, I put it as an answer to my own question for technical reasons, see below. –  Axel Sep 22 '12 at 7:49

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).
I see what may have been confusing you. As I said (p. 136 original version, 134 of the second edition), you can start with cochain complexes $Y$ and $Y'$ that don't necessarily arise by dualizing chain complexes, and then the Kunneth theorem does apply directly. The finite type question arises from the dualization. –  Peter May Sep 22 '12 at 12:55