Let $X,Y$ be CW complexes. By Kunneth formula, we have a group isomorphim $$ H^n(X\times Y;G) \cong \oplus_{p+q=n} H^p(X;H^q(Y;G))$$ Is there a natural map realizing this isomorphism?
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$\begingroup$ what happened to the other answer which was below? I'd be interested to know whether the argument was flawed, or any reason why it was deleted. (please?) $\endgroup$– bananastackSep 9, 2014 at 14:21
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$\begingroup$ Isn't it the eilenberg-zilber map? On the level of chains we have $C^n(X \times Y) \rightarrow (C^*(X) \otimes C^*(Y))^n = \oplus_{p+q = n} C^p(X) \otimes C^q(Y)$. The latter is a chain complex that computes the $p$-th cohomology of $X$ with coefficients in the $q$-th cohomology (just like how $C^*(X) \otimes G$ computes cohomology of $X$ with coefficients in $G$) $\endgroup$– Elden ElmantoSep 9, 2014 at 16:49
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$\begingroup$ @user125763 I believe the argument in that previous answer has a flaw, because the Serre's spectral sequence only gives a filtration instead of a direct sum structure. Maybe the author deleted it. $\endgroup$– Boyu ZhangSep 9, 2014 at 21:16
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$\begingroup$ @EldenElmanto Why is the cohomology of $\oplus _{p+q=n}C^p(X)\otimes C^q(Y)$ the $p$-th cohomology of $X$ with coefficient $H^q(Y)$? This is exactly what I need. Is it a standard result? Do you have a reference for it? Thanks a lot! $\endgroup$– Boyu ZhangSep 9, 2014 at 21:22
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$\begingroup$ I see. Maybe considering the filtrations coming from both projections turns them into a direct sum decomposition? $\endgroup$– bananastackSep 10, 2014 at 0:37
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1 Answer
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I came across this question in my 1961 DPhil Thesis; this was written up in two papers which are available from my Publication List,
[3]. ``Cohomology with chains as coefficients'', Proc. London Math. Soc. (3) 14 (1964), 545-565.
[4]. ``On K\"{u}nneth suspensions'', Proc. Camb. Phil. Soc. 60 (1964) 713-720.
It is shown in the first paper that the isomorphism could be chosen to be natural with respect to maps of $X$ but not with respect to maps of either $Y$ or of $G$. The naturality was important for the second paper, which was poorly titled: it should have been something like "$k$-invariants of function spaces". Paper [3] also contains as an Appendix formulae for this isomorphism in some special cases
The aim of this work was to investigate the Postnikov system of $Y^X$ by induction on the Postnikov system of $Y$. Michael Barratt suggested this as another way of determining in some cases the extensions involved in his work on exact sequences of track groups; such a determination in his paper "Track Groups II" used Whitney tube systems!
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answered Sep 9, 2014 at 13:45