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Let $k$ be a field, and let $X$ and $Y$ be CW-complexes of finite type (although the question makes sense for $k$ a ring and for more general chain complexes of finitely generated free abelian groups). The Künneth and Universal Coefficient Theorems for cohomology give $$ H^n(X\times Y;k) \cong \bigoplus_{i+j=n} H^i\big(X;H^j(Y;k)\big)\supseteq\bigoplus_{i+j=n}\operatorname{Ext}\big(H_{i-1}(X);H^j(Y;k)\big). $$ (Here and elsewhere, where coefficients are omitted we mean $\mathbb{Z}$ coefficients.) Meanwhile, the Künneth Theorem for homology gives $$ H_n(X\times Y) \supseteq \bigoplus_{i+j=n}\operatorname{Tor}\big(H_{i-1}(X);H_j(Y)\big). $$ Now, we have a pairing of cohomology and homology $$ H^n(X\times Y;k)\otimes H_n(X\times Y)\to k, $$ and morally it should be true that this pairing restricted to the Ext and Tor terms should be given by the usual pairing of Ext and Tor, followed by the cohomology-homology pairing for $Y$: $$ \operatorname{Ext}\big(H_{i-1}(X);H^j(Y;k)\big)\otimes \operatorname{Tor}\big(H_{i-1}(X);H_j(Y)\big) \to H^j(Y;k)\otimes H_j(Y)\to k. $$

My question is, does this statement appear in the literature somewhere?

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    $\begingroup$ Is this pairing depending on the embedding of $Ext$ and $Tor$ into cohomology and homology respectively? Originally we get Ext and Tor terms as splitting quotient, so for thinking them as a subgadget, we need to pick up a non-cannonical embedding and it seems that this may affect the result. If I am saying something stupid or it does not make sense(it happens a lot) please just ignore me. $\endgroup$ Feb 21, 2015 at 1:15
  • $\begingroup$ The pairing itself is canonical, but you are right that the way Tor sits in the homology depends on a choice of splitting. I think the statement could still be true, but would also accept counter-examples as answers! $\endgroup$
    – Mark Grant
    Feb 21, 2015 at 6:52
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    $\begingroup$ A stupid question: what happens if we take $Y$ to be a point, so that we only have Universal coefficient theorems? Is this case trivial? $\endgroup$
    – user43326
    Feb 22, 2015 at 16:11
  • $\begingroup$ @user43326: Good point. Well, it's not trivial to me! But I would expect (if true) it should be proved somewhere already. $\endgroup$
    – Mark Grant
    Feb 22, 2015 at 21:40
  • $\begingroup$ So, along with the line of @MingcongZeng what happens if we take Y to be a point, X to be $\Sigma ^nMz/2\vee \Sigma ^{n-1}MZ/2$? Of course, one can choose the embedding of Ext and Tor so that we have the compatibility at the end, but Is there anything canonical about this embedding? $\endgroup$
    – user43326
    Feb 26, 2015 at 20:50

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