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I would like to know if there is a reference for the fact that the following diagram commutes: $$ \begin{array}{ccccccccc} 0 & \to & H_*(X) \otimes H_*(Y) & \to & H_*(X\times Y) & \to & \mathrm{Tor}_1^{\mathbb Z}(H_*(X),H_*(Y)) & \to & 0 \\ && \downarrow && \downarrow && \downarrow & \\ 0 & \to & H_*(Y) \otimes H_*(X) & \to & H_*(Y\times X) & \to & \mathrm{Tor}_1^{\mathbb Z}(H_*(Y),H_*(X)) & \to & 0, \\ \end{array} $$ where both rows are from the Künneth formula (essentially copied from Wikipedia), and the vertical arrows are the natural isomorphisms obtained by flipping the factors.

$H_*$ means singular homology - but actually this part isn't so important; a reference for a version with any homology or cohomology theory is all I want, provided that it is not one where the Tor term trivially vanishes.

I have a proof for K-theory (in fact, C*-algebra K-theory) but I imagine that it is not the first proof.

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    $\begingroup$ This should be migrated to MSE. $\endgroup$
    – Sean Tilson
    Commented May 5, 2015 at 11:20
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    $\begingroup$ Is this a trend on MO of people who did "alg top in grad school" deeming questions at that level off-topic even when asked by non-alg-top people? Also, the question is a reference request, not a request for a proof $\endgroup$
    – Yemon Choi
    Commented May 5, 2015 at 20:30
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    $\begingroup$ There were votes to close? I can see why that would be alarming. I don't have enough reputation to see such things I guess. I didn't think the comment was too censorious. I just feel like I should be careful on mediums like this and I wanted to clarify that I didn't mean anything rude or condescending. Electronic communication allows people to attach their own tone to things, and I like to get rid of ambiguity. Cheers! $\endgroup$
    – Sean Tilson
    Commented May 5, 2015 at 23:33
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    $\begingroup$ The 5 lemma definitely doesn't do it. The thing to prove is that the diagram commutes (which is a hypothesis for the 5 lemma). Also, it doesn't follow from naturality of the maps in the Künneth formula, because this naturality is in each variable and not specifically about swapping the variables. $\endgroup$
    – Aaron Tikuisis
    Commented May 6, 2015 at 6:57
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    $\begingroup$ Also, all you need is the right hand square to commute and that implies the result. I believe this follows by looking at what happens in chain complexes. The map $H_*(C_*)\otimes H_*(D_*) \to H_*(C_* \otimes D_*)$ is nice enough that it commutes with the twist map (up to sign as observed above). Then there is the map $S_*(X) \otimes S_*(Y) \to S_*(X\times Y)$ which commutes with the twist (up to sign) after taking homology. This is definitely a more delicate issue than I had thought. I would recommend looking at Switzer's Algebraic Topology book, chapter 13. (Let me know if you need a copy) $\endgroup$
    – Sean Tilson
    Commented May 6, 2015 at 9:41

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Larry Taylor just posted a preprint to the arXiv,

http://arxiv.org/abs/1506.02063

called Splitting the Künneth formula. Commutativity of your diagram is stated as Corollary 7.2, suggesting that the author is not aware of a published proof.

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