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This comes from page 48/49 in Bott and Tu's book on Differential forms in Algebraic Topology.

Let $M$ and $N$ be two manifolds and consider two projections $\pi_M : M \rightarrow N$ and $\pi_N : M \times N \rightarrow N$. With these maps we can bring forms from both $M$ and $N$ to $M \times N$: $$\pi_M^* : \Omega^i(M) \rightarrow \Omega^i(M \times N),$$and $$\pi_N^* : \Omega^j(N) \rightarrow \Omega^j(M \times N).$$With these then we can edge them to get a new $i+j$ form on $M \times N$, $$\Omega^i(M) \times \Omega^j(N) \rightarrow \Omega^{i+j}(M \times N),$$given by $(\omega,\eta) \mapsto \pi_M^*\omega \wedge \pi_N^*\eta.$

This map is clearly bilinear since everything is linear. By the universal mapping property for bilinear maps we have an unique induced map in the their tensor product such that $$\Omega^i(M) \times \Omega^j(N) \rightarrow \Omega^i(M) \otimes \Omega^j(N) \rightarrow \Omega^{i+j}(M\times N),$$corresponds to $$(\omega,\eta) \mapsto \omega \otimes \eta \mapsto \pi_M^*\omega \wedge \pi_N^*\eta.$$Now the authors claim this gives rise to a map in cohomology $$\psi : H^*(M) \otimes H^*(N) \rightarrow H^*(M \times N).$$

This is the part that I am not understanding: I believe we need to check that the map given by the universal property is a cochain map?

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  • $\begingroup$ probably the authors believe the check is not that hard $\endgroup$
    – FShrike
    Commented Jan 14 at 12:45
  • $\begingroup$ @FShrike I guess, but is it? And how would one proceed? We compute $d(\pi_M^* \omega \wedge \pi_N^*\eta) = \pi_M^*d\omega \wedge \pi_N^*\eta + (-1)^i\pi_M^*\omega \wedge \pi_N^*d\eta$. Now, what is $d(\omega \otimes \eta)$? $\endgroup$
    – user57
    Commented Jan 14 at 12:47
  • $\begingroup$ Unfortunately I don't know enough about the subject to help you at the moment :) all I can do is vote $\endgroup$
    – FShrike
    Commented Jan 14 at 12:49

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I think what the authors mean is that it is clear that $\Omega^i(M) \times \Omega^j(N) \to \Omega^{i+j}(M \times N)$ sends cocycles to cocycles and coboundaries to coboundaries, so it seems natural by restricting and taking quotients that it defines a linear map $H^i(M) \otimes H^j(N) \to H^{i+j}(M \times N)$. I think the easiest thing to do here is to check that $$H^i(M) \times H^j(N) \to H^{i+j}(M \times N),~~ \left([\omega], [\eta]\right) \mapsto [\pi^*_M \omega \wedge \pi^*_N \eta]$$ is well defined and is bilinear so that you can pass it to the tensor product.

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  • $\begingroup$ To see that $\Omega^i(M) \times \Omega^j(N) \rightarrow \Omega^{i+j}(M \times N)$ sends cocycles to cocycles is in fact easy to see. For if $\omega,\eta$ are cocycles then $d(\pi_M^*\omega \wedge \pi_N^*\eta) = 0$ easily. Now, for the coboundaries, am I supposed to assume that both $\omega$ and $\eta$ are coboundaries? because if so then I am having an hard time doing it since it goes to $\pi_M^*\omega \wedge \pi_N^*\eta = d(\pi_M^*\omega) \wedge d(\pi_N^*\eta)$ but I was supposed to get a single $d$ "outside" $\endgroup$
    – user57
    Commented Jan 14 at 14:13
  • $\begingroup$ Remember that $d(\alpha \wedge \beta) = d\alpha \wedge \beta + (-1)^{a}\alpha \wedge d\beta$ with $a = \deg(\alpha)$ and $d^2\beta = 0$ in general. Here we have $d\left(\pi^*_M \omega \wedge d\pi^*_N \eta\right) = d\pi^*_M \omega \wedge d\pi^*_N\eta$. $\endgroup$
    – NaNoS
    Commented Jan 14 at 14:32
  • $\begingroup$ I am sorry but I am not understanding how that adds on the problem. That's exactly the problem I have. I need to write $\pi_M^*\omega \wedge \pi_N^*\eta = d(something)$, for it to be a coboundary in $\Omega^{i+j}(M \times N)$. $\endgroup$
    – user57
    Commented Jan 14 at 14:36
  • $\begingroup$ If $\omega$ and $\eta$ are coboundaries, write $d\alpha = \omega$ and $d\beta = \eta$. Then I claim that $\pi_M^*\omega \wedge \pi_N^* \eta$ is a coboundary because $\pi_M^*\omega \wedge \pi_N^* \eta = d\pi_M^*\alpha \wedge d\pi_N^* \beta = d(\pi^*_M\alpha \wedge d\pi_N^* \beta)$. $\endgroup$
    – NaNoS
    Commented Jan 14 at 14:42
  • $\begingroup$ Oh nice, you putted the $d$ on the $d\beta$! ok thank you a lot! $\endgroup$
    – user57
    Commented Jan 14 at 14:50

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