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?
Best Answer
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.