# Künneth formula from de Rham cohomology (Nakahara)

Nakahara, in his book "Geometry, Topology and Physics", states the following proof leading up to the Künneth formula. Let $M_1, M_2$ be two smooth manifolds, then the wedge of any pair of non-trivial representatives $\omega_1$ of $H^p(M_1)$ and $\omega_2$ of $H^q(M_2)$, is a non-trivial representative of $H^{p+q}(M_1\times M_2)$. Proving the inclusion $\bigoplus_{p+q=r} H^p(M_1)\otimes H^q(M_2)\subset H^r(M_1\times M_2)$.

It is clear that $[\omega_1\wedge\omega_2]\in H^{p+q}(M_1\times M_2)$, it remains to show non-triviality. Suppose on the contrary that $\omega_1\wedge\omega_2$ were exact, then $$\omega_1\wedge\omega_2=d(\alpha\wedge\beta+\gamma\wedge\delta)=d\alpha\wedge\beta+(-1)^{p-1}\alpha\wedge d\beta+d\gamma\wedge\delta+(-1)^p\gamma\wedge d\delta\quad(*)$$ for some $\alpha\in\Omega^{p-1}(M_1),\beta\in\Omega^{q}(M_1),\gamma\in\Omega^{p}(M_1),\delta\in\Omega^{q-1}(M_1)$. Nakahara claims that by comparing the two sides of $(*)$ we should find $\alpha=0$, $\delta=0$ which provides the necessary contradiction.

This is what I have so far. Since $\Omega^s(M_1)\wedge\Omega^{t}(M_2)$ and $\Omega^{s'}(M_1)\wedge\Omega^{t'}(M_2)$ are linearly dependent only when $s=s'$ and $t=t'$, we can separate $(*)$ into three parts $$\omega_1\wedge\omega_2=d\alpha\wedge\beta+(-1)^p\gamma\wedge d\delta\quad\text{with}\quad \alpha\wedge d\beta=0,\quad d\gamma\wedge\delta=0$$ To satisfy the last two equations we distinguish four cases:

1. $\alpha=0,\delta=0$;
2. $d\beta=0,\delta=0$;
3. $\alpha=0,d\gamma=0$;
4. $d\beta=0,d\gamma=0$.

For the first three cases the contradiction with the non-exactness of $\omega_1$ and $\omega_2$ is immediate after substitution. In the fourth case however, we retain $\omega_1\wedge\omega_2=d\alpha\wedge\beta+(-1)^p\gamma\wedge d\delta$ with both $\beta\in\ker d$ and $\gamma\in\ker d$ and I am unsure how to proceed from there.

## 1 Answer

Note that you have a few typos: $\beta\in\Omega^q(M_2)$ and $\delta\in\Omega^{q-1}(M_2)$.

So let's think about integrating $\omega_1\wedge\omega_2$ over a typical generating $(p+q)$-cycle $\sigma\times\tau$ in $M_1\times M_2$. We'll have $$\int_{\sigma\times\tau}\omega_1\wedge\omega_2 = \left(\int_\sigma\omega_1\right)\left(\int_\tau \omega_2\right).$$ On the other hand, $$\int_{\sigma\times\tau} d\alpha\wedge\beta = \left(\int_\sigma d\alpha\right)\left(\int_\tau \beta\right) = 0,$$ since the first integral on the right-hand side is $0$ by Stokes's Theorem. Similarly, $\displaystyle\int_{\sigma\times\tau} \gamma\wedge d\delta = 0$. It follows that either $\displaystyle\int_\sigma\omega_1=0$ or $\displaystyle\int_\tau\omega_2=0$. Since $\sigma$ and $\tau$ are arbitrary cycles, we conclude that either $\omega_1$ or $\omega_2$ must represent a trivial cohomology class. (If not, we could choose $\sigma$ and $\tau$ with both integrals nonzero.)