# Product of inexact differential forms is inexact

Suppose we have a product manifold $M = M_1 \times M_2$. Let $\omega$ be a closed but inexact form on $M_1$ and $\eta$ a closed but inexact form on $M_2$. Then the claim is that $$\omega \wedge \eta$$ is inexact on the manifold $M$. I $\require{enclose}\enclose{horizontalstrike}{\mathrm{suspect}}$ know that the general statement "a product of inexact forms is inexact" is false, but I don't know how to go about proving the above claim.

The context of this question is in de Rham cohomology: I'm trying to prove Künneth's formula, and as a prelude I want to show that differential forms of the form above correspond to non-trivial elements of the cohomology groups of $M$.

• For a counterexample to the general case, it's not too hard to find inexact $\omega,\eta$ such that $\omega \wedge \eta = 0$. – Daniel Fischer May 6 '16 at 13:06
• Ah that's obvious now, thank you – gj255 May 6 '16 at 13:21

I think the key here is de Rham's theorem, (part of) which can be pertinently paraphrased:

A closed $k$-form $\omega$ is exact if and only if $\int_c \omega = 0$ for every closed $k$-chain $c$.

The fact that

$$\int_{c_1 \times c_2} \omega \wedge \eta = \int_{c_1} \omega \int_{c_2} \eta$$

is then all you need.