Wedge product descend to the cohomology

Question

I found this statement in Raoul Bott "Differential Forms in Algebraic Topology": "Because the wedge product is an antiderivation, it descends to cohomology." Apparently this meant to be really obvious but I'm not sure how. So I thought of showing that for given $\omega, \omega_0 \in \Omega^p(M), \tau, \tau_0 \in \Omega^q(M)$ and $\omega_1 \in \Omega^{p-1}(M), \tau_1 \in \Omega^{q - 1}(M)$ such that $\omega = \omega_0 + d\omega_1, \tau = \tau_0 + d\tau_1$ then $\omega \wedge \tau = \omega_0 \wedge \tau_0 + d(\mbox{something...})$, but LHS is \begin{eqnarray} \omega \wedge \tau &=& \omega_0 \wedge \tau_0 + \omega_0 \wedge d\tau_1 + d\omega_1 \wedge \tau_0 + d\omega_1 \wedge d\tau_1 \\ &=& \omega_0 \wedge \tau_0 + \omega_0 \wedge d\tau_1 + d\omega_1 \wedge \tau_0 + d(\omega_1 \wedge d\tau_1) - (-1)^{p-1}\omega_1\wedge d^2\tau_1 \\ &=& \omega_0 \wedge \tau_0 + \omega_0 \wedge d\tau_1 + d\omega_1 \wedge \tau_0 + d(\omega_1 \wedge d\tau_1) \end{eqnarray} But I don't know how to put the rest in the total differential form. Am I even trying to show the right thing?

Answer 1

If $\omega$ and $\tau$ are closed forms (which you have not used), then

$$d(\omega \wedge \tau) = d\omega \wedge \tau + (-1)^{p-1}\omega \wedge d\tau = 0,$$

hence $\omega \wedge \tau$ is also closed, hence it corresponds to a class in cohomology. Let us see that this class depends only on the classes of $\omega$ and $\tau$. Let us suppose that $\omega = d\eta$ is exact, so that its cohomology class is trivial, and let us show that the cohomology class of $\omega \wedge \eta$ is also trivial. We have:

$$d\eta \wedge \tau = d(\eta \wedge \tau) - (-1)^{p-1}\eta \wedge d\tau$$

$$ = d(\eta \wedge \tau)$$

is also exact.