In the proof of the relation between Lusternik-Schnirelman-category and Cup length (of de Rham Cohomology) for smooth manifolds from this note (theorem 2) the argument goes: Let the given manifold $M$ be covered by $U_1,...,U_l$, each $U_i$ contractible in $M$. Let $f_i$ be the contractions. Now let $\omega_j$, $j=1,...,l$ be closed forms. We need to show their wedge product is exact. By the Poincaré lemma, $(f_i^*\omega_i)|U_i=0$.
Now this: Since, for each $i$, $f_i$ is homotopic to the identity, there exists $\theta_i$ such that $\omega_i=(f_i^*\omega_i) + d\theta_i$. As I previously had little exposure to differential forms, I cannot see why pulling back with such $f_i$ amounts to adding an exact form.
Because $f_i$ is homotopic to the identity, the induced map on cohomology is the identity. So $[f_i^* \omega] = [\omega]$. To say that two forms are cohomologous means precisely that they differ by an exact form, so that $\omega = f_i^*\omega + d\theta$.