My definition of bump form is:
An $n$-form $\omega = f dx^1 \wedge \ldots \wedge dx^n$ on an $n$-dimensional smooth manifold $M$ is a bump form if it is compactly supported on a coordinate domain $(U_i, \varphi_i)$ and if $\left | \int_U \omega \right | = \pm 1$.
Now, when we consider the compactly supported cohomology $H_c^n(M)$, we know that if two compactly supported forms are cohomologous, then they have the same integral. That is because two cohomologous $n$-forms differ by an exact $(n-1)$-form (still compactly supported), and we can use Stokes' theorem to prove the claim.
In my lectures, in order to prove that the top degree cohomology of some manifold is trivial we proved that there exists a bump form in the manifold cohomologous to the $0$-form. Then the claim follows because any $n$-form can be written as a sum of bump forms, and any two bump forms in the manifold are cohomologous up to a sign.
My question is: how can a bump form be cohomologous to the $0$-form? If they were cohomologous, they would have the same integral! What am I missing?
My thoughts are: maybe the proof of the last claim is by contradiction? i.e. we are proving that there is no bump form in the manifold, otherwise such a bump form would be cohomologous to $0$. And that would be a contradiction. However, this line of reasoning didn't convince me.