Top de Rham cohomology of a closed connected oriented manifold is real

By [F2], every smooth form on $M$ has compact support, because its support is closed in compact $M$. Consequently ${\mathrm{\Omega }}_c^k(M)={\mathrm{\Omega }}^k(M)$ in each degree, with the same differential. This includes the degree $n-1$ primitive space, so both kernels and images defining the degree-$n$ quotients agree. Thus $H_c^n(M)=H_{\mathrm{d}\mathrm{R}}^n(M)$ by the identity on representatives.

The other hypotheses are exactly those of [F1]. Its isomorphism sends the common class of $\omega$ to ${\int }_M\omega$, so under step 1.1 it is the asserted isomorphism on ordinary cohomology. In particular an ordinary exact top form has a compact primitive here, and a zero-integral form is exact. Surjectivity is witnessed by scalar multiples of the normalized bump in [F1].

At $n=0$ the manifold is a single oriented point as proved in [F1]; the map is multiplication by its orientation sign and the negative-degree image is zero. At $n=1$ primitives are ordinary smooth functions, all compactly supported because $M$ is compact. Zero forms have zero image. Nonemptiness is required for surjectivity, and boundarylessness for [F1]; there are no manifold boundary endpoints to omit. No partition or new choice is used in this identity of complexes.