Nonzero total integral obstructs exactness on a closed manifold

If $\omega =d\eta$, compactness of $M$ makes $\eta$ compactly supported. Take the finite nonnegative chart cutoffs ${\chi }_i$ from [F1] near $\mathrm{supp}\eta$. They satisfy $\eta ={\sum }_i{\chi }_i\eta$ globally, so [F3] gives $d\eta ={\sum }_{i}d({\chi }_i\eta )$. Each ${\chi }_i\eta$ has compact support inside a boundaryless chart. Pull it into that chart and extend it by zero to a smooth compactly supported $(n-1)$-form on ${\mathbb{R}}^n$; its support stays away from the artificial chart edge. Chart-pullback naturality and [F2] make the signed chart integral of $d({\chi }_i\eta )$ zero. By the single-chart formula and finite linearity of [F1], ${\int }_M\omega ={\sum }_i{\int }_{M}d({\chi }_i\eta )=0$. Thus a nonzero integral excludes exactness.

Let $\omega$ be positive on nonempty $M$. It is then a nonzero nonnegative compactly supported top form, so the strict-positivity clause of [F1] gives ${\int }_M\omega >0$. Step 1.1 excludes exactness. On the empty manifold every integral is zero, so the nonzero-integral hypothesis cannot hold; the positive-form conclusion explicitly assumes nonempty $M$.