How to show that a locally exact form is globally closed

Question

Every closed differential form is locally exact. That's quite obvious, as it follows (Poincare lemma) from the vanishing $k$-th cohomology groups ($k>0$) of contractible open subsets of $\mathbb R^n$, which is what the local description is. But I have some trouble showing the other direction. Say for $k=1$. Given a differential 1-form $\omega$, which is locally exact: \begin{equation} \omega(x)=\omega_i(x)dx^i=\frac{\partial f}{\partial x^i}(x)dx^i \end{equation} for some function $f$ and on some local coordinates $x^i$. Then how do I see that this form is globally closed?

Answer 1

If $\omega$ is locally exact, then $d \omega$ is locally equal to zero (because the differential commutes with the restriction operator, and an exact form is closed). But the only form which is locally equal to zero is the zero form, hence $d\omega = 0$ and $\omega$ is closed.