Prove that a (smooth) $1$-form $\omega$ on $S_1$ such that $\int_{S_1}\omega=0$ is the differential of some $f:S^1\to\mathbb{R}$.
Hint: Let $h:\mathbb{R}\to S^1$ defined by $h(t)=(\cos(t),\sin(t))$. And let $g(t)=\int_0^th^*\omega$. Prove that $g(t+2\pi)=g(t)$, hence $g=f\circ h$ for some $f$. Show that $df=\omega$.
I proved that if $t\in\mathbb{R}$ then $\int_t^{t+2\pi}h^*\omega=\int_{S_1}\omega =0$, so $g(t+2\pi)=g(t)$. Then clearly $f:S^1\to\mathbb{R}$ given by $f(\cos(t),\sin(t))=g(t)$ is well defined and $g=f\circ h$.
So, my question is how we prove $df = \omega$? This means $df_p=\omega(p)$ for $p\in S_1$.
However, if $h(t)=p$ by the chain rule $dg_t=df_{p}\circ dh_t$. I think we can compute $dg_t$ easily enough by the Calculus Fundamental Theorem. But is $dh_t$ invertible? (Then we would have $df_p=dg_t\circ dh_t^{-1}$.)
Any hint would be appreciated. Thank you.
you only have to prove the result locally. $h$ defines a coordinate chart locally and actually defines a local diffeomorphism. So if you look at the whole thing in the right conceptual way it's obvious.