 Solution

The integral of an exact differential form over the closed curve $S^1$ is zero by Stokes theorem, so the map is well defined on de Rham cohomology. It is surjective because the angular form $d\theta$ has integral $2\pi$. If a closed one-form $\alpha$ has zero integral, defineThe zero period makes this definition $2\pi$-periodic, and $dF=\alpha$. Thus the kernel is zero andis an isomorphism.