= 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, define
$$
F(e^{i\theta})=\int_0^\theta\alpha.
$$
The zero period makes this definition $2\pi$-periodic, and $dF=\alpha$. Thus the kernel is zero and
$$
H^1_{\mathrm{dR}}(S^1)\xrightarrow{\int_{S^1}}\mathbb R
$$
is an isomorphism.
Back to article page
