I want to compute the first DeRham cohomology group of the circle. In symbols $H^1_{dR}(S^1)$. Let $p(x)=e^{ix}$ the map from $\mathbb{R}$ to the circle $S^1$, $\Omega^1(S^1)$ the set of all $1$-forms on $S^1$. Then for $\alpha\in\Omega^1(S^1)$ we have $p^*\alpha=f(t)dt$ for some function $f$.
I want to deduce an isomorphism $\psi:H^1_{dR}(M)\rightarrow\mathbb{R}$. Therefore define $\tilde{\psi}:\Omega^1(S^1)\rightarrow\mathbb{R}$ by $\tilde{\psi}(\alpha)=\int_0^{2\pi}f(t)$. The claim is that this determines an isomorphism $\psi:H^1_{dR}(S^1)\rightarrow\mathbb{R}$. I already show that this map $\psi$ is well-defined and $\mathbb{R}$ linear. I now have to show that if $\psi(\alpha)=0$ that this implies that $\alpha$ is exact.
I think you have to consider the function $p^*g(t)=\int_0^tf(\tau)d\tau$ but I don't see how. Can someone say me how to get the conclusion I want? Does all this imply directly that we have the isomorphism? Can i also do the same argument for $H^1_{dR}(S^1\times S^1)$?
HINT: Think about why $\int_0^t f(\tau)d\tau$ gives you a well-defined potential function down on $S^1$. This is path-independence for the integral on the circle.
Yes, the same technique works for $H^1_{dR}(S^1\times\dots\times S^1)$.