Showing de Rham cohomology $H^1(S^n)$ is zero I'm trying to find an elementary way to see that the 1st de Rham cohomology of the n-sphere is zero for $n>1$, $H^1(S^n) = 0$.
This is part of an attempt to find the de Rham cohomology of the n sphere generally. I have shown that $H^k(S^n) = H^{k-1}(S^{n-1})$ (for $k>1$) but to find the de Rham cohomology generally I need to show that $H^1(S^n) = 0$ and I am struggling to do this.
Is it possible to take a closed one form $\alpha$ on a sphere and explicitly find a function $f$ on $S^n$ that has $df = \alpha$, I've tried writing it out in coordinates and explicitly integrating but I haven't managed to get very far with this!

I think the function you are looking for is $f(x)=\int_{\gamma_x} \alpha$ where $\gamma_x$ is a path from some chosen base point $x_0$ to $x$. But then for this to be well defined you have to say that all paths between $x$ and $x_0$ are homotopic, i.e. that your sphere is simply connected, so the whole argument gets a bit circular...
P.S. Traying to write things out in coordinates will not get you far because a closed form not being exact is a "global" thing, not a local one
