# Can I use de Rham Cohomology to prove exactness of 1-forms on the $2$ sphere?

The exercise ask to prove that for every 1-form closed on the $2$ sphere there was a function defined on the sphere such that its differential is the form. Then it asks if this function is unique.

Well, I use the fact that the $1$ de Rham cohomology on the $2$ sphere is trivial to claim the existence of such function.

Then, I use Stoke's theorem with the fact the the sphere is a compact regular domain to claim it uniqueness.

Is there anything wrong with this argumentation?

The first part is fine. By definition, $H^1(S^2) = 0$ means that any element in $\ker(d : \Omega^1(S^2) \to \Omega^2(S^2))$, i.e. any closed $1$-form, is actually in $\operatorname{im}(d : \Omega^0(S^2) \to \Omega^1(S^2))$, i.e. it is exact, i.e. it is the differential of some function.
Suppose that you have a closed $1$-form $\alpha$. You know it's exact, so $\alpha = df$ for some $f$. Now if you have another function $g$ such that $\alpha = dg$, then you get $df = dg \implies d(f-g) = 0$. The question is does $f=g$, or equivalently does $f-g = 0$. So now you are essentially asking: is the zero function the only closed $0$-form (function) on $S^2$?
By definition $H^0(S^2) = \ker(d : \Omega^0(S^2) \to \Omega^1(S^2))$, and you don't quotient by $\operatorname{im}(d)$, because there is no such thing as $\Omega^{-1}$. So you are asking if $H^0(S^2) = 0$. The answer is no, because $H^0(S^2) = \mathbb{R}$, which basically follows from the fact that $S^2$ is (path-)connected. In fact it's easy to see that $H^0(S^2)$ is spanned by the constant functions on $S^2$.
More concretely, given some $f$ such that $df = \alpha$, you can always add a constant $c \in \mathbb{R}$ to $f$, and you still get $d(f+c) = df + dc = df = \alpha$.