Sorry, this question may seems a bit pedantic and trivial.
In the language of differential forms, if $f$ is a zero-form, and I get $df = 0$ with $d$ being the exterior derivative. Then we know $f = C$ where C is some constant number. I was wondering if there is a particular proposition or theorem to justify this. I am not talking about basic calculus because that is trivially true. However, in the language of differential forms, do we need a few more steps to jump from $df=0$ to $f = C$?
Thanks a lot.
Actually $f$ constant on each connected component. Prove the claim for functions on connected manifolds first. Suppose $f : M \to \mathbb{R}$ is a smooth map on a connected manifold with $df=0$. Fix a point $p \in M$, you want to show that the set $$ \mathcal{C} =\{ q\in M \mid f(q)=f(p) \} $$ is open and closed in $M$. Hence $\mathcal{C} = M$ which means that $f$ is constant on $M$. The set $\mathcal{C}$ is closed trivially. Show $\mathcal{C}$ is open by using charts and result in elementary calculus.