Let $M$ be an abstract manifold, and let $f \in C^{\infty}(M)$. Assume that $df = 0$.
Prove that each level set $\{p \in M | f(p) = c\}$ is both closed and open. Conclude that $f$ is constant on each component of $M$, and show that the dimension of the kernel of $d$ in $C^{\infty}(M)$ is the number of connected components (which is either a finite number or infinite).
I consider $d: C^{\infty}(M) \to \Omega^1(M)$
$dim(N (d)) =$ number of connected components of $M$
if $df =0$, then $f$ is constant in each connected component of $M$. If $f \in N (d)$, $f = (c_i)$ such that
$i \in$ {connected components of $M$ }
If $f \in N (d)$; $f = \sum c_i f_i$ where $f_i: M \to \Bbb R.$
I think the road is good, but I don't know how to continue
$f$ is continuous therefore the preset of a closed set ($\{c\}$ in our case) is closed.
If $(U_p,\phi_p)$ is a chart containing $p$ then clearly $df=0$ implies $f_{U_p}$ is constant, hence our set $f^{-1}(c) = \cup_{p \in f^{-1}(c)} U_p $ is an union of open sets.