Let $M$ be a smooth manifold, $f:M \to \mathbb{R}$ be a $C^{\infty}$ map and $f(p)=0$.
**My question:**Does there exist a neighborhood $U$ of $p$ in $M$ such that $f(U)=0$?
i know by coordinate system $(V,x)$ of $M$, $f\circ x^{-1}: \mathbb{R}^n \to \mathbb{R}$ is continuous and $W:= (f\circ x^{-1})^{-1}{(0)}$ is closed (not open subset). Is there any theorem that can help me?
If so, $f$ would be constant (on each connected component of $M$). To see this, assume that $M$ is connected, take some $p\in M$ and put $g := f - f(p)$. Then $g(p) = 0$. Hence, $g(x) = 0$ for all $x\in M$ in a neighborhood of $p$. Thus, $g$ is locally constant and so is $f$. But $f$ is continuous and therefore constant.