Regular level set theorem proof, manifolds Let $F:N\rightarrow M$ be a smooth man between manifolds of dimension $n$ and $m$ respectively. A non-empty regular level set $F^{-1}(c)$ where $c\in M$ is a submanifold of $N$ of dimension equal to $n-m$.
The proof starts of by something like this:
Choose a chart $(V,\psi)=(V,y^1,.....,y^m)$ centred about $c$. By continuity of smooth maps, $F^{-1}(V)$ is an open set in $N$ that contains $F^{-1}(c)$. In $F^{-1}(V)$, $F^{-1}(c)=(\psi \circ F)^{-1}(\textbf{0})$.
However the part I don't understand is the following:
In $F^{-1}(V)$, $F^{-1}(c)$ is the common zero set of the functions $r^i \circ (\psi \circ F)$, where $r^i$ is the standard coordinates on euclidean space.
How do I actually prove that?

Upgrading my comment to an answer:
We wish to show that $F^{-1}(c)$ is the common zero set of the functions $r^i \circ \psi \circ F$ for $i \in \{1, \dots, m\}$. In other words, we want
$$F^{-1}(c) = \bigcap_{i = 1}^m (r^i \circ \psi \circ F)^{-1}(0).$$
We already know that $F^{-1}(c) = (\psi \circ F)^{-1}(\mathbf{0})$, and of course $\{\mathbf{0}\} = \bigcap_{i=1}^m (r^i)^{-1}(0)$. Thus,
$$F^{-1}(c) = (\psi \circ F)^{-1}(\{\mathbf{0}\}) = (\psi \circ F)^{-1}\left(\bigcap_{i=1}^m (r^i)^{-1}(0)\right) = \bigcap_{i=1}^m (\psi \circ F)^{-1}((r^i)^{-1}(0)) = \bigcap_{i=1}^m (r^i \circ \psi \circ F)^{-1}(0),$$
as desired.
