On the proof that the inverse value set of a regular value is a submanifold

On the proof that the inverse value set of a regular value is a submanifold

I have a doubt on the proof of the following, well-known theorem:
Let $f:M^m\rightarrow N^n$ ($m\geq n$) be a $C$ map, $r\geq 1$.If $y\in
f(M)$ is a regular value, then $f^{-1}(y)$ is a $C$ submanifold of $M$
The proof for this uses the Local Form of Submersions, that is, $f$ can be
representated, in open sets A and B around $x\in f^{-1}(y)$ and $y$,
respectively, as $$
\tilde{f}:V\underset{open}\subset\mathbb{R}^n\times\mathbb{R}^{m-n}\cong
\mathbb{R}^m\longrightarrow W\underset{open}\subset \mathbb{R}^n $$ such
that $\tilde{f}(x,y)= x$ in adequate open sets $V$ and $W$, where $y\in B$
corresponds to $0\in W$. The set $f^{-1}(y)\cap A$ is the diffeomorphic
(by the local chart) to $\tilde{f}^{-1}(0)\cap V=\{(x,y)\in V:x=0\}$
Then, if
$\pi:\mathbb{R}^n\times\mathbb{R}^{m-n}\rightarrow\mathbb{R}^{m-n}$ is the
projection $\pi(x,y)=y$, then $\left(\pi\vert_{\tilde{f}^{-1}(0)\cap V},
\tilde{f}^{-1}(0)\cap V\right)$ is a local chart for
$\tilde{f}^{-1}(0)\cap V$, which is open in $\tilde{f}^{-1}(0)$ since $V$
is open.
Using the local character of submanifolds and its invariance under
diffeomorphisms, and noting that $\tilde{f}^{-1}(0)\cap V$ is an
isomorphism, we have that $f^{-1}(y)\cap A$ is a submanifold of $M$.
Is this correct?