If $f: A \subset \mathbb{R}^m \to \mathbb{R}^n$ is of class $C^1$ and in $a \in A$ rank of $f$ is $p$ there is an embedding $\phi : V \to A$, of class $C^{\infty}$ such that $f\circ \phi$ is an embedding.
My attempt was: Let $\mathbb{R}^m = \mathbb{R}^p\times \mathbb{R}^{m-p}.$ Consider $\pi :A \to \mathbb{R}^p$ defined as $\pi(x,y) = x$. Then $\pi(A)$ is open. Take $V = \pi(A)$.
Then let $\phi : V \to A$ defined as $\phi(x) = (x,0)$. Note that $\phi$ is injective and it is an immersion. Then it is an embedding. Now $f\circ \phi(x) = (x,0)$ locally by the rank theorem, then $f\circ \phi$ is an embedding.
It is right?