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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?

L.F. Cavenaghi
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  • Is A open? 2) by rank of $f$, you mean rank of $df_a$, right? If so, you just need to embed a small $p$-disc centered at $a$ whose tangent is transverse to $\ker df_a$.
    • – Llohann May 18 '16 at 11:53
  • @LlohannDallagnolSperança, I am sorry: 1) A is open! 2) Yes, I mean the rank of the linear transformation $df_a$. What dou you mean by embed a small p-disc centered at $a$ whose tangent is transverse to $\ker df_a?$ – L.F. Cavenaghi May 18 '16 at 14:44
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    thining on $\ker df_a$ as a $n-p$ dimensional subspace, you can choose any $p$-dimensional complement to it. If you take a small disc in such complement and translate it to $a$, you can prove that the restriction of $f$ to this disc has rank $p$ and is injective. – Llohann May 18 '16 at 14:51