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Let $f:U\rightarrow \Bbb{R}^{m}$ be a function of class $C^{k\geq 1}$, where $U$ is an open subset of $\Bbb{R}^n$. If $f'(a)$ has rank $p$ for some $a\in U$, there exists an embedding (immersion that is homeomorphism into its image) $\phi:V\rightarrow U$ of class $C^\infty,$ where $V\subset\Bbb{R}^{p}$ such that $f\circ \phi:V\rightarrow\Bbb{R}^{m}$ is an embedding.

I have seen this question, and I did understand the intuition, but I'm having problems to write it down.

By a Lemma, since $\dim(\textrm{Im}(f'(a)))$, there exists a decomposition $\Bbb{R}^m=\Bbb{R}^p\bigoplus\Bbb{R}^{m-p}$ such that the projection $$\pi:\Bbb{R}^{m}\rightarrow\Bbb{R}^{p},(x,y)\mapsto x$$ applies $\textrm{Im}(f'(a))$ isomorphically.

Question: How do I construct the map $\phi$?

Maybe I can use this projection to define my $\phi$, but I don't know exactly how to do it.

Matheus Manzatto
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Mateus Rocha
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2 Answers2

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It is not needed anything much more substantial than what Llohann said in the post you link.

Take a complement $E$ of $\ker f'(a)$. If $\iota: E \to \mathbb{R}^n $ is the inclusion, then $f \circ \iota$ is an immersion near $a$. (We've just killed the kernel of $f'(a)$.) Locally, it is then an embedding. Now you just pick any isomorphism $\Psi: \mathbb{R}^p \to E$ and let $\phi = \iota \circ \Psi$, of course restricted to the necessary neighbourhood.

Aloizio Macedo
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By a proposition (LIMA, Curso de Análise, vol.2), since $f'(a)$ has rank $p$, there exists a neighbourhood $A$ where $f$ has constant rank $p$. Because $f'(a)$ has rank $p$, we also have that there is a subspace $E\subset \mathbb{R}^n$ $p$-dimensional such that $f'(a)\big|_E$ is injective and $\mathbb{R}^p\overset{\varphi}{\rightarrow} E$. If we define $V_1=\varphi^{-1}(A\cap E)$ and $\phi=(\varphi^{-1})\big|_{V_1}\rightarrow U$, we obtain an embedding of class $C^\infty$. As we consider $V_1\subset A$, where $f$ has rank $p$, it's possible to show that $f \circ \phi$ is an immersion and we can find an open subset $V\subset V_1$ such that $(f \circ \phi)\big|_V$ is an embedding of class $C^k$.

Laura Estivalez
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    "(...) since $f'(a)$ has rank $p$, there exists a neighbourhood $A$ where $f$ has constant rank $p$." This is false. The simplest example is perhaps $f(x)=x^2$. $f'(0)$ has rank $0$ and $f$ does not have constant rank around $0$. This bogs down to rank not being continuous, but lower semi-continuous. What is true is that if $f'(a)$ has rank $p$, then there exists a neighbourhood around $a$ where $f$ has rank greater than or equal to $p$. In particular, if $f$ has full rank at $a$, then there exists a neighborhood around $a$ where it has full rank. Maybe this is the proposition you mention. – Aloizio Macedo Apr 22 '20 at 17:45
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    You're absolutely right. What the proposition actually garantees is that we can find an open set $A$ where $f$ has constant rank in each of its connected components. Even more, $A=\cup_p A_p$, where $A_p$ is the interior of the subset of $U$ where $f$ has rank $p$. What happens in the counterexample you've presented is that $f'(x)$ has rank $0$ only for $x=0$, so $A_0= \emptyset$. – Laura Estivalez Apr 22 '20 at 21:28