11
$\begingroup$

In a lecture we stated the theorem as follows:

Let $\Omega\subseteq\mathbb{R}^n$ be an open set and $f:\Omega\to\mathbb{R}^n$ a $\mathscr{C}^1(\Omega)$ function. If $|J_f(a)|\ne0$ for some $a\in\Omega$ then there exists $\delta>0$ such that $g:=f\vert_{B(a,\delta)}$ is injective and ...

This only is a sufficient condition, so is there any function whose jacobian has determinant $0$ at every point but still is injective? If the determinant only vanished on one single point something similar to $f(x)=x^3$ at $x=0$ in $\mathbb{R}$ would do the trick, but if $|f'(x)|=0$ for every $x\in\Omega\subseteq\mathbb{R}$ then $f$ is constant and not injective. Does the same hold in $\mathbb{R}^n$?

Thanks

$\endgroup$

1 Answer 1

12
$\begingroup$

Actually, this is not possible in $\mathbb{R}^n$ either.

Indeed, if you have any $\mathscr{C}^1$ injective function $f: \Omega \rightarrow \mathbb{R}^n$, then $f$ is open and a homeomorphism on its image (invariance of domain : https://en.m.wikipedia.org/wiki/Invariance_of_domain ).

From Sard’s theorem (https://en.m.wikipedia.org/wiki/Sard%27s_theorem ), the set of critical values has null measure in $\mathbb{R}^n$, thus has empty interior, thus the set of critical points has no interior as well.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .