10
$\begingroup$

In my class 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

$\endgroup$
10
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.