# “Counterexample” for the Inverse function theorem

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

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.