Can a function f be locally invertible while having a Jacobian with det = 0?

Question

I have learnt the Inverse Function Theorem, which states that if $f:\mathbb{R}^n \rightarrow \mathbb{R}^n$ is of class $C^1$, and $x_0 \in \mathbb{R}^n$,

then if $det(Jac_f(x_0)) \neq 0$ i.e. the Jacobian is invertible, $f$ admits local inverse $g$, and $Jac_g(f(x_0)) = (Jac_f(x_0))^{-1}$.

In an exercise though I was asked to check if the function $f(x,y) = (\cos x+\cos y, \sin x+\sin y)$:

A. Has local $C^1$ inverse in the points $(x, y)$ such that $x-y\in\pi\mathbb{Z}$. (I have checked that the Jacobian in those points is NOT invertible).

B. Has local inverse in the points $(x, y)$ such that $x-y\in\pi\mathbb{Z}$.

So my question is: is there a way to check if a function has local inverse without this function satisfying the hypothesis of the theorem?

Answer 1

Yes, this can happen even in case of $n=1$, for example take $f(x) = x^3$. The inverse theorem makes a claim in the other direction, that is if the determinant of the Jacobian is not zero, then there is ... .

Answer 2

It depends if you were asked for an inverse function, or for a $C^1$ inverse function. If the former, you can check by working directly from the definition. Every neighbourhood of $(x_0,y_0)$ contains an $\epsilon$-ball around $(x_0,y_0)$. So show that $f$ is not in invertible in any ball around $(x_0,y_0)$.

If the latter, then no, a zero derivative means the inverse doesn’t have a finite derivative.