As I understand it, the inverse function theorem says:
Let $h$ be a continuously differential map from an open subset $U\subset \mathbb{R}^n$ into $\mathbb{R}^n$. If the Jacobian $Jh$ has nonzero determinant at $a\in U$, then there exist neighborhoods $V\ni a$ and $W \ni h(a)$ such that the restriction $\widehat{h}:V\rightarrow W$ is bijective; in which case, its unique inverse function $g:W\rightarrow V$ exists and is continuously differentiable, and its derivative satisfies the matrix equation $Jg = (J\widehat h)^{-1}$.
This is essentially the defining condition for a smooth function to be locally invertible, saying that if $h$ is smooth and its Jacobian is invertible at point $a$, then $h$ is locally invertible at point $a$. I am curious how the properties of smoothness, invertible Jacobians, and locally invertible functions relate to each other:
There exist functions that are locally invertible but not smooth; for example, this one:$$Z(x) = \begin{cases}x & x \in \mathbb{Q}\\\\ x+1 & x \notin \mathbb{Q}\end{cases}$$
My conjecture, which I'm trying to prove or find a counterexample for, is:
If a function $g:\mathbb{R}^n\rightarrow \mathbb{R}^n$ is locally invertible at point $a$, and $g$ is also continuously differentiable at point $a$, then its Jacobian exists and is nonzero at point $a$.
So far, I have constructed the example $Z(x)$, showing that it is possible to have a discontinuous, locally invertible function. I have struggled to construct a counterexample to my conjecture, which would amount to a locally invertible smooth function that nonetheless has pathological partial derivatives such that its Jacobian is noninvertible. Having enough of the partial derivatives be zero or otherwise linearly dependent would zero out the Jacobian, but I haven't found whether this is possible while still preserving local invertibility.