I am looking for a regular surface such that a point on the surface, the parametrization of local coordinates is differentiable with exactly continuous inverse. For $f:\mathbb{R} \to \mathbb{R}, f(x)=x^3$ is a differentiable function with exactly continuous inverse, $f^{-1}:\mathbb{R} \to \mathbb{R}, f^{-1}(y)=y^{1/3}$. I am looking for something like this for surfaces.
I am using Do Carmo's definition of a regular surface, from Differential Geometry on Curves and Surfaces:
DEFINITION $1$. A subset $S \in \mathbb{R}^3$ is a regular surface if, for each $p \in S$, there exists a neighborhood $V$ in $\mathbb{R}^3$ and a map $x: U \to V \cap S$ of an open set $U \in R^2$ onto $V \cap S \in \mathbb{R}^3$ such that
- $\bf{x}$ is differentiable. This means that if we write $\textbf{x}(u, v) = (x(u,v), ~y(u,v), ~z(u,v)), ~(u,v) ∈ U,$ the functions $x(u,v),y(u,v),z(u,v)$ have continuous partial derivatives of all orders in $U$.
- $\bf{x}$ is a homeomorphism. Since $\bf{x}$ is continuous by condition 1, this means that $\bf{x}$ has an inverse $\textbf{x}^{−1} : V \cap S \to U$ which is continuous.
- (The regularity condition.) For each q ∈ U, the differential $d\textbf{x}_q: \mathbb{R}^2 → \mathbb{R}^3$ is one-to-one.