I am going through Elementary Differential Geometry by O'Neil, and I am at Theorem 3.2 on page 151. O'Neil comments that a rigorous proof of this theorem requires the methods of advanced calculus, and did not give a proof. I would love to see a proof of it, or get some hints as to how to construct the proof. The difficulty is in the definition of proper patches, that their inverse functions need only be continuous - not differentiable.
Following the Theorem I have included a lot of details in case some of the definitions are interpreted differently.
$\mathbf{Theorem 3.2:}$ Let $M$ be a surface in $\mathbb{R}^3$. If $F:\mathbb{R}^n \to \mathbb{R}^3$ is a (differentiable) mapping whose image lies in M, then considered as a function $F:\mathbb{R}^n \to M$ into $M$, $F$ is differentiable (as defined above).
The following paragraph is what (as defined above) refers to:
For a function $F:\mathbb{R}^n \to M$, each patch $\mathbf{x}$ in $M$ gives a coordinate expression $\mathbf{x}^{-1}(F)$ for F. Evidently this composition function is defined only on the set O of all points $\mathbf{p}$ of $\mathbb{R}^n$ such that $F(\mathbf{p})$ is in $\mathbf{x}(D)$. We define $F$ to be $\underline{differentiable}$ provided all its coordinate expressions are differentiable in the usual Euclidean sense.
The other required definitions are as follows:
A $\underline{mapping}$ is defined as a differentiable function $F:\mathbb{R}^n \to \mathbb{R}^m$.
A $\underline{coordinate patch}$ $\mathbf{x}: D \to \mathbb{R}^3$ is defined as a one-to-one regular mapping of an open set $D$ of $\mathbb{R}^2$ into $\mathbb{R}^3$. A $\underline{proper patch}$ as a patch for which the inverse function $\mathbf{x}^{-1}: \mathbf{x}(D) \to D$ is continuous (that is, has continuous coordinate functions).
Finally, a $\underline{surface}$ in $\mathbb{R}^3$ is a subset $M$ of $\mathbb{R}^3$ such that for each point $\mathbf{p}$ of $M$ there exists a proper patch in M whose image contains a neighborhood of $\mathbf{p}$ in $M$.
Any help is much appreciated.
All the best,
Nick