From Do Carmo:
Let $S_1$, $S_2$ be regular surfaces. Suppose $S_1\subset V\subset \mathbb{R}^3$
and $\varphi:V\rightarrow \mathbb{R}^3$ is a differentiable map such that $\varphi(S_1)\subset S_2$. Then the restriction
\begin{align*}
\varphi\big|_{S_1}:S_1\rightarrow S_2
\end{align*}
is differentiable.
The statement seems obvious but I am having trouble proving it. Perhaps I am thinking about it too hard. I am comfortable with teh statement: 'The restriction of a smooth map is also smooth' – but this is in terms of smooth maps from Euclidean space to Euclidean space.
In order to show $\hat{\varphi}:=\varphi|_{S_1}$ is differentiable at some point $p\in S_1$ (when considered as a map between two regular surfaces), we must show that given parameterisations
\begin{align*}
\textbf{x}_1:U_1\subset\mathbb{R}^2&\rightarrow S_1\\
\textbf{x}_2:U_2\subset\mathbb{R}^2&\rightarrow S_2
\end{align*}
such that $p\in\textbf{x}_1(U_1)$ and $\varphi(p)\in\textbf{x}_2(U_2)$, that the map
\begin{align*}
\textbf{x}_2^{-1}\circ {\varphi}\circ \textbf{x}_1:U_1\rightarrow\mathbb{R}^2
\end{align*}
is a differentiable map between Euclidean space at $\textbf{x}_1^{-1}(p)$.
So how would one show this? The immediate 'argument' I think of is chain rule – 'the composition of differentiable maps is differentiable' — We have that $\textbf{x}_1$ is a differentiable map between Euclidean space by definition, as is $\varphi$. However, the map $\textbf{x}_2^{-1}$ is differentiable in the sense of a map between two regular surfaces ($S_2$ and the $xy$-plane). So can we really employ the chain rule here given that $\textbf{x}_2^{-1}$ is defined as only a map from a regular surface to $\mathbb{R}^2$?
What am I missing?
Edit: From the advice below, I have tried to adapt the argument on Pg 70 of Do Carmo: is this a suitable justification that
\begin{align*}
\textbf{x}_2^{-1}\circ\varphi\circ\textbf{x}_1
\end{align*}
is differentiable at $\textbf{x}_1^{-1}(p )$?
Given that $\varphi(\textbf{x}_1(U_1))\subset\textbf{x}_2(U_2)$, it follows there exists $q\in U_2$ such that $\varphi(p )=\textbf{x}_2(q)$.
Extend the map $\textbf{x}_2=(x_2(u,v),y_2(u,v),z_2(u,v))$ to a map $\tilde{\textbf{x}}_2:U_2\times \mathbb{R}\rightarrow \mathbb{R}^3$ defined by
\begin{align*}
\tilde{\textbf{x}}_2:(u,v,t)\rightarrow (x_2,y_2,z_2+t).
\end{align*}
Then $\tilde{\textbf{x}}_2$ is differentiable and $\tilde{\textbf{x}}_2|_{U_2\times\{0\}}=\textbf{x}_2$. The determinant of $d\tilde{\textbf{x}}_2(q)\neq 0$. Thus, by the inverse function theorem, there exists a neighbourhood $M$ ($\subset\mathbb{R}^3$) of $\textbf{x}_2(q)=\varphi(p )$ such that $\tilde{\textbf{x}}_2^{-1}$ exists AND is differentiable on $M$ (as a map between Euclidean spaces). Define $N:=\textbf{x}_2(U_2)\cap M$. Then $\tilde{\textbf{x}}_2^{-1}|_N$ is smooth (as a map between a subset of $\mathbb{R}^3$ and $U_2\times \mathbb{R}$ (restriction of smooth map is smooth)). If $\pi$ is the projection on to the first two factors, then $\pi$ is smooth and thus so is $\pi\circ \tilde{\textbf{x}}_2^{-1}|_N=\textbf{x}_2^{-1}$. Thus we have
\begin{align*}
\textbf{x}_2^{-1}\circ\varphi\circ\textbf{x}_1=\pi\circ\tilde{\textbf{x}}_2^{-1}\circ\varphi\circ\textbf{x}_1
\end{align*}
This map on the right hand side is a composition of smooth maps between Euclidean spaces. Thus, by the chain rule, it is smooth.
Best Answer
I'm assuming you are working with the definitions given in Do Carmo's "Curves and Surfaces" book and that $V \subseteq \mathbb{R}^3$ is an open set. First, let us write precisely the domains and ranges of the maps involved:
As you have noted, $\mathbf{x}_2^{-1} \colon \mathbf{x}_2(U_2) \rightarrow U_2$ is not classically differentiable but is differentiable in the sense of a map between a regular surface $\mathbf{x}_2(U_2)$ and an open set in $\mathbb{R}^2$ (if you want, you can embed $\mathbb{R}^2$ in $\mathbb{R}^3$ and consider $U_2$ as a regular surface in the sense of Do Carmo and then $\mathbf{x}_2^{-1}$ will be differentiable in the sense of a map between two regular surfaces but you won't gain much from it).
Thus, you have two options: