Differential geometry: restriction of differentiable map to regular surface is differentiable

Solution 1:

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:

1. The map $\mathbf{x}_1 \colon U_1 \rightarrow \mathbb{R}^3$ is classically differentiable. Here, $U_1 \subseteq \mathbb{R}^2$ is an open set and $\mathbf{x}_1(U_1) \subseteq S_1$.
2. The map $\varphi \colon U \rightarrow \mathbb{R}^3$ is classically differentiable. Here, $U \subseteq \mathbb{R}^3$ is an open set and $\varphi(S_1) \subseteq S_2$.
3. The map $\mathbf{x}_2 \colon U_2 \rightarrow \mathbb{R}^3$ is classically differentiable. Here, $U_2 \subseteq \mathbb{R}^2$ is an open set and $\mathbf{x}_2(U_2) \subseteq S_2$.

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:

1. Prove that $\varphi \circ \mathbf{x}_1|_{U_1 \cap \mathbf{x}_1^{-1}(\varphi^{-1}(\mathbf{x}_2(U_2)))} \colon U_1 \cap \mathbf{x}_1^{-1}(\varphi^{-1}(\mathbf{x}_2(U_2))) \rightarrow \mathbf{x}_2(U_2)$ is a differentiable map between an open subset in $\mathbb{R^2}$ and a regular surface. Then, if you already proved this, apply the chain rule for maps between open subsets and regular surfaces to deduce that $\mathbf{x}_2^{-1} \circ \varphi \circ \mathbf{x}_1|_{U_1 \cap \mathbf{x}_1^{-1}(\varphi^{-1}(\mathbf{x}_2(U_2)))}$ is smooth as a map between open subsets (or as a map between regular surfaces - the two notions concide).
2. Show that one can extend the map $\mathbf{x}_2^{-1}$ to a map $\hat{\mathbf{x}}_2^{-1} \colon V \rightarrow \mathbb{R}^2$ where $\mathbf{x}_2(U_2) \subseteq V$ and $V$ is an open subset of $\mathbb{R}^3$ such that $\hat{\mathbf{x}}_2^{-1}$ is differentiable. Then, use the $\mathbb{R}^n$ version of the chain rule to deduce that the composition of $\varphi \circ \mathbf{x}_1|_{U_1 \cap \mathbf{x}_1^{-1}(\varphi^{-1}(\mathbf{x}_2(U_2)))} \colon U_1 \cap \mathbf{x}_1^{-1}(\varphi^{-1}(\mathbf{x}_2(U_2))) \rightarrow V$ and $\hat{\mathbf{x}}_2^{-1} \colon V \rightarrow \mathbb{R}^2$ is smooth, thus showing that $\mathbf{x}_2^{-1} \circ \varphi \circ \mathbf{x}_1|_{U_1 \cap \mathbf{x}_1^{-1}(\varphi^{-1}(\mathbf{x}_2(U_2)))}$ is also smooth. This is a standard argument done using the inverse function theorem and you can see an example of it in page 70 of Do Carmo's book.