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

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.

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.
• Thanks for your response. 'The chain rule for maps between open subsets and regular surfaces' - where am I able to find this? – beedge89 Aug 24 '15 at 9:42
• I have added an edit above and have tried to answer using option 2. Would this suffice? – beedge89 Aug 24 '15 at 10:42
• Looks good. Note that you don't want to restrict $\tilde{\mathbf{x}}_2^{-1}$ to $N$ because $N$ is not an open subset of $\mathbb{R}^3$ and then you can't use the regular chain rule. Also, you need to make the open subset on which $\mathbf{x}_1$ is defined possibly smaller so that the image of $\varphi \circ \mathbf{x}_1$ will land in $M$. – levap Aug 25 '15 at 9:11
• But N is the intersection of two open sets and is therefore open? – beedge89 Aug 26 '15 at 1:34
• $\mathbf{x}_2(U_2)$ is an open subset of the regular surface $S_2$, while $M$ is an open subset of $\mathbb{R}^3$. The intersection is an open subset of $S_2$ but not an open subset of $\mathbb{R}^3$. – levap Aug 26 '15 at 11:48