Question about the definition of differentiability of a map between surfaces

Question

In his book on Curves and Surfaces, Do Carmo defines what a differentiable map between surfaces means as follows:

A continuous map $\varphi:V_1 \subset S_1 \to S_2$ of an open set $V_1$ of a regular surface $S_1$ to a regular surface $S_2$ is said to be differentiable at $p \in V_1$ if, given parametrizations $$\mathbf{x}_1: U_1 \subset \mathbb{R}^2 \to S_1, \quad \mathbf{x}_2: U_2\subset \mathbb{R}^2 \to S_2$$ with $p \in \mathbf{x}_1(U_1)$ and $\varphi(\mathbf{x}_1(U_1))\subset \mathbf{x}_2(U_2),$ the map $$\mathbf{x}_2^{-1}\circ \varphi \circ \mathbf{x}_1:U_1 \to U_2$$ is differentiable at $q=\mathbf{x}_1^{-1}(p).$

My question is the following: Why do we need to assume that $\varphi$ is a continuous map? If we remove that word from the definition, is there an example of a map $\varphi: S_1 \to S_2$ between surfaces $S_1$ and $S_2$ that satisfies this new definition but that is not continuous?

Thanks in advance.

Answer 1

In standard multivariable calculus we consider functions $g : U \to V$ between open subsets $U \subset \mathbb R^m$ and $V \subset \mathbb R^n$. In fact, one can easily see that differentiability in $x \in U$ implies continuity in $x$.

do Carmo defines the concept of differentiability of a function $\varphi : V \to S_2$ by reducing it to the differentiability of the functions $\mathbf{x}_2^{-1}\circ \varphi \circ \mathbf{x}_1:U_1 \to U_2$ in the multivariable sense. In my opinion is not absolutely clear what he means by

given parametrizations $$\mathbf{x}_1: U_1 \subset \mathbb{R}^2 \to S_1, \quad \mathbf{x}_2: U_2\subset \mathbb{R}^2 \to S_2$$ with $p \in \mathbf{x}_1(U_1)$ and $\varphi(\mathbf{x}_1(U_1))\subset \mathbf{x}_2(U_2)$.

Does he mean that we have to consider some such pair of parametrizations of all such pairs? Anyway, one can show that this does not really matter. In fact, do Carmo writes

The proof that this definiton does not depend on the choice of parametrizations is left as an exercise.

Whatever the "correct" interpretation of the definition may be, the continuity of $\varphi$ at $p$ is needed to assure the existence of a pair of parametrizations such that $\varphi(\mathbf{x}_1(U_1))\subset \mathbf{x}_2(U_2)$. Note that $p \in \mathbf{x}_1(U_1)$ and $\varphi(\mathbf{x}_1(U_1))\subset \mathbf{x}_2(U_2)$ is only possible if $\varphi(p) \in \mathbf{x}_2(U_2)$. That is, the minimal requirement is that we take parameterizations $\mathbf{x}_1$ around $p \in S_1$ and $\mathbf{x}_2$ around $\varphi(p) \in S_2$.

We can of course always consider the subset $\varphi^{-1}(\mathbf{x}_2(U_2)) \subset S_1$, but if $\varphi$ is not continuous at $p$, then we cannot be sure that $\varphi^{-1}(\mathbf{x}_2(U_2))$ is an open neigborhood of $p$ in $S_1$. It is even possible that $\varphi^{-1}(\mathbf{x}_2(U_2)) = \{p\}$. Thus there may not exist a parameterization $\mathbf{x}_1$ around $p \in S_1$ such that $\mathbf{x}_1(U_1) \subset \varphi^{-1}(\mathbf{x}_2(U_2))$ which is the same as $\varphi(\mathbf{x}_1(U_1))\subset \mathbf{x}_2(U_2)$.

You should compare this with Definition 1 on p. 72. Here do Carmo considers functions (no continuity assumption!) $f: V \to \mathbb R$ and defines differentiability via a parameterization around $p$. Continuity is not required here because $f$ maps each open neigborhood of $p$ in $S$ to $\mathbb R$. It is easy to see that differentiability implies continuity in this case.

Update:

Here is an example of "poor behavior" of non-continuous maps.

Take $S_1 = U \times \{0\} \subset \mathbb R^3$ with $U = \{(x,y) \in \mathbb R^2 \mid x^2 + y^2 < 1\}$ and $S_2 = S^2$ = unit sphere in $\mathbb R^3$. Define $$\varphi: S_1 \to S_2, \varphi(x,y, 0) = \begin{cases} (x,y, \sqrt{1 - x^2-y^2}) & (x,y) \ne (0,0) \\ (0,0,-1) & (x,y) = (0,0) \end{cases}$$ Consider the point $p = (0,0,0) \in S_1$. Around $\varphi(p)$ we take do Carmo's parameterization $\mathbf{x_2} : U \to \mathbb R^3$ defined on p.56. Then $\varphi^{-1}(\mathbf{x}_2(U)) = \{p\}$ and therefore no parameterization $\mathbf{x_1} : U_1 \to \mathbb R^3$ of $S_1$ around $p$ has the property $\varphi(\mathbf{x}_1(U_1))\subset \mathbf{x}_2(U_2)$. You can nevertheless consider the function $\mathbf{x}_2^{-1}\circ \varphi \circ \mathbf{x}_1$, but its domain is only $\{(0,0)\}$. It does not make sense to call such a function differentiable at $q = (0,0)$.

Answer 2

As Paul explained, without continuity, there is no way to guarantee that the domain of the composition $\psi=x_2^{-1}\circ \phi \circ x_1$ is open in $R^2$. Of course, one can generalize the definition of a differentiable map so that the domain of the map need not be open:

Given a subset $E\subset {\mathbb R}^n$, a function $f: E\to {\mathbb R}^m$ and a point $p\in E$, one says that $f$ is differentiable at $p$ if there exists a linear map $L: {\mathbb R}^n\to {\mathbb R}^m$, such that $$ f(x)=f(p) + L(x-p) + \eta(x), $$
where $\eta(x)=o(|x-p|)$ as $x\to p$ in $E$.

However, this definition is always satisfied if $p$ is an isolated point in $E$, which leads to trouble and, when defining differentiability, one normally assumes that this is not the case. In your situation, there is no way to guarantee that the domain of $\psi$ has no isolated points (if you do not assume continuity of $\phi$). Here is another troublesome example:

Consider $S_1={\mathbb R}^2$ and $S_2= {\mathbb R}^2\times \{0, 1\}$ (two copies of the plane). Consider the discontinuous function $\phi: S_1\to S_2$,
$$ \phi(x,y)= (x,y)\times \{0\} $$ if $(x,y)\in {\mathbb Q}^2$ and $$ \phi(x,y)= (x,y)\times \{1\} $$ if $(x,y)\notin {\mathbb Q}^2$.

Take the identity map as the parameterization of $S_1$. Choose the two "obvious" charts on $S_2$, which are $$ \varphi_0: (x,y)\times \{0\}\mapsto (x,y) $$ $$ \varphi_1: (x,y)\times \{1\}\mapsto (x,y). $$ Then the compositions $\varphi_i\circ \phi$ are both differentiable everywhere in the sense of the above definition. However, $\phi$ is discontinuous.

By tinkering further with the definitions, one can eliminate such examples, but why bother? It is much easier to require continuity of the map to begin with.

Answer 3

After thinking about this question, I concluded that the word "continuous" on the original definition may be omitted. In other words, I propose the following new definition that differs on the original one only at the word "continuous":

Definition 1. A map $\varphi:V_1 \subset S_1 \to S_2$ of an open set $V_1$ of a regular surface $S_1$ to a regular surface $S_2$ is said to be differentiable at $p \in V_1$ if, given parametrizations $$\mathbf{x}_1:U_1 \subset \mathbb{R}^2 \to S_1, \quad \mathbf{x}_2:U_2 \subset \mathbb{R}^2 \to S_2$$ with $p \in \mathbf{x}_1(U_1)$ and $\varphi(\mathbf{x_1}(U_1)) \subset \mathbf{x}_2(U_2),$ the map $$\mathbf{x}_2^{-1}\circ \varphi \circ \mathbf{x}_1 : U_1 \to U_2$$ is differentiable at $q=\mathbf{x}_1^{-1}(p)$.

Now I assert that we can deduce at least the continuity of $\varphi$ at $p$ from this new definition. That is,

Proposition Let $\varphi: V_1 \subset S_1 \to S_2$ a map between regular surfaces defined on an open set $V_1$ of $S_1$.If $p \in V_1$ and $\varphi$ is differentiable (in the sense of Def. 1) at $p$, then $\varphi$ is continuous at $p$.

Proof. By the Definition 1, we can choose parametrizations $$\mathbf{x}_1:U_1 \subset \mathbb{R}^2 \to S_1, \quad \mathbf{x}_2:U_2 \subset \mathbb{R}^2 \to S_2$$ with $p \in \mathbf{x}_1(U_1)$ and $\varphi(\mathbf{x_1}(U_1)) \subset \mathbf{x}_2(U_2),$ such that the map $$\bar{\varphi}=\mathbf{x}_2^{-1}\circ \varphi \circ \mathbf{x}_1 : U_1 \to U_2$$ is differentiable at $q=\mathbf{x}_1^{-1}(p)$.

Observe that by the conditions given in Def 1., we can actually compose $\mathbf{x}_2^{-1}\circ \varphi \circ \mathbf{x}_1$ without any problem. Since $\bar{\varphi}$ is differentiable (in the usual sense) at $q$, then it is continuous at $q$. And hence, the map $\mathbf{x}_2\circ \bar{\varphi} \circ \mathbf{x}_1^{-1}$ is continuous at $p$. Furthermore, $$\varphi|_{\mathbf{x}_1(U_1)}=\mathbf{x}_2\circ \bar{\varphi} \circ \mathbf{x}_1^{-1}.$$

Hence, $\varphi|_{\mathbf{x}_1(U_1)}$ is continuous at $p$. But, since $\mathbf{x}_1(U_1)$ is an open subset of $S_1$, then, in fact, $\varphi$ is continuous at $p$. The proposition follows. $\blacksquare$

Observe that the previous proposition shows the continuity of $\varphi$ at $p$, but not the continuity of $\varphi$ over all $V_1,$ as the original definition requires. This procedure actually guarantees that $\varphi^{-1}(\mathbf{x}_2(U_2))$ contains an open neighbourhood of $p$ at $S_1$ since $\varphi$ is continuous at $p$ and $\varphi(p) \in \mathbf{x}_2(U_2).$ This makes me think that we don't actually need to assume continuity on the original definition. Am I right?