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.
Best Answer
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
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
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 $q = (0,0)$.