Do Carmo Differential Geometry book: Proposition 4, Chapter 2.2. Mistake in the proof?

Question

In Proposition 4 of Chapter 2.2 in Do Carmo's differential geometry book, the author claims that if we already know that a set $S\subseteq \mathbb{R}^3$ is a regular surface, and we have a candidate $\textbf{x}: U \subset \mathbb{R}^2 \rightarrow \mathbb{R}^3 $ for a parametrization, we do not have to check that $\textbf{x}^{-1}$ is continuous, provided that the other conditions (for $\textbf{x}$ being a parametrization) hold. I think there is an error in the proof. I will first state the relevant definition:

Definition 1:

A subset $S\subset \mathbb{R}^3$ is a regular surface if, for each $p \in S$, there exists an open set $V$ in $\mathbb{R}^3$, and a map $\textbf{x}: U \rightarrow V \cap S$ of an open set $U \subseteq \mathbb{R}^2$ onto $V \cap S$ such that

1. $\textbf{x}$ is infinitely differentiable.
2. $\textbf{x}$ is homeomorphism.
3. For each $q \in U$, the differential $d\textbf{x}_q : \mathbb{R}^2 \rightarrow \mathbb{R}^3$ is one-to-one.

Proposition 4, Chapter 2.2:

First, I think that at the statement, he should also mention that $U$ is open and that $x(U)=V\cap S$, for some open set $V\subset \mathbb{R}^3$. My main issue is that in the proof (attached below), he does not use the fact that $S$ is a regular surface. So, if the proof is correct, then this would imply that Condition 2 in Definition 1 can be replaced by a requirement that $\textbf{x}$ is one-to-one. I think this is not possible to do. Here is the proof in the book:

I think that the mistake is this: To argue that $\textbf{x}^{-1}$ is continuous at some point $w\in \textbf{x}(U)$, it suffices to show that for some $A\ \subseteq S$ that is open in $S$ and contains $w$, $\textbf{x}^{-1}$ restricted to $A$, is continuous. In the proof, this set $A$ is $\textbf{x}(V_1)$, but we don't know that this is open in $S$. From the inverse function theorem, we only know that the projection of this set in the $xy$ plane (the set $V_2$ in the proof) is open.

The issue that I am describing had been raised here, but there hasn't been any clarifying answer.

Answer 1

Consider the commutative diagram

where the open nbhs have been chosen in such a way that $\varphi^{-1}$ exists and is smooth (we can apply the Inverse Function Theorem to $\bar\varphi$ because $D\bar\varphi(q)=\pi\circ D\bar{\mathbf x}(q)$ for $q\in U$).

From this diagram we deduce that $(\pi\circ\iota)\circ\mathbf x = j\circ\varphi$. Hence, $j\circ\varphi\circ\mathbf x^{-1}=\pi\circ\iota$ is continuous. And given that $j\colon V\hookrightarrow\mathbb R^2$ is the open inclusion, this implies that $\varphi\circ\mathbf x^{-1}$ is continuous. Now the continuity of $\varphi^{-1}$ implies the continuity of $\mathbf x^{-1}$.

Answer 2

You are correct. While the statement of the proposition is true, in the versions of Do Carmo that I have (the most recent is from 2016), the proof failed to establish the truth. The method of the proof works perfectly for an injective, regular parametrization of the figure-eight curve in the plane and does indeed show that the parametrization defines homeomorphisms from smaller open intervals to pieces of the curve. The reason you cannot deduce the global inverse is continuous (and the way it is different from what happens when the image is inside a regular surface) is that these pieces of the figure-eight curve are not open sets in the topology of the figure-eight curve. This is the part of the proof that is lacking: Do Carmo never shows that the parametrization in question is an open mapping (in terms of the surface). Although this fact can be established under the hypothesis of Proposition 4, it is not necessarily true for an injective, regular parametrized surface (e.g. a cylinder over the figure-eight curve) in terms of the global image. Don't get me wrong, I think Do Carmo has written an outstanding book, but considering how much it has been used, I'm surprised how long this flaw has survived.

Answer 3

This is my attempt at building on Tim Bratten's response to show that $\mathbf{x}(V_1)$ is necessarily open.

By Proposition 3 of Chapter 2.2, for each $q' \in V_1$, we can find a neighborhood $W$ in $S$ of $\mathbf{x}(q')$ such that $W$ is the graph of a differentiable function of the form $z = f(x, y)$. As $\mathbf{x}$ is continuous, we can find an open set $V_1' \subset V_1$ such that $\mathbf{x}(V_1') \subset W$. Then $$ \mathbf{x}(V_1') = \{ ( x(u, v), y(u, v), f( x(u, v), y(u, v) ) ): (u, v) \in V_1' \} $$ and $V_2' = (\pi \circ \mathbf{x})(V_1')$ is open from which it follows that $\mathbf{x}(V_1')$ is open. We can then express $\mathbf{x}(V_1)$ as a union of such sets and so it is open.