Doubt on the definition of a regular surface - do Carmo's Differential Geometry

Question

The definition I am given is the follwing:

Definition. $S \subset \mathbb{R^3}$ is a regular surface if, $\forall p \in S$, $\exists V \in \mathbb{R^3}: V \cap S \subset \mathbb{R^3}$ is the codomain of a regular parametrized surface.

My problems understanding this. From my understanding I concluded the following:

Let $\phi (u_1,u_2) = (x(u_1,u_2),y(u_1,u_2),z(u_1,u_2))$, $\phi:U \subset \mathbb{R^2}\rightarrow \mathbb{R^3}$ be a regular parametrized surface. If $S=\phi(U)$ then S is a regular surface.

But does the definition also let me conclude something for $S$ for the cases when $S \subset \phi(U)$ and $\phi(U) \subset S$? i.e., can I conclude if $S$ is a regular surface (or not) in the cases where $S\subset \phi(U)$ and $\phi(U)\subset S$?

Thanks for your help.

Answer 1

In more plain language, do Carmo's definition says $S$ is a surface if every point $p$ in $S$ has a neighborhood $V$ in space (note subset, not element) such that $V \cap S$ is the image of an immersion of a non-empty plane open set.

As you say, $\phi(U)$ is a regular surface with your notation. The uncertainty appears to result from the order that various data are given to us:

1. We start with a subset $S$ in Euclidean three-space.

2. We pick an arbitrary point $p$ of $S$, and ask if

3. There exists an open set $V \subseteq\mathbf{R}^{3}$ containing $p$ and

4. An open set $U$ in the plane and

5. An immersion (regular mapping) $\phi:U \to V \cap S$.

If yes, then $S$ is a surface.

The choices in each item depend on the choices of preceding items. In programming terms, $V$, $U$, and $\phi$ are "locally scoped variables".

Note that by definition we have $\phi(U) = V \cap S \subseteq S$.

By contrast, the question "[Can we conclude $S$ is a surface] when $S \subset \phi(U)$[...]?" is strange-sounding because $S$ is being chosen after $U$ and $\phi$. In the programming analogy, the question treats $U$ and $\phi$ as having meaning "outside their scope". That's not how the definition works.

In the hope this clarifies: If $U$ is a non-empty open set in the plane and if $\phi:U \to \mathbf{R}^{3}$ is an immersion, then by do Carmo's definition the image $S = \phi(U)$ is a surface. In fact, if $V$ is an open set containing $\phi(U)$, then our mapping $\phi:U \to \phi(U)$ satisfies the definition, and therefore demonstrates $S$ is a surface.

(Technical caution: Many authors assume more in the definition of a regular surface, namely that in points 4. and 5., $\phi$ is proper, i.e., preimages of compact sets are compact, so the inverse mapping $\phi^{-1}:\phi(U) \to U$ is continuous.)