Equivalent definitions of a surface do Carmo Differential Geometry of Curves and Surfaces defines a regular surface as per the below post.
Lee Introduction to Smooth Manifolds defines an embedded or regular surface to be an embedded or regular submanifold of $\mathbb{R}^3$ of codimension 1, namely a subset $S\subset\mathbb{R}^3$ that is itself a smooth $2$-dimensional manifold whose topology is the subspace topology and whose smooth structure ensures the inclusion map $\iota:S\hookrightarrow\mathbb{R}^3$ is an embedding.
Question:  Are these definitions equivalent?  If so can someone present or point to in the literature a detailed proof.
 A: Let's look at what do Carmo actually writes for the definition of a regular surface:

A subset $S \subset \mathbb{R}^3$ is a regular surface if, for each $p \in S$, there exists a neighborhood $V \subseteq \mathbb{R}^3$ and a map ${\bf x} : U \to V \cap S$ of an open set $U \subseteq \mathbb{R}^2$ onto $V \cap S \subset \mathbb{R}^3$ such that:
  
  
*
  
*${\bf x}$ is differentiable.  This means that if we write
  $${\bf x}(u, v) = (x(u,v), y(u,v), z(u,v)), \qquad u, v \in U$$
  the functions $x, y, z$ have continuous partial derivatives of all orders in $U$.
  
*${\bf x}$ is a homeomorphism. [...]
  
*(The regularity condition) For each $q \in U$, the differential
  $d{\bf x}_q$ is [injective].
The mapping $\bf x$ is a parametrization or a system of (local) coordinates in (a neighborhood of $p$.  The neighborhood $V \cap S$ of $p$ in $S$ is called a coordinate neighborhood.

This is section 2-2, definition 1 in do Carmo.
Notice a couple of things.  Since $V \cap S$ doesn't have a a smooth manifold structure yet, we can't strictly speaking talk about ${\bf x}$ being differentiable, and if you look at the definition do Carmo gives what he's really saying is that the composition
$$U \xrightarrow{\bf x} S \cap V \hookrightarrow \mathbb{R}^3$$
is differentiable.  Similarly, when he talks about $d{\bf x}_q$, he's talking about the differential of this same map, since again we can't talk about ${\rm T}_q (S \cap V)$ yet as it hasn't been defined.

Okay, now a smooth embedding, according to Lee, is an injective immersion which is a homeomorphism onto its image.  Since $i$ the inclusion of a subspace, the topological conditions are already satisfied, so it remains only to check that $i$ is an immersion, i.e. that $i_\ast : {\rm T}_P S \to {\rm T}_P \mathbb{R}^3$ is injective for each $P \in S$.
Take a local parametrization ${\bf x} : U \to S$ at $P$, say with ${\bf x}(0) = P$.  By definition, the composition
$$U \xrightarrow{\bf x} S \xrightarrow{i} \mathbb{R}^3$$
is an immersion.  Since $(i \circ {\bf x})_\ast = i_\ast \circ {\bf x}_\ast$, the composition
$${\rm T}_0 U \xrightarrow{{\bf x}_\ast} {\rm T}_P S \xrightarrow{i_\ast} {\rm T}_P \mathbb{R}^3$$
is injective.  Above, you've said you believe that $S$ is a 2-manifold.  Given this, $\dim {\rm T}_0 U = \dim {\rm T}_P S = 2$ and $\dim {\rm T}_P \mathbb{R}^3 = 3$.  So by elementary linear algebra the only way for the composition to be injective is for ${\bf x}_\ast$ to be an isomorphism and $i_\ast$ to be injective.
A: The answer to my question is yes, and is given by Theorem 5-2 in Spivak Calculus on Manifolds combined with Theorem 5.8 in Lee Introduction to Smooth Manifolds.  See also here.
