While I was reading Do Carmo's differential geometry, I had several questions about the definition of acregular surface.
From condition 2, the author said : "... $x^{-1}:V \cap S \rightarrow U$ which is continuous; that is, $x^{-1}$ is the restriction of a continuous map $F:W \subset \mathbb{R}^3 \rightarrow \mathbb{R}^2 $ defined on an open set $W$ containing $V \cap S$." I know that if $x^{-1}$ is the restriction of a continuous map $F:W \subset \mathbb{R}^3 \rightarrow \mathbb{R}^2$, then $x^{-1}$ is continuous w.r.t. the subspace topology of $V \cap S$. "
However, how can we prove the converse? I.e. if we already know $x^{-1}$ is continuous, how can we show that it is an restriction of a continuous function $F$ which is defined on an open set $W \subset \mathbb{R}^3$?
My second question is concerned with the definition of "differentiable" of condition 1. From MathWorld, it requires $x$ is differentiable – does that 'differentiable' mean $x \in C^{\infty}$?