About the definition of regular surface in do Carmo

Question

According to do Carmo, the definition of regular surface requires us to check $X^{-1}$ to be continuous (where $X$ is a local parametrization). But doesn't it infer from other conditions (as shown in his Prop. 4 of Sec 2-2, see photo attached here !!!)? Although he stated regular surface has to be known in advance (prop. 4), actually it seems that he doesn't use it in the proof. He simply uses $X$ to be differentiable and regular and inverse function theorem to conclude $X^{-1}$ is continuous.

Answer 1

$\newcommand{\Param}{\mathbf{x}}$It's difficult to answer fully since "Conditions 1 and 3 of Definition 1" are on a different page than shown in your photograph, but the usual reason for requiring $\Param^{-1}$ to be continuous is to avoid examples such as $$ \Param(u, v) = (\sin u, \sin 2u, v),\quad -\pi < u < \pi,\ 0 < v < 1, $$ an injective regular map whose image is not a surface: