Despite the title, I'm actually dealing with the proof of the criterion concerning orientability of a regular surface. Here is the proposition to be proved:
As the red circle indicates, do Carmo, the author, used connectedness to prove the sufficiency. What if $S$ cannot be covered with a family of connected coordinate neighborhoods? Maybe I should paste the definition of a regular surface:
$\mathbf{X}(U)$ is the so-called coordinate neighborhood at $p$. Thanks.
Best Answer
Any coordinate neighborhood of $p \in S$ comes along with a map $x : U \to V \cap S$ as in your definition. Now let $p' = x^{-1}(p)$. Since $U$ is open in $\mathbb R^2$, there exists an open disk $U' = U(p',r) = \{ q \in \mathbb R^2 \mid \lVert q - p' \rVert < r \}$ which is contained in $U$. This is a connected set (it is even path connected). Since $x$ is a homeomorphism, $x(U')$ is open in $V \cap S$. Hence there exists an open $V' \subset \mathbb R^3$ such that $V' \cap S = x(U')$. Then define $x' : U' \to V' \cap S$ as the restricton of $x$. It is obviuos that properties 1. - 3. are satisfied. That is, we found a connected coordinate neighborhood $x'(U')$ around $p$.