If $U$ is an open in $\mathbb{R}^2$ and $f:U\to \mathbb{R}$ is $C^{\infty}$ and $c\in \mathbb{R}$ is such that $\bigtriangledown f(p)\neq0$ for all $p\in f^{-1}(c)$, then $S=f^{-1}(c)$ is a surface.
My definition of surface is as follows: A surface is a presuperficie $X$ (related topological space such that for all $p\in X$, there exists an open environment $U$ of $p$ together with a homeomorphism $\phi:U\to \phi(U)\subset\mathbb{R}^2$) whose underlying topological space is Haudorff and second countable, equipped with a $A=\{(U_{\alpha},\phi_{\alpha})\}_{\alpha\in\Lambda}$ collection of neighborhoods in $X$ such that
$X=\bigcup_{\alpha\in\Lambda}U_{\alpha}$
and for all $\alpha, \beta\in\Lambda$ with $U_{\alpha}\cap U_{\beta}\neq\phi$, the transition functions:
$\phi_{\beta}\circ\phi_{\alpha}^{-1}:\phi_{\alpha}(U_{\alpha}\cap U_{\beta})\to\phi_{\beta}(U_{\alpha}\cap U_{\beta})$
$\phi_{\alpha}\circ\phi_{\beta}^{-1}:\phi_{\beta}(U_{\alpha}\cap U_{\beta})\to\phi_{\alpha}(U_{\alpha}\cap U_{\beta})$
they are applications $C^{\infty}$.
My question would be what would be the letters in this example, could it be $\{(U_{\alpha},\phi_{\alpha})\}$? Why? And how do I show what follows? Thank you very much.