Hi, everyone. My question is:
Given a point $p$ in the part of $S^2$ above the $xy$-plane, how do I find a neighborhood $V$ of $p$ (an open set that contains $p$) and an open subset $U'$ of $\mathbb{R}^2$ so that $\mathbf{x}_1$ maps $U'$ onto $V\cap S^2$ as the definition says? Thanks.
Let $p=(x_0,y_0,z_0)$. You know that $z_0>0$. Take $r>0$ such that$$\{(x,y)\in\mathbb{R}^2\,|\,\|(x,y)-(x_0,y_0)\|<r\}\subset\{(x,y)\in\mathbb{R}^2\,|\,\|(x,y)\|<1\}$$and define$$U'=\{(x,y)\in\mathbb{R}^2\,|\,\|(x,y)-(x_0,y_0)\|<r\}$$and$$A=\{(x,y,z)\in\mathbb{R}^3\,|\,(x-x_0)^2+(y-y_0)^2<r^2\wedge z>0\}.$$Then $A$ is an open set and $\mathbf{x}_1$ maps $U'$ homeomorphically onto $A\cap S^1$.