Prove that $\mathbf S^2$ is a regular surface in $\mathbb R^3$

differential-geometrygeneral-topologygeometrysurfaces

I want to prove that the unit sphere $\mathbf{S}^2 = \{(x,y,z): x^2 + y^2 + z^2 = 1\}$ is a regular surface in $\Bbb R^3$, using strictly only the following definition. The task is to produce open sets $U,V$ and a map $\varphi$ satisfying the properties below.

(Surface.) $S\subset \Bbb R^3$ is called a regular surface if for any $p\in S$, there exists an open set $U\subset \Bbb R^3$ containing $p$, an open set $V\subset \Bbb R^2$, and a differentiable$\color{red}{^1}$ map $\varphi: V\to U$ such that:

  1. The restriction map $\varphi:V\to U\cap S$ is a homeomorphism.
  2. For all $(x,y)\in V$, $D\varphi_{(x,y)}: \Bbb R^2\to\Bbb R^3$ (the derivative map of $\varphi$) is injective.

My efforts:
I am not sure, but I think we might be able to take the help of the stereographic projection at some stage. Firstly, I think it suffices to consider just $p = (0,0,1)$ due to the rotational symmetry of $\mathbf S^2$. We must find an open set $U \subset \Bbb R^3$ containing $p$, which satisfies the conditions above. My idea is to consider an open ball $B$ around $p$ (as in the figure below), and intuitively convince myself that the patch $B\cap \mathbf S^2$ is homeomorphic to a rectangle $V$ in $\Bbb R^2$. Hopefully, injectivity of $D\varphi_{(x,y)}$ for all $(x,y)\in V$ will also hold. However, it is difficult to explicitly pin these details down. Is there any other way out? If not, could I get some help in completing this attempt?

enter image description here

Footnotes:
$\color{red}{1.}$ $\varphi$ is differentiable in the sense that if $\varphi(u,v) = (\mathbf x(u,v), \mathbf y(u,v), \mathbf z(u,v))$ then $\mathbf x, \mathbf y, \mathbf z: \Bbb R^2\to\Bbb R$ have partial derivatives of all orders.

Best Answer

I found a sketch of the proof here, which I am adding as an answer.

It suffices to consider $p = (0,0, 1)$, due to the rotational symmetry of the sphere. Let $\mathbb H = \{(x,y,z)\in \mathbb S^2: z > 0\}$ and $\mathbb D= \{(x,y): x^2 + y^2 < 1\}$. Define $\pi: \mathbb H\to \mathbb D$ by $\pi: (x,y,z)\mapsto (x,y)$. $\pi$ is a continuous bijection with inverse $\varphi$ given by $$\pi^{-1}(x,y) = \varphi(x,y) = (x,y,\sqrt{1-x^2-y^2})$$ That $\varphi: \mathbb D\to \mathbb H$ is a homeomorphism is immediate from construction, since $\pi$ and $\varphi$ are both continuous and $\pi \circ \varphi = \text{id}_{\mathbb D}$ and $\varphi\circ\pi = \text{id}_{\mathbb H}$. Also, $$D\varphi_{(x,y)}: (h,k) \mapsto \left(h, k, \frac{-xh-yk}{\sqrt{1-x^2 - y^2}} \right)$$ is injective.

Related Question