Representation of sphere in parametrized form

Question

Let$\newcommand{\Vec}[1]{\mathbf{#1}}\newcommand{\Reals}{\mathbf{R}}$ $V = \{(\theta, \varphi); 0 < \theta < \pi, 0 < \varphi < 2\pi\}$ and let $\Vec{x}:V \to \Reals^{3}$ be given by $$ \Vec{x}(\theta, \varphi) = (\sin\theta \cos\varphi, \sin\theta \sin\varphi, \cos\theta). $$ Clearly, $\Vec{x}(V) \subset S^{2}$. We shall prove that $\Vec{x}$ is a parametrization of $S^{2}$.

Why is $\theta$ not equal to zero?

Answer 1

In order for a smooth function of two variables to constitute a "regular parametrization" at a point, the derivative must have rank two, i.e., have linearly independent columns.$\newcommand{\Vec}[1]{\mathbf{#1}}$

When $\theta = 0$ or $\theta = \pi$ (i.e., when $\Vec{x}(\theta, \varphi) = (0, 0, \pm 1)$ regardless of $\varphi$), the derivative matrix $D\Vec{x}(\theta, \varphi)$ does not have rank two.

Answer 2

At $\theta =0$ we are at the poles and tangent direction is undefined and depends on how you approach the poles.