Proving that surface is not elementary

Question

I need help with proving or disproving that the following surface $x^2+y^2+z^2=a^2,a>0$ is elementary. If I had another condition $z>0$ then the surface would be elementary because $rank g'=2$ where $g(x,y)=(x,y,\sqrt{a^2-x^2-y^2})$. But here I know that solution is that this set is not elementary, but not sure how to prove that using this definition with rank. Every help is appreciated.

Answer 1

Here's the beginning of an approach:

Suppose we had such a homeomorphism $g : U \to \mathbb R^3$. Then, the component functions would satisfy $$x(u, v)^2 + y(u, v)^2 + z(u, v)^2 = a^2,\text{ for all } u, v \in U.$$

The derivative $g'$ is of the form $$g'(u, v) = \begin{pmatrix}x_u(u, v) & x_v(u, v) \\ y_u(u, v) & y_v(u, v) \\ z_u(u, v) & z_v(u, v) \end{pmatrix}, $$ and we'd like to show that for all points $(u, v)$ this matrix cannot have rank 2. Note that differentiating the first equation with respect to $u$ and $v$ leads to two equations: $$2 x x_u + 2y y_u + 2z z_u = 0,$$ $$2 x x_v + 2y y_v + 2z z_v = 0.$$

I claim that these two equations lead to the rank of $g'$ being smaller than 2. Perhaps you can show that there is at least one point $(u, v)$ where one of the 2 by 2 minors $$x_u y_v - x_v y_u,\ x_u z_v - x_v z_u,\ y_u z_v - y_v z_u$$ vanishes.