Smoothness of streographic projection

Question

Let $\pi:S^2\to \mathbb{R}^2$ be the streographic projection from the north pole $(0, 0, 1)$. i.e., $$\pi(x_1,x_2,x_3)= \left(\frac{x_1}{1-x_3},\frac{x_2}{1-x_3}\right)$$ for any $x=(x_1,x_2,x_3)\in S^2-\{(0,0,1)\}$ (thus, $x_1^2+x_2^2+x_3^2=1$).

The question is, how to prove that this $\pi$ is smooth?

I should find every $n$-th derivative of $\pi$ and show that they are continuous on an open neighborhood in $S^2$ to get the smoothness, but can't find/calculate the derivative(even the first one).

Should I calculate the Jacobian of $\pi$? It would be a $3\times2$ metrix. Then what about the second derivative?

Answer 1

For the question to make sense one needs to have a differentiable structure on the sphere. One common way to equip the sphere with a manifold structure is to fix an atlas given by the stereographic projection from the north and south pole.

With this atlas, the question becomes trivial, because the map $\pi$ becomes the identity on the local chart.

Answer 2

The sphere $S^2$ is covered by the six open half spheres $x_i>0$ and $x_i<0$ $(1\leq i\leq3)$. For each of these the two other coordinate functions $x_k$, $x_l$ form a chart. As an example, denote by $D$ the open unit disc in the $(x_1,x_3)$-plane. Then the map $$\psi:\quad D\to S^2,\quad (x_1,x_3)\mapsto\bigl(x_1,\sqrt{1-x_1^2-x_3^2}, \>x_3\bigr)$$ serves as a chart on $\bigl\{(x_1,x_2,x_3)\in S^2\,\bigm|\,x_2>0\bigr\}$. In terms of this chart the stereographic projection appears as $$\pi:\quad (x_1,x_3)\mapsto\left({x_1\over 1-x_3},\>{\sqrt{1-x_1^2-x_3^2}\over 1-x_3}\right)\ ,$$ and this is obviously $C^\infty$ on $D$.