A natural function from the unit sphere to the complex plane is given by $$\phi:\mathbb{S}^2\to\mathbb{C}:(x,y,z)\mapsto \left(\frac{x}{1-z}\right) + i\left(\frac{y}{1-z}\right).$$
A short intuition for what the above function is doing is given at the end of the post. I've been able to understand why the function corresponds with the intuition, but I'm struggling in finding its inverse.
This post points to the function $$\psi:\mathbb{C}\to\mathbb{S}^2:z\mapsto \frac{(2 \operatorname{Re} z,2\operatorname{Im}z, |z|^2-1)}{|z|^2+1}.$$
Is $\psi$ the inverse of $\phi$? If so, why?
In this answer, we can walk through deriving the inverse formulas, not just accepting them from the benevolent geometry gods and verifying that they work.
Let $z = u + iv \in \mathbb{C}$. Now suppose that $z = \phi(x, y, z)$ for some $(x, y, z) \in \mathbb{S}^2 \subset \mathbb{R}^3$. In other words, $$ \left\{ \begin{aligned} u &= \frac{x}{1-z} \\[3pt] v &= \frac{y}{1-z} \end{aligned} \right. $$ How do we recover the coordinates $(x, y, z)$ from $(u, v)$? It will help to remember that the sphere is a surface, hence has only $2$ degrees of freedom, which is expressed in the defining relation $$ x^2 + y^2 + z^2 = 1. $$ Notice: $$ u^2 + v^2 = \frac{x^2 + y^2}{(1-z)^2} = \frac{1 - z^2}{(1-z)^2} = \frac{1+z}{1-z}. $$ Since
$$ w = \frac{1+z}{1-z} \quad\iff\quad z = \frac{w-1}{w+1}, $$ we have $$ z = \frac{u^2 + v^2 - 1}{u^2 + v^2 + 1}. $$ and $$ 1-z = \frac{2}{u^2 + v^2 + 1}. $$ Consequently, $$ x = (1-z)u = \frac{2u}{u^2 + v^2 + 1} $$ and $$ y = (1-z)v = \frac{2v}{u^2 + v^2 + 1}. $$
Putting these all together, $$ \left\{ \begin{aligned} x &= \frac{2u}{u^2 + v^2 + 1} \\[3pt] y &= \frac{2v}{u^2 + v^2 + 1} \\[3pt] z &= \frac{u^2 + v^2 - 1}{u^2 + v^2 + 1}, \end{aligned} \right. $$
Finally, recall that $u = \operatorname{Re}(z)$ and $v = \operatorname{Im}(z)$, so $u^2 + v^2 = |z|^2$, which allows us to define the inverse map $\psi$.