If $\phi : S^n - \{e_{n+1}\} \to \mathbb{R}^n$ is the stereographic projection, how to compute $\phi^{-1}$? If $\phi : S^n - \{e_{n+1}\} \to \mathbb{R}^n$ is the stereographic projection, how to compute $\phi^{-1}$?
If $$\phi(x) = \frac{1}{1 - x_{n+1}}(x^1,\ldots,x^n),$$
how to compute $\phi^{-1}(y)?$

Let $y = \phi(x) = (y_1, \ldots, y_n)$. As $x \in S^n$, we have 
$$ |y|^2 = \frac 1{(1-x_{n+1})^2}\sum_{i=1}^n x_i^2 = \frac{1 - x_{n+1}^2}{(1-x_{n+1})^2} = \frac{1 + x_{n+1}}{1-x_{n+1}} \iff x_{n+1} = \frac{|y|^2-1}{|y|^2 + 1} $$
Hence,
\begin{align*}
   \frac 1{1- x_{n+1}} &= \frac {|y|^2 + 1}{|y|^2 + 1 - (|y|^2 - 1)}\\
     &= \frac 12(|y|^2 + 1) 
\end{align*}
Hence, for $1 \le i \le n$:
$$ x_i = (1 - x_{n+1})y_i = \frac{2y_i}{|y|^2 + 1} $$
That is 
$$ x = \phi^{-1}(y) = \left(\frac{2y}{|y|^2 + 1}, \frac{|y|^2 - 1}{|y|^2 + 1}\right). $$
