If $\phi : S^n - \{e_{n+1}\} \to \mathbb{R}^n$ is the stereographic projection, how to compute $\phi^{-1}$?

Question

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 $\phi^{-1}(y)?$

Answer 1

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 $$ x = \phi^{-1}(y) = \left(\frac{2y}{|y|^2 + 1}, \frac{|y|^2 - 1}{|y|^2 + 1}\right). $$

Answer 2

If $y=(y_1,\dots, y_n)$ and $x=\phi^{-1}(y)\in \Bbb R ^{n+1}$, then $$ x_i = \frac {2 y_i}{1+y_1^2+\cdots + y_n^2} \qquad I<n+1 \\ x_{n+1} = \frac {-1+y_1^2+/cdots + y_n^2}{1+y_1^2+\cdots + y_n^2} $$