How to solve $y = \frac{x}{\sqrt{1-|x|^2}}$ for x

Question

I was reading about an example for a diffeomorphism from $B^n$ to $\mathbb{R}^n$, where $B^n := \{x \in \mathbb{R}^n : |x|<1\}$ is the unit ball in $\mathbb{R}^n$, in the example it was given that the diffeomorphism has the form $F:B^n\to\mathbb{R}^n$, $F(x) = \frac{x}{\sqrt{1-|x|^2}}$ with inverse given by $G(y) = \frac{y}{\sqrt{1+|y|^2}}$. I cannot find the inverse by solving for $y$ in $F$, can someone please tell me how to do that?

Answer 1

Note that for each component $1\leq i \leq n$, $$y_i^2 = \frac{x_i^2}{1-|x|^2} \Rightarrow |y|^2 = \sum_i y_i^2 = \frac{|x|^2}{1-|x|^2} \Rightarrow 1+|y|^2 = \frac{1}{1-|x|^2}.$$ So we have that $$y_i = \frac{x_i}{\sqrt{1-|x|^2}}= x_i\sqrt{1+|y|^2} \Rightarrow x_i = \frac{y_i}{\sqrt{1+|y|^2}} \Rightarrow x = \frac{y}{1+|y|^2},$$ as required.