Inverse of a function $B_r(0) \to \mathbb{R}^n$

Question

Definition. Let $B_r(0) := \{x \in \mathbb{R}^n \mid \|x\| < r\}$ be the open ball in $\mathbb{R}^n$ with radius $r>0$ and center $0:=(0,\ldots,0) \in \mathbb{R}^n$.

Let $s_r : B_r(0) \to \mathbb{R}^n$ be the map, defined by $$s_r(x) = \frac{r}{\sqrt{r^2-\|x\|^2}} \cdot x $$

Does this function have an inverse? If so, how does $s_r^{-1} : \mathbb{R}^n \to B_r(0)$ look like?

Answer 1

So, for injectivity, let $x,y \in B_r$ with $s_r(x)=s_r(y)$. Then, it follows that $$ \begin{alignat*}{3} && s_r(x) &= s_r(y) \\ &\iff& \frac{rx}{\sqrt{r^2-\|x\|^2}} &= \frac{ry}{\sqrt{r^2-\|y\|^2}} \\ &\iff& \underbrace{\frac{1}{\sqrt{r^2-\|x\|^2}}}_{\text{scalar}} \underbrace{x}_{\text{vector}} &= \underbrace{\frac{1}{\sqrt{r^2-\|y\|^2}}}_{\text{scalar}} \underbrace{y}_{\text{vector}} \end{alignat*} $$ Since vectors in $\mathbb{R}^n$ are equal iff all components are equal, it suffices to prove that the scalar factors are the same: $$ \begin{alignat*}{3} &\iff& \frac{1}{\sqrt{r^2-\|x\|^2}} &= \frac{1}{\sqrt{r^2-\|y\|^2}} \\ &\iff& \frac{1}{r^2-\|x\|^2} &= \frac{1}{r^2-\|y\|^2} \\ &\iff& r^2-\|x\|^2 &= r^2-\|y\|^2 \\ &\iff& \|x\|^2 &= \|y\|^2 \\ &\iff& \|x\| &= \|y\| \end{alignat*} $$ This should prove the injectivity of $s_r$.

Answer 2

Let $y=s_r(x)$. Take norm on both sides and solve the equation for $\|x\|$ to get $\|x\|=\frac {r\|y\|} {\sqrt {r^{2}+|\|y\|^{2}}}$. Can you write down $x$ in terms of $y$ now?