Let $f: B^n \to \Bbb R^n$ be a map from the unit disc given by $x \mapsto \dfrac{x}{\sqrt{1-\|x\|^2}}$. Show that this map is a diffeomorphism.
So I'm trying to find a way to derive the inverse for this map, but I find it quite difficult. I've managed to solve the $1$-dimensional case with the following:
Consider $f:B^1 \to \Bbb R$ given by $x \mapsto \dfrac{x}{\sqrt{1-|x|^2}}$. To find the inverse of $f$ we set $y = \dfrac{x}{\sqrt{1-|x|^2}}$ from where we can solve to get $x^2+y^2|x|^2 = y^2$. And now the crucial point, since $|x|^2 = x^2$ we get $x^2+y^2x^2 = y^2$ and so $x= \dfrac{y^2}{\sqrt{1+y^2}}$ i.e. the inverse $f^{-1}$ is given by $y \mapsto \dfrac{y^2}{\sqrt{1+y^2}}$.
The crucial point here is that due to $x$ being a real number I have $|x|^2 = x^2$, but $x^2$ is ambiguous when $x$ is a vector so this method wont generalize.
A similar approach will only lead to $$\begin{align*} y &= \frac{x}{\sqrt{1-\|x\|^2}} \\ y(\sqrt{1-\|x\|^2})&=x \\ y^2(1-\|x\|^²)&=x^2 \\ y^2-y^2\|x\|^2&=x^2 \end{align*}$$
so $$x^2+y^2\|x\|^2=y^2$$ but this is as far as I can get as I don't know anything I could to the term $\|x\|^2$. What could I do here?
Your function has the form $x \mapsto \tau(|x|) \cdot x$ (that $\cdot$ being scalar multiplication). The function therefore takes rays to rays, stretching each vector by a scalar depending only on the radius coordinate $r=|x|$ that scalar being $$\tau(r)=\frac{1}{\sqrt{1-r^2}} $$
The inverse function therefore also takes rays to rays, stretching each vector by a scalar depending only on $|x|$, i.e. it has the form $x \mapsto \sigma(|x|) \cdot x$.
Composing the two we obtain $\tau(\sigma(|x|) \cdot |x|) \cdot (\sigma(|x|) \cdot x) = x$ and therefore $$\biggl(\tau\bigl(\sigma(r)\cdot r\bigr) \cdot \sigma(r)\biggr) \cdot r = r $$ $$\frac{\sigma(r)}{\sqrt{1 - \sigma(r)^2 r^2}} = 1 $$ and one can solve for $$\sigma(r) = \frac{1}{\sqrt{1 + r^2}} $$ The inverse map is therefore $$x \mapsto \frac{x}{\sqrt{1+|x|^2}} $$ Smoothness everywhere of both maps is evident (thanks to @peek-a-boo for unblocking my mind-block): the norm is smooth, $1-|x|^2$ is positive smooth on the open ball and $1+|x|^2$ is positive and smooth everywhere; $\sqrt{\cdot}$ is smooth on positive numbers; and so on.