Finding the inverse of a (challenging) function

Question

I just started reading Differential Topology by Guillemin and Pollak and one of the very first exercises is the following:

Let $B_a$ be the open ball $\{x : \vert x \vert ^2 < a \}$ in $\mathbb{R}^k$. $\left(\vert x \vert^2 = \sum x_i^2\right)$. Show that the map $$ x \mapsto \frac{ax}{\sqrt{a^2-\vert x \vert^2}}$$

is a diffeomorphism of $B_a$ onto $\mathbb{R}^k$. Hint: Compute its inverse directly.

I tried different ways to compute the inverse mapping but unfortunately without success.

My very first naive approach was writing it as $$ y = \frac{ax}{\sqrt{a^2-\vert x \vert^2}}$$ and trying to get $x$ separated. Didnt manage to do so. My next attempt was by using the composition $f(f^{-1}(x)) = x$ to get the identity but that attempt did not succeed either.

I am clearly missing something. Can anyone help me with a few hints or tell me the secret of finding the inverse mapping?

Answer 1

Consider the scalar case. Then we see that \begin{align} |y|^2= \frac{a^2|x|^2}{a^2-|x|^2} \ \ \implies \ \ |x|=\frac{a|y|}{\sqrt{a^2+|y|^2}}. \end{align} Hence an appropriate candiate is given by \begin{align} f^{-1}(y) = \frac{ay}{\sqrt{a^2+|y|^2}}. \end{align}

Indeed, by direct calculation we see that \begin{align} f^{-1}(f(x)) = f^{-1}\left( \frac{ax}{\sqrt{a^2-|x|^2}}\right) = \frac{a^2x}{\sqrt{a^2-|x|^2}}\left(a^2+\frac{a^2|x|^2}{a^2-|x|^2}\right)^{-1/2} =y. \end{align}

Answer 2

First, I would separate the scalars from the vectors:

$$ f(x) = \frac a {\sqrt{a^2 - |x|^2}} x $$

Notice that this function just scales $x$: it doesn't change the "direction" of $x$, only its magnitude.

In particular, it maps magnitudes $m \to m'$ as

$$ m'(m) = \frac {am} {\sqrt{a^2 - m^2}} $$

Inverting the above function is straightforward; that will tell you how to "undo" the scaling of the original function. By plugging in the inverse $m$ map into something like the following, you would get your overall inverse:

$$ f^{-1}(x) = \frac {m(|x|)} {|x|} x $$

Answer 3

Let's look at the case when $x$ and $y$ are scalars.

The trick to finding the inverse of $y=f(x)$ is to write $x=f(y)$. So, $$x= \frac{ay}{\sqrt{a^2-{\mid y\mid}^2}}$$. Now solve for $y$ in terms of $x$...

Get $$x^2=\frac{a^2y^2}{a^2-{\mid y\mid}^2}$$. So, $$x^2=\frac{a^2y^2}{a^2-y^2}\implies x^2(a^2-y^2)=a^2y^2\implies a^2x^2=(a^2+x^2)y^2\implies y=\frac{ax}{\sqrt{a^2+x^2}}\implies y=\frac{ax}{\sqrt{a^2+{\mid x\mid}^2}}$$ is our inverse. You should check it...