Prove that a map $f:B(0,1)\subset \mathbb R^n\to \mathbb R^n$ is a diffeomorphism

Question

Let $B(0,1)$ be the open ball given by $x\in \mathbb R^n$ such that $\|x\|<1$ and consider $f:B(0,1)\to\mathbb R^n$ given by: $$f(x)=\frac{x}{\sqrt{1-\|x\|^2}}$$ In order to prove that $f$ is a diffeomorphism, define $g:\mathbb R^n\to \mathbb R^n$ given by $g(y)=\frac{y}{\sqrt{1+\|y\|^2}}$.

1. How can I show that $g(\mathbb R^n)=B(0,1)$?
2. It is easy show that $f\circ g$ is the identity map, do I have to prove that $g\circ f$ is also the identity map?

Many thanks!

Answer 1

I will write some technical details that explain what you have done and what needs to be done to finish the exercise.

We have $f\colon B(0,1) \to \mathbb R^n$ and $g\colon\mathbb R^n\to \mathbb R^n$. Before we can even consider composition $f\circ g$, we need to know that $g(\mathbb R^n)\subseteq B(0,1)$. You have probably seen it implicitly just by calculating $(f\circ g)(y)$, you get square root of $1-\frac{\|y\|^2}{1+\|y\|^2}$ in denominator which needs to be positive for the expression to make sense, and that is equivalent to $\|g(y)\| < 1$.

Now that you have $\|g(y)\| < 1$ and you proved $(f\circ g)(y) = y$, you can conclude that $g$ is injective and $f$ is surjective, but we still don't have that $g$ is surjective (or $f$ injective).

If you can prove $g(\mathbb R^n) = B(0,1)$, then it follows that $g$ is surjective onto $B(0,1)$ and therefore $g\colon \mathbb R^n \to B(0,1)$ is a bijective map. $g$ being bijective means that it has both left and right inverse and they must be equal, and since we know that the left inverse is $f$, $f$ is the inverse of $g$, so we get $(g\circ f)(x) = x$ for free.

If you can prove $g\circ f$ is identity, you will get that $f$ is injective and $g$ is surjective, so they are both bijective and inverse to each other. You automatically get $g(\mathbb R^n) = B(0,1)$.

So, you need to prove either 1. or 2., and you will get the other one automatically. Since you can show $(f\circ g)(y) = y$, I recommend that you show that $(g\circ f)(x) = x$ since it is completely analogous.

In the end, you still need to show that both $f$ and $g$ are smooth to finish the proof that they are diffeomorphisms.

Answer 2

1. If $y\in\Bbb R^n$, then$$\|g(y)\|=\left\|\frac y{\sqrt{1+\|y\|^2}}\right\|=\frac{\|y\|}{\sqrt{1+\|y\|^2}}<1.$$And if $p\in B(0,1)$, then $p=g\bigl(f(p)\bigr)$ (I will prove this below).
2. If $\require{cancel}p\in B(0,1)$, then\begin{align}g\bigl(f(p)\bigr)&=g\left(\frac p{\sqrt{1-\|p\|^2}}\right)\\&=\frac{\frac p{\sqrt{1-\|p\|^2}}}{\sqrt{1+\left\|\frac p{\sqrt{1-\|p\|^2}}\right\|^2}}\\&=\frac p{\sqrt{1-\|p\|^2+\cancel{\left(1-\|p\|^2\right)}\frac{\|p\|^2}{\cancel{1-\|p\|^2}}}}\\&=p.\end{align}