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

Solution 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.