Prove that $\phi$ is a homeomorphism Let $B_{a}$ be the open ball centered at 0 of radius $a$ in $\Bbb {R}^{n}$, that is $B_{a}=\{x\in \Bbb {R}^{n}:\left\lVert x\right\rVert\lt a\}$. Define a map $\phi:B_{a}\to \Bbb {R}^{n}$ by $\phi (x)=\frac{ax}{\sqrt{a^{2}-{\left\lVert x\right\rVert}^{2}}}$. Prove that $\phi$ is a homeomorphism.
So I have to find the contiuous inverse function of $\phi$?
 A: In fact the best approach is to find a continuous inverse function of $\phi$. This is not hard. Write
$$y = \frac{ax}{\sqrt{a^2-{\left\lVert x\right\rVert}^{2}}} .$$
Take the norm on both sides, square and resolve for $\lVert x\rVert$. This gives
$$\lVert x\rVert = \frac{a\lVert y\rVert}{\sqrt{a^2 + {\left\lVert y\right\rVert}^{2}}}.$$
Since $y = \lambda x$ with $\lambda > 0$, we get for $x \ne 0$
$$\frac{y}{\lVert y\rVert} = \frac{x}{\lVert x\rVert}$$
and therefore
$$x = \lVert x\rVert \frac{x}{\lVert x\rVert} = \lVert x\rVert \frac{y}{\lVert y\rVert} = \frac{a\lVert y\rVert}{\sqrt{a^2 + {\left\lVert y\right\rVert}^{2}}} \frac{y}{\lVert y\rVert} = \frac{a y}{\sqrt{a^2 + {\left\lVert y\right\rVert}^{2}}}  .$$
The map
$$ \psi : \mathbb R^n \to B_a, \psi(y) = \frac{a y}{\sqrt{a^2 + {\left\lVert y\right\rVert}^{2}}} $$
is now easily verified to be the desired inverse of $\phi$. Note that $\psi(0) = 0$ and, for $y \ne 0$, $\lVert \psi(y) \rVert = \frac{a \lVert y \rVert}{\sqrt{a^2 + {\left\lVert y\right\rVert}^{2}}}  < \frac{a \lVert y \rVert}{\sqrt{ {\left\lVert y\right\rVert}^{2}}} = a$, i.e. $\psi(\mathbb R^n) \subset B_a$.
