Let $a > 0$, and set $B_a = \{x \in \mathbb{R}^n : |x|^2 < a \}$. Let $\phi : B_a \to \mathbb{R}^n$ be given by $\phi(x) = \frac{ax}{\sqrt{a^2 - |x|^2}}$. Prove that $\phi$ is a diffeomorphism of $B_a$ onto $\mathbb{R}^n$.
I am stucked in proofing that the above function is 1-1 and onto because in the definition of the function the scalar part depends on the norm of x as a vector, so if I equate 2 functions say one of them of y and the other of x there will be a difference in the scalar value on both sides so what shall I do?
related links: Show that the map is a diffeomorphism
Hint: Show that the map $\psi : \Bbb R^n \to B_a$ defined by the equation $$\psi(y) = \frac{ay}{\sqrt{a^2 + \lvert y\rvert^2}}$$ is the inverse of $\phi$.