Prove $\Phi$ is a diffeomorphism

Question

Let $d \in \mathbb N$.

Define $\Phi:]0, \infty[ \times \{ x \in \mathbb R^{d-1}: |x|<1\}\to \{y \in \mathbb R^d:y_{1}>0\}, (r,x) \mapsto r(\sqrt{1-|x|^2},x)$

Show that $\Phi$ is a diffeomorphism.

Ideas:

$1.$ Bijectivity:

$1.1$ Injectivity: Let $x,y \in \mathbb R^d$ and $r,s \in ]0, \infty[$ whereby $x \neq y$ or $r \neq s$

if $x \neq y$ is clear that $\Phi(r,x)\neq \Phi(r,y)\Rightarrow$ injectivity

So let $s \neq r$ and $\Phi(s,x)=s(\sqrt{1-|x|^2},x) \neq\Phi(r,x)=r(\sqrt{1-|x|^2},x) \Rightarrow$ injectivity

$1.2$ Surjectivity: Let $z \in \{y \in \mathbb R^d:y_{1}>0\}$ it is clear that there is an $r \in ]0, \infty[$ so that $r \sqrt{1-|x|^2}=z_{1}\Rightarrow$ surjectivity

$\Rightarrow$ Bijectivity

On the issue of Differentiability of $\Phi$ as well as Differentiability $\Phi^{-1}$ I am lost, as this is the first time doing this...

Any ideas, corrections, tips?

Answer 1

Hint: The easiest way is to find the inverse function. For this we compute $$ \vert \Phi(r,x) \vert = r $$ as $$ \vert \Phi(r,x) \vert = \vert r \vert \sqrt{(1-\vert x \vert^2)+ x_1^2 + \dots + x_{d_1}^2} = \vert r \vert \sqrt{1- \vert x \vert^2 + \vert x \vert^2} = \vert r \vert = r. $$ Hence, the inverse function is (for $y=(y_1, \dots, y_d)$) $$ \Psi (y_1, \dots, y_d) = \left(\vert y \vert, \frac{1}{\vert y \vert} y_2, \dots, \frac{1}{\vert y \vert} y_d \right).$$