Let $\Omega:=\mathbb{R}^n\setminus\overline{B}_R(0)$ with $R>0$ and $n>1$. We call $\phi\colon\Omega\to G:=B_R(0)\setminus\left\{0\right\}$ with $$ y=\phi(x):=\frac{R^2}{\lVert x\rVert^2}x\text{ for }\lVert x\rVert > R $$ the Kelvin transformation (relating to the sphere $S_R(0)$). Show, that the Kelvin transformation $\phi\colon\Omega\to G$ is a diffeomorphism and find the Inverse of $\phi$. Hint: clever use of n-dimensional ball coordinates
Edit Now I show you my idea.
Here is my idea how to show that:
The transformation does only change the distance from $x$ to the origin.
So when I use ball coordinates, only the first coordinate (which is the radius) changes:
If the radius is $R'>R$ before the transformation, it is $R^2/R'$ after the transformation.
So one can write the Kelvin-transformation in n-dim. ball coordinates as follows
$$ \phi(\varphi(R',\Phi_1,\ldots,\Phi_{n-1})):=\varphi(R^2/R',\Phi_1,\ldots,\Phi_{n-1}) $$ where $R'>R$ and $x=\varphi(R',\Phi_1,\ldots,\Phi_{n-1})$i.e $\varphi$ describes the transformation to n-dim- ball coordinates.
I think with this it is not that difficult to show that the Kelvin-transformation is a diffeomorphism and to find its inverse.
It is
$$ \phi(\varphi(R^2/R',\Phi_1,\ldots,\Phi_{n-1}))=\varphi(R',\Phi_1,\ldots,\Phi_{n-1}) $$
so $\phi=\phi^{-1}$ and therefore $\phi$ is bijective.
Then $\phi=\varphi\circ T$ with
$$ T\colon (R',\Phi_1,\ldots,\Phi_{n-1})\longmapsto (R^2/R',\Phi_1,\ldots,\Phi_{n-1}). $$
Because all component functions of $T$ are continiously differentiable, and the transformation $\varphi$ is continiously differentiable,too, $\phi=\varphi\times T$ as a composition of two functions that are continiously differentiable, is also continiously differentiable.
So $\phi$ is a diffeomorphism and $\phi=\phi^{-1}$.
Am I right?
Notice that: $$\phi( \phi(x))= \frac{R^2}{||\phi(x)||^2} \phi(x)= \frac{R^2}{||\frac{R^2}{||x||^2}x||^2} \cdot \frac{R^2}{||x||^2}x = \frac{R^2||x||^4}{R^4 ||x||^2} \cdot \frac{R^2}{||x||^2}x = x$$