Diffeomorphism from $\{(x, y) \in \mathbb R^2 : 0 < x^2 + y^2 < 1\}$ to $\{(x, y) \in \mathbb R^2 : x^2 + y^2 > 1\}$

Question

Give an example of a diffeomorphism of the class $C^1$ that takes the set $\{(x, y) \in \mathbb R^2 : 0 < x^2 + y^2 < 1\}$ on $\{(x, y) \in \mathbb R^2 : x^2 + y^2 > 1\}$.

I did this task but I need to make sure it's correct and if it is a sufficient explanation of the problem:

$K=\{(x, y) \in \mathbb R^2 : 0 < x^2 + y^2 < 1\}$
$L=\{(x, y) \in \mathbb R^2 : x^2 + y^2 > 1\}$

$\Phi: (0,1) \times ( 0, 2\pi] \to K$
$\Phi(r, \alpha) = (r\cos \alpha, r \sin \alpha)$ - diffeomorphism of the class $C^1$
$\Phi \left( (0,1) \times ( 0, 2\pi] \right) =K$
$\Phi^{-1}(K)=(0,1) \times (0, 2\pi]$
$\Phi^{-1}$ is also diffeomorphism of the class $C^1$

$f:(0,1)\times (0,2\pi] \to (1,+\infty) \times (0,2\pi]$
$f(r, \alpha) = (\frac 1r, \alpha)$ - diffeomorphism of the class $C^1$
$f \left( (0,1) \times (0, 2\pi] \right) = (1, +\infty) \times (0, 2\pi]$

$g: (1, +\infty) \times (0, 2\pi] \to L$
$g(r, \alpha) = (r\cos \alpha, r \sin \alpha)$ - diffeomorphism of the class $C^1$
$g \left( (1,+\infty)\times (0, 2\pi] \right) =L$

So $g \circ f \circ \Phi^{-1}$ is searched diffeomorphism of the class $C^1$

Answer 1

An easier diffeomorphism takes points $(r_1,\phi_1)$ for $0<r<1$ and $0\le \phi<2\pi$ from domain to points $(r_2,\phi_2)$, where $r_2=1/r_1$ and $\phi_2=\phi_1$. Therefore, $$ x_2=r_2\cos\phi_2=\frac{\cos\phi_1}{r_1}=\frac{r_1\cos\phi_1}{r_1^2}=\frac{x_1}{x_1^2+y_1^2}. $$ Similarly, $y_2=\frac{y_1}{x_1^2+y_1^2}$. Hence, the mapping $\Phi(x_1,y_1)=\left(\frac{x_1}{x_1^2+y_1^2},\frac{y_1}{x_1^2+y_1^2}\right)$ defines a diffeomorphism from $\{(x,y):0<x^2+y^2<1\}$ to $\{(x,y):x^2+y^2>1\}$.

Addendum

@Ted Shifrin mentioned a good remark that this approach (which is originally inspired from Riemann's sphere mapping to plane) can be extended to any dimensions. For those who are curious, a vectorial mapping between multi-dimensional spaces is $$ \Phi(\bar x)=\frac{\bar x}{\bar x\cdot\bar x}, $$ where $\cdot$ represents inner product.

Answer 2

Passing through polar coordinates is the right intuition to have, but the thing is that you function $\Phi$ is not exactly a diffeomorphism: its domain is a rectangle while its codomain is an annulus, and it is clear that these two figures are not even homeomorphic.

Due to the circular symmetry of both $K$ and $L$, one can expect to find a circularly symmetric solution. In polar coordinates, it would read $ (r,\theta) \mapsto (f(r), \theta) $ with $f\colon (0,1)\to (1,+\infty)$ a diffeomorphism. Let us just translate this intuition into Cartesian coordinates, to make sure everything is well defined. Recall that we have the relations \begin{align} r &= \sqrt{x^2+y^2}, & \cos \theta &= \frac{x}{\sqrt{x^2+y^2}}, & \sin \theta &= \frac{y}{\sqrt{x^2+y^2}}, \\ x &= r\cos \theta, & \text{and } y &= r \sin \theta. \end{align} The Cartesian coordinates of a point of the form $(f(r),\theta)$ are then $(f(r)\cos\theta, f(r)\sin\theta)$, and it follows that the particular solutions we were looking for would look like $$ (x,y) \in K \longmapsto \left(f\left(\sqrt{x^2+y^2}\right) \frac{x}{\sqrt{x^2+y^2}}, f\left(\sqrt{x^2+y^2}\right) \frac{y}{\sqrt{x^2+y^2}} \right), $$ with $f \colon (0,1) \to (1,+\infty)$ any diffeomorphism. For instance, $f_0(t)= \frac{1}{t}$, $f_1(t) = 1 - \ln t$ and $f_2(t) = \frac{1}{1-t}$ are such functions.