Is function invertible?

Question

Reflection on the unit circle:

Let $E=\mathbb R ^{2} - \left\{0,0\right\} $ be perforated plane and $f: E \mapsto E$ defined by $f\left(x,y\right)=\left(\frac{x}{x^{2}+y^{2} } , \frac{y}{x^{2}+y^{2} } \right) $

Show with Jacobian matrix that $f$ is in all points local invertible. Show that $f$ is also global invertible. Find $f^{-1}$ and explain mapping geometrically.

What I did:

I started with determinante of Jacobian matrix to show that function is local invertible, but I got the following result.

Is that enough for showing that function is local invertible? How do I do rest?

$Df=\frac{-x^{4}-y^{4}-2x^{2}y^{2} }{\left(x^{2}+y^{2} \right) ^{2} } $

Answer 1

See the comment on the question for local invertibility.

Now, we prove that $f$ is invertible by showing that $f$ is one-one and onto. Given $(x_1, y_1), (x_2,y_2) \in \Bbb R^2 - \{(0,0)\}$, we have:

$$f(x_1,y_1) = f(x_2, y_2) \implies \begin{cases} \frac{x_1}{x_1 ^2 + y_1^2} = \frac{x_2}{x_2 ^2 + y_2^2} \\ \frac{y_1}{x_1 ^2 + y_1^2} = \frac{y_2}{x_2 ^2 + y_2^2}\end{cases}$$

Squaring the equations and adding them to each other, we get $x_1^2 + y_1^2 = x_2^2 + y_2^2$, which then given $x_1 = x_2$ and $y_1 = y_2$.

Next, given $(X,Y) \in \Bbb R^2 - \{(0,0)\}$, we want to find $(x,y)$ such that $f(x,y) = (X,Y)$, so we are solving for $x$ and $y$:

$$\begin{cases} X = \frac{x}{x^2 + y^2} \\ Y = \frac{y}{x^2 + y^2} \end{cases}$$

Notice then that $X^2 + Y^2 = \frac1{x^2 + y^2}$

Edit

Now, on one hand, $X = \frac{x}{x^2 + y^2}$ gives (after cross multiplying) $x = X(x^2 + y^2)$, but $x^2 + y^2 = \frac1{X^2 + Y^2}$ (by taking reciprocals in the equation right above "Edit"), therefore:

$$x = X \times \frac{1}{X^2 + Y^2} = \frac{X}{X^2 + Y^2}$$

Similarly, we get:

$$y = \frac{Y}{X^2 + Y^2}$$

Answer 2

You can see $f$ as a map $\mathbb{C}\setminus\{0\}\to\mathbb{C}\setminus\{0\}$, with $$ f(z)=\frac{z}{|z|^2}=\frac{1}{\bar{z}} $$ by considering $z=x+iy$ and so $$ f(f(z))=f(\bar{z}^{-1})=z $$ Therefore $f$ is the inverse of itself.

The map is the circular inversion: the points $O(0,0)$, $P$ and $P'=f(P)$ are aligned and $OP\cdot OP'=1$.

The computation of the Jacobian is easy. Let $f_1(x,y)=\frac{x}{x^2+y^2}$ and $f_2(x,y)=\frac{y}{x^2+y^2}$; then $$ \frac{\partial f_1}{\partial x}=\frac{y^2-x^2}{(x^2+y^2)^2} \qquad \frac{\partial f_1}{\partial y}=-\frac{2y}{(x^2+y^2)^2} \\ \frac{\partial f_2}{\partial x}=-\frac{2x}{(x^2+y^2)^2} \qquad \frac{\partial f_2}{\partial y}= \frac{x^2-y^2}{(x^2+y^2)^2} $$ so the Jacobian is $$ -\frac{1}{(x^2+y^2)^4}(x^4+2x^2y^2+y^4)=-\frac{1}{(x^2+y^2)^2} $$

Answer 3

The mapping $f$ is indeed reflection in the unit circle. In polar coordinates, the point $(x,y)=(r\cos\theta, r\sin\theta)$ is mapped by $f$ to the new point $$ \left(\frac{r\cos\theta}{r^2},\frac{r\sin\theta}{r^2}\right)=\left(\frac1r\cos\theta,\frac1r\sin\theta\right).$$ Geometrically, $f$ sends $(x,y)$ to a new point with the same argument (angle), but distance $\frac1r$ from the origin instead of $r$. Geometrically it is clear that this map is invertible, because you can recover the original point by applying $f$ once more. So $f^{-1}=f$.