Is function invertible? 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} } $
 A: 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:
$$y = \frac{Y}{X^2 + Y^2}$$
A: 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
$$
-\frac{1}{(x^2+y^2)^4}(x^4+2x^2y^2+y^4)=-\frac{1}{(x^2+y^2)^2}
$$
A: 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 $f^{-1}=f$.
