Bijection from $[-1,1]\times[-1,1] \rightarrow \{(x,y) \in \mathbb{R}: \sqrt{x^2+y^2} \leq 1\}$ I am trying to find and prove a bijection from the square $[-1,1]\times[-1,1]$ to the unit circle. Given a point $(x,y)$ in the square, my function maps it to the point $(r, \theta)$ in the circle where 
$r = \frac{\sqrt{x^2+y^2}}{\sqrt{1+(y/x)^2}}$
and
$\theta = \begin{cases} \arctan(\frac{y}{x}) &\mbox{if } x > 0 \\ 
\arctan(\frac{y}{x})+\pi & \mbox{if } x < 0 \\
\pi/2 & \mbox{if } x = 0\ \mbox{and}\ y > 0\\
-\pi/2 & \mbox{if } x = 0\ \mbox{and}\ y < 0 \end{cases}$
In other words, the angle of the point on the circle is the same as the 'angle' of the point on the square, and the radius is the 'radius' of the point on the square, scaled down to fit inside the circle.
I'm trying to prove the injection, which means I need to prove that 
$f(a, b) = f(c, d) \implies a = c\ \mbox{and}\ b = d$
or 
$\frac{\sqrt{a^2+b^2}}{\sqrt{1+(b/a)^2}} = \frac{\sqrt{c^2+d^2}}{\sqrt{1+(d/c)^2}}\ \mbox{and}\ \arctan(b/a) = \arctan(d/c) \implies a = c\ \mbox{and}\ b = d$
How can I do this?
Edit: I realize that my formula for $r$ is slightly incorrect, because when $|y| > |x|$ then the denominator is $\sqrt{1+(x/y)^2}$.
 A: A different approach could be to stay in Cartesian coordinates by defining
$$
f(x,y)=\left(\frac x{\sqrt{1+(y/x)^2}},\frac y{\sqrt{1+(y/x)^2}}\right)
$$
when we are in the triangular region $0\leq y < x \leq 1$ which bijectively maps $1/8$ of the square to $1/8$ of the circle, defining $f(0,0)=(0,0)$, and then extending to the rest of the shapes by re-using the same idea as for this triangular region. Then we only need to prove bijectivity for this region.

So suppose $f(a,b)=f(c,d)\neq(0,0)$. Then we have the system of equations:
$$
\begin{align}
\frac a{\sqrt{1+(b/a)^2}}&=\frac c{\sqrt{1+(d/c)^2}}\\
&\text{and}\\
\frac b{\sqrt{1+(b/a)^2}}&=\frac d{\sqrt{1+(d/c)^2}}
\end{align}
$$
Squaring all of these expressions and simplifying we see that
$$
\begin{align}
\frac {a^4}{a^2+b^2}&=\frac {c^4}{c^2+d^2}\\
&\text{and}\\
\frac {a^2b^2}{a^2+b^2}&=\frac {c^2d^2}{c^2+d^2}\\
\end{align}
$$
From the second equation we have $b=0\iff d=0$ since we have assumed $a,c>0$. In that case the first equation implies $a=c$.
If $b,d>0$ we see that
$$
\frac{a^4}{c^4}=\frac{a^2b^2}{c^2d^2}\implies\frac ac=\frac bd \iff\frac ab=\frac cd
$$
So we must have $(c,d)=(ka,kb)$ for some $k>0$. Thus
$$
\begin{align}
\frac a{\sqrt{1+(b/a)^2}}&=\frac c{\sqrt{1+(d/c)^2}}\\
&=\frac{ka}{\sqrt{1+(kb/(ka))^2}}\\
&=\frac{ka}{k\sqrt{1/k^2+(b/a)^2}}\\
&=\frac{a}{\sqrt{1/k^2+(b/a)^2}}
\end{align}
$$
implying $k^2=1$ so $k=1$ and therefore $(c,d)=(a,b)$.
A: Yet another (and much cleaner) approach is to define
$$
f(x,y)=
\begin{cases}
(x,y)&\text{if }x=0\vee y=0\\
(qx,qy)&\text{otherwise}
\end{cases}
$$
where $q=\max\{|x|,|y|\}/\sqrt{x^2+y^2}$. This function has an explicit inverse, namely:
$$
g(x,y)=
\begin{cases}
(x,y)&\text{if }x=0\vee y=0\\
(\lambda x,\lambda y)&\text{otherwise}
\end{cases}
$$
where $\lambda=\sqrt{x^2+y^2}/\max\{|x|,|y|\}$.
I claim that $q$ and $\lambda$ only depend on the angular position of the point in question, which renders them reciprocal to each other. Let us prove this.

This is much simpler to prove. It suffices to show that $q$ and $\lambda$ are invariant under scaling from which it then follows that they must be reciprocal. So let some factor $k>0$ be given. Then
$$
q=\frac{\max\{|x|,|y|\}}{\sqrt{x^2+y^2}}=\frac{k\max\{|x|,|y|\}}{k\sqrt{x^2+y^2}}=\frac{\max\{|kx|,|ky|\}}{\sqrt{(kx)^2+(ky)^2}}
$$
so the points $(x,y)$ and $(kx,ky)$ will result in the same $q$-value. A completely similar argument shows the same for $\lambda$.
So since all $f$ and $g$ do to points is to scale them by some factors, and since by the just established arguments these factors are reciprocal, they must be inverse functions of each other.
