TITLE: Finding a bijective function between an open disk and the open square
QUESTION [5 upvotes]: How can I find a bijective function between these two sets?
  $$\{(x,y)\in\mathbb{R}^2 \,|\, x^2+y^2<1\}, \quad (-1,1) \times (-1,1) .$$

I already thought of first writing between 2nd and set of real numbers, but then I find myself stuck with finding between the reals and the 1st set. Any help would be appreciated.

REPLY [10 votes]: One domain is $\{\rho_1(x,y)<1\}$, the other is $\{\rho_2(x,y) <1\}$, where $\rho_1(x,y) =\sqrt{x^2+y^2}$, and $\rho_2(x,y) = \max(|x|,|y|)$. Find a map $F\colon (x,y) \mapsto (x',y')$ so that $\rho_1(x,y) = \rho_2(x',y')$. It should be linear on each line through the origin. One can take:
$$F(x,y) = \frac{\rho_1(x,y)}{\rho_2(x,y)} \cdot (x,y)$$
that is
\begin{eqnarray}
F(x,y) = \frac{\sqrt{x^2+y^2}}{\max(|x|,|y|)} \cdot (x,y)
\end{eqnarray}
with inverse 
\begin{eqnarray}
F^{-1}(x,y) = \frac{\max(|x|,|y|)}{\sqrt{x^2+y^2}} \cdot (x,y)
\end{eqnarray}
$F$ provides a homeomorphism from the disk to the square.