Maps circle to "edge" of square region I was read a book and then I try to solve this problem:
The map that sends the interval in $S^1$ from $x_i$ to $x_{i+1}$ onto the intervals in the square from $y_i$ to $y_{i+1}$ defines a homeomorphism from the circle to the square. 

If $\{ (x,y) : x^2 + y^2 = 1 \}$ is the circle and $\{ (x,y) : x = \pm 1, -1 \leq y \leq 1 \text{ or } -1 \leq x \leq 1, y = \pm 1 \}$ is the square then explicit homeomorphism are givem by
circle $\to$ square
$(x,y) \to (x/m, y/m)$
and
square $\to$ circle
$(x,y) \to (x/r, y/r)$
where $m = \max (|x|,|y|)$ and $r = \sqrt{x^2 + y^2}$. Intuitively we just twist or bend circle to square.
In first try to construct these maps, I used the trigonometric fuctions (like https://arxiv.org/abs/1509.06344 , page 8), but this is not that problem do. My question is, how can I construct the map of circle to square and square to circle using the problem information?
Thanks.
 A: You already have explicit homeomorphisms that work. 
Let me be somewhat more general and denote $\|(x,y)\|_\infty=\max\{|x|,|y|\}$ and $\|(x,y)\|_2=\sqrt{x^2+y^2}$. Then closed disk is described as $\{x\in\Bbb R^2\,|\, \|x\|_2\leq 1\}$ and circle as $\{x\in\Bbb R^2\,|\, \|x\|_2=1\}$, while closed square as $\{x\in\Bbb R^2\,|\, \|x\|_\infty\leq1\}$ and its boundary is $\{x\in\Bbb R^2\,|\, \|x\|_\infty=1\}$. Your maps can be written as
$$x\mapsto x/\|x\|_\infty,\quad x\mapsto x/\|x\|_2$$
Quick check shows that $\|x/\|x\|_\infty\|_\infty = 1$ and $\|x/\|x\|_2\|_2 = 1$ so these really do map circle to boundary of square and vice versa.
More generally, you can consider mapping whole disk to square and square to disk. In that case you would define maps with:
$$x\mapsto\frac {\|x\|_2}{\|x\|_\infty}x,\quad x\mapsto \frac {\|x\|_\infty}{\|x\|_2}x $$
(except at $x=0$ where this is not well defined - extend them by sending $0$ to $0$ in both cases)
Now, for $\|x\|_2\leq 1$ you have $$\left\|\frac {\|x\|_2}{\|x\|_\infty}x\right\|_\infty = \|x\|_2 \leq 1$$ and for $\|x\|_\infty\leq 1$ you have $$\left\|\frac {\|x\|_\infty}{\|x\|_2}x\right\|_2 = \|x\|_\infty \leq 1$$ so everything checks out.
If you restrict those maps to boundaries, you will get your original maps: for $\|x\|_2 = 1$, $\frac {\|x\|_2}{\|x\|_\infty}x = x/\|x\|_\infty$ and similarly for $\|x\|_\infty = 1$.
What these maps do is scale appropriate vectors. Make sure that you understand why both are continuous and confirm that they are inverses.
