Is it true that $[0,1]\times [0,1]\cong \overline{B}(0,1)$? I think this is true, can anyone give me an example of a homeomorphism between $[0,1]\times [0,1]$ and $\overline{B}(0,1)$?
Of course, $\overline{B}(0,1)$ is the closed ball $\{x\in\mathbb{R}^2:\left\|{x}\right\| \le 1\}$.
Thanks.
 A: I do not provide a formula but it should be possible this way: 
First move the center of the square to the center of the ball ($[-0.5,0.5] \times [-0.5, 0.5]$). Then you look at a ray from the centerpoint outwarts. Take a point on that ray and consider it as a vector. Multiply the vector by the inverse of the distance on the ray from the centerpoint to the border of the square.
I hope that helps.
A: First, write a homeomorphism from the unit square to the unit circle, for instance the map $f$ given by $f(x,0)=(\cos(\frac{\pi}{2} x),\sin(\frac{\pi}{2} x)), f(1,y)=(\cos(\frac{\pi}{2} y+\frac{\pi}{2}),\sin(\frac{\pi}{2} y+\frac{\pi}{2}))$ etc. Then extend by scaling: map the square $[r,1-r]\times [r,1-r]$ to the circle of radius $1-2r$ in the same way. This gives a well defined map; the inverse image of an open subset $S$ of the disc not intersecting $[(-1,0),(1,0)]\cup [(0,-1),(0,1)]$ is a union of horizontal or vertical intervals. If for instance the intervals are the lower edges of squares then their $y$-coordinate varies over $\{1/2(1-\sqrt{ x_1^2+x_2^2}):(x_1,x_2)\in S\}$, which is open.$^*$ Similarly for an open subset $T$ intersecting the axes we can write its inverse image as a union of corners over an open range of radii.
$^*$ If you're not convinced that this range is open, it's enough to suppose $S$ is an open disc about $(p_1,p_2)$ of radius $s$, in which case this range is $(\sqrt{p_1^2+p_2^2}-s,\sqrt{p_1^2+p_2^2}+s)$.
