Prove a square is homeomorphic to a circle $s:=\{|x|\le 1,|y|\le 1\} $
$c:=\{{x}^{2}+{y}^{2}\le1\}$
Prove $\overset{\circ}{s}  \cong \overset{\circ}{c}$

ok... not to sure what to do.
I think $\overset{\circ}{s}  \to\overset{\circ}{c}$ is something like:
$$(x,y)\rightarrow\left(\frac{x}{\sqrt{{x}^{2}+{y}^{2}}},\frac{y}{\sqrt{{x}^{2}+{y}^{2}}}\right)$$
What is the inverse for this? Do i need an inverse? Do i just prove the function is continous and the inverse is continuous?
Please help...
I think the following more general result is true and the argument is also quite simple:


  Let $V$ be a real normed vector space and $||\cdot ||_1$, $||\cdot ||_2$ two equivalent norms on $V$, i.e. there are positive constants $C_1, C_2$ such that $||x||_1\le C_1 ||x||_2$ and $||x||_2\le C_2 ||x||_1$. Note that each norm gives rise to the same topology in $V$ because they are equivalent. Define the balls unit balls with respect to each norm as $B_1=\{x\in V:||x||_1\le 1\}$ and $B_2=\{x\in V:||x||_2\le 1\}$. Then $B_1\approx B_2$.


Define $\phi:D_1\to D_2$ as $\phi(x)=\frac{||x||_1}{||x||_2} x$ if $x\neq 0$ and $\phi(0)=0$. It's clear that $\phi$ is continuous in $V-0$ because the norms and the scalar product are continuous functions and continuity at $0$ is also easy to verify using the fact that $||\cdot ||_1$ and $||\cdot ||_2$ are equivalent norms. The inverse is given by $\psi:D_2\to D_1,\psi(x)=\frac{||x||_2}{||x||_1} x$ if $x\neq 0$ and $\psi(0)=0$. So $\phi$ is the desired homeomorphism.

The problem here is a special case of this taking $V=R^2$, the norms $||\cdot||_2$ and $||\cdot||_\infty$ (remembering that all norms in $R^2$, a finite dimensional vector space, are equivalent) and noting that my homeomorphism $\phi$ maps the boundary of the circle to the boundary of the square and so you get an homeomorphism from the interiors by restricting the domain.
You don't need to find the inverse because these two spaces are compact.