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.

Note that if we draw any ray outward from the origin, then for any $r\ge 0$ it will intersect precisely one point of the set $$C_r:=\left\{(x,y)\in\Bbb R^2:\sqrt{x^2+y^2}=r\right\}$$ and exactly one point of the set $$S_r:=\bigl\{(x,y)\in\Bbb R^2:\max(|x|,|y|)=r\bigr\}.$$ Every point of each $C_r$ and every point of each $S_r$ is hit by some such ray, and the only point that is hit by more than one ray is the origin, which is the single point of $S_0=C_0$.

Furthermore, $\overset{\circ}{s}$ is the disjoint union of $S_r$ for $0\le r<1$ and $\overset{\circ}{c}$ is the disjoint union of $C_r$ for $0\le r<1$.

A natural candidate for a homeomorphism is to take any point of $\overset{\circ}{s}$, find a ray from the origin that it lies on and the appropriate $S_r$ containing the point, and map the point to the point of $C_r$ on the ray. This is readily a well-defined bijection by the discussion above, as is its inverse (which is basically the same, but maps $C_r$ points to their corresponding $S_r$ points). You need only show that they are both continuous.

It would be simpler to show continuity if we had formulas, though. We already know that the origin will be mapped to the origin. Take $(x,y)\in\overset{\circ}s$ with $(x,y)\ne (0,0)$. Putting $r=\max(|x|,|y|),$ we have $(x,y)\in S_r$. We must map $(x,y)$ to the point $(x',y')$ such that $\sqrt{(x')^2+(y')^2}=r$ lying (not necessarily strictly) between $(x,y)$ and $(0,0)$. That is, we will have $(x',y')=t(x,y)$ for some $0<t\le 1$. In particular, since we need $$r=\sqrt{(tx)^2+(ty)^2}=t\sqrt{x^2+y^2},$$ then we need $$t=\frac{r}{\sqrt{x^2+y^2}}=\frac{\max(|x|,|y|)}{\sqrt{x^2+y^2}},$$ so the map $\overset{\circ}s\to\overset{\circ}c$ is given by $$(x,y)\mapsto\frac{\max(|x|,|y|)}{\sqrt{x^2+y^2}}(x,y)=\frac{|x|+|y|+\bigl||x|-|y|\bigr|}{2\sqrt{x^2+y^2}}(x,y)$$ for $(x,y)\ne0$ and $(0,0)\mapsto(0,0)$. Now, in each variable, this map is continuous for $(x,y)\ne (0,0)$ (as a quotient of a continuous function over a positive continuous function), and so the map is continuous on $\overset\circ s\smallsetminus\{(0,0)\}$. We can show without too much difficulty that $$\frac{|x|+|y|+\bigl||x|-|y|\bigr|}{2\sqrt{x^2+y^2}}(x,y)\to(0,0)$$ as $(x,y)\to(0,0)$, so the map is continuous there, too.

Now take $(x,y)\in\overset{\circ}c$ with $(x,y)\ne (0,0)$. Put $r=\sqrt{x^2+y^2}$. We must map $(x,y)$ to the point $(x',y')$ such that $\max(|x'|,|y'|)=r$ and such that $(x,y)$ lies (not necessarily strictly) between $(x',y')$ and $(0,0)$. Then we need $(x',y')=t(x,y)$ for some $t\ge1,$ so $$r = \max(|x'|,|y'|)= \max(|tx|,|ty|)= \max(t|x|,t|y|)= t\cdot\max(|x|,|y|),$$ so $$t=\frac{r}{\max(|x|,|y|)}=\frac{\sqrt{x^2+y^2}}{\max(|x|,|y|)},$$ and so the map $\overset{\circ}c\to\overset{\circ}s$ is given by $$(x,y)\mapsto\frac{\sqrt{x^2+y^2}}{\max(|x|,|y|)}(x,y)$$ for $(x,y)\ne(0,0)$ and $(0,0)\mapsto(0,0)$. Once again, the map is continuous away from the origin, and using the fact that $$\frac{\sqrt{x^2+y^2}}{\max(|x|,|y|)}\le\sqrt2$$ for all $(x,y)\ne (0,0)$ we can see that the map is continuous at the origin, too.

Hint: The isomorphism is given by $$\phi\colon\overset{\circ}{c}\to\overset{\circ}{s}\colon (x,y)\mapsto\begin{cases} (0,0) & x=y=0 \\\left(\frac{x}{|x|}\sqrt{x^2+y^2},\frac{y}{|x|}\sqrt{x^2+y^2}\right) & |x|\geq|y| \\ \left(\frac{x}{|y|}\sqrt{x^2+y^2},\frac{y}{|y|}\sqrt{x^2+y^2}\right) & |x|<|y|\end{cases}$$

You can, in fact, prove that any bounded convex open subset of $R^n$ is homeomorphic to $R^n$, thus any two bounded convex open subsets of $R^n$ are homeomorphic to each other. The interiors of a circle and a square are bounded convex open subsets of $R^2$, so this general answer also answers the specific question. My proof sketch goes as follows, although it's possible there might be a mistake. Let $A$ be a bounded convex open subset of $R^n$.

1. WLOG $0 \in A$, since if not, we can just translate.

2. Let $B(A)$ be the boundary of $A$. For any nonzero $x \in R^n$ define $L(x) := \{cx:c \in R^+\}$ i.e. $L(x)$ is the ray in the direction of $x$. Prove: $B(A) \cap L(x)$ is a singleton for all $x$. As a corollary, if $\alpha \in B(A) \cap L(x)$, then $cx \in A$ if $|cx| < |\alpha|$ and $cx \not\in A$ if $|cx| \geq |\alpha|$.

3. Define $g:R^n\backslash\{0\} \rightarrow R^n$ by $g(x)$ is the element in $B(A) \cap L(x)$. Prove $g$ is continuous.

4. Define $h:A \rightarrow R$ by $h(0) = 1$, and for $x$ nonzero, $h(x) = \frac{|g(x)|}{|g(x)|-|x|}$. Then define $f:A \rightarrow R^n$ by $f(x) = h(x)x$. Prove $h,f$ are continuous.

5. Finally, we prove that $f$ is the desired homeomorphism by finding the inverse function and proving it's continuous. The inverse can be found algebraically. You can use the corollaries from part 2 to prove that the inverse does indeed map into $A$.

Note that this proof sketch assumes a decent undergraduate level understanding of analysis.