Homeomorphism from square to unit circle Can we find a homeomorphism from the square $Q_2$ of side length $2$ centered on the origin and the unit circle $S^1$?
We can easily define a map $r:Q \longrightarrow S^1$ by
$$(x,y) \mapsto  \bigg(\frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}} \bigg)$$ which is a radial projection onto the unit circle, but how can we define $r^{-1}$ and show that it is continuous? 
Thoughts
I think we may define the inverse map as 
$$(x,y) \mapsto \bigg(\frac{x}{\sqrt2\max{│x│, │y│}} , \frac{y}{\sqrt2\max {│x│, │y│}}\bigg)$$At least intuitively this maps to a square, and the $\frac{1}{\sqrt2}$ term scales appropriately. However, how can we demonstrate these are continuous maps, thus demonstrating that $Q_2$ and $S^1$ are homeomorphic?
Any help would be appreciated. Regards, MM.
 A: I'm not sure why you need algebraic formulas. Define each map $f$ to be radial projection onto the appropriate curve (from the origin) and find continuity by the property "$\forall$ neighborhood $U$ of $f(x)$ $\exists$ a neighborhood of $x$ whose image is contained in $U$", which is easy to check by eye.
A: It seems off to me. For example, in your second map, the point $(1,0)$, which is on the unit circle, gets mapped to $(\frac1{\sqrt2},0)$, which is not on the square.
If I were you, I would write out the four different maps for each side of the square - I think it simplifies the process, especially since once you remove the max terms, the function is easier to work with.
A: Let me outline a proof; I shall leave the details of this proof as exercises.
Exercise 1: Prove that the unit circle $S^1$ is homeomorphic to the space obtained from the unit interval $[0,1]$ by identifying the endpoints $0$ and $1$. (Hint: the exponential mapping $\theta\to e^{i\theta}$ can be composed with a linear mapping to give a homeomorphism.) 
Let $C$ be the image of a simple closed curve in the plane, that is, let $C$ be the image of a continuous function $f:[0,1]\to \mathbb{R}^2$ such that $f:(0,1)\to\mathbb{R}^2$ is injective and $f(0)=f(1)$. 
Exercise 2: Prove that $C$ is homeomorphic to the unit circle $S^1$. (Hint: show that $f:[0,1]\to C$ induces a continuous bijection $\tilde{f}:S^1\to C$ using Exercise 1 and then show that $f$ is an open mapping using the compactness of $S^1$ and the fact that the plane is Hausdorff.)
We now need an elementary lemma of point-set topology:
Exercise 3: Let $X$ be a topological space and let $X=A\cup B$ where $A$ and $B$ are closed subspaces of $X$. Prove that if $g:A\to Y$ and $h:B\to Y$ are continuous functions into a topological space $Y$ such that $g(x)=h(x)$ for all $x\in A\cap B$, then there is a unique continuous function $f:X\to Y$ such that $f(a)=g(a)$ and $f(b)=h(b)$ for all $a\in A$ and $b\in B$.
Finally, we can prove the result of your question:
Exercise 4: Prove that the unit square is the image of a simple closed curve in the plane and conclude that it is homeomorphic to the unit circle. (Hint: you can use Exercise 3 to "glue" continuous mappings together.)
Exercise 5: Give examples of images of simple closed curves in the plane (using Exercise 3) and conclude (using Exercise 2) that they are homeomorphic to the unit circle $S^1$.
I hope this helps!
A: If you choose to go about it in a computational manner, you can prove that both of the maps are continuous by noting that they are compositions of continuous functions.
A: Following in line with @Amitesh Datta's answer above, rather than proving the homeomorphism directly there are simpler ways that one can arrive at the desired result. 
This following technique uses a fairly high-powered result (in the sense that once you know it, you can prove a lot of things are homeomorphic quite easily), but I think it's a very useful result that anyone doing topology should be aware of.

Definition: An open $n$-cell is any topological space that is homeomorphic to the open unit ball $\mathbb{B}^n$. A closed $n$-cell is any topological space homeomorphic to $\overline{\mathbb{B}^n}$

This is the high-powered result that I mentioned earlier.

Theorem: If $D \subseteq \mathbb{R}^n$ is a compact convex subset with nonempty interior, then $D$ is a closed $n$-cell and it's interior is an open $n$-cell. In fact given any point $p \in \operatorname{Int}(D)$, there exists a homeomorphism $F: \overline{\mathbb{B}^n} \to D$ that sends $0$ to $p$, $\mathbb{B}^n$ to $\operatorname{Int}(D)$ and $\mathbb{S}^{n-1}$ to $\operatorname{Bd} D$.

Now note that if we let $A \subseteq \mathbb{R}^2$ be the (solid) square of side length $2$, then $A$ is easily seen to be a convex subset, and is thus a closed $2$-cell by the theorem above. Moreover the square $Q_2$ is the topological boundary of the solid square $A$ and the theorem above then immediately shows that $\operatorname{Bd}(A) = Q_2 \cong S^1$ as desired.
