## Homeomorphisms

Given two topological spaces, $(X,T_X)$ and $(Y,T_Y),$ we say that a map $\varphi \colon X \to Y$ is a homeomorphism if it satisfies the three properties below:

1. It is a bijection (that is, both injective and surjective),
2. it is continuous, and
3. it has a continuous inverse $\varphi^{-1} \colon Y \to X.$

Two spaces for which there is such a homeomorphism are called homeomorphic.  As it turns out, homeomorphic spaces have the same topological properties.

In what follows, we will construct several interesting homeomorphisms to train our skills:

### Disks and squares

Consider the unit disk $D_2$ and the unit square $\square_2$ defined by

$\begin{array}{l} D_2 = \{ x=(x_1,x_2) \in \mathbb{R}^2 : x_1^2 + x_2^2 \leq 1 \}\\ \square_2 =\{ x=(x_1,x_2) \in \mathbb{R}^2 : \lvert x_1 \rvert \leq 1, \lvert x_2 \rvert \leq 1 \} \end{array}$

We construct a homeomorphism $\varphi \colon \square_2 \to D_2$ as follows:

$\varphi(x_1,x_2) = \begin{cases} 0 &\text{if } (x_1,x_2) = (0,0),\\ \displaystyle{\frac{\text{max}(\lvert x_1 \rvert, \lvert x_2 \rvert )}{\sqrt{x_1^2 + x_2^2}}} ( x_1,x_2)&\text{otherwise.} \end{cases}$

This function maps for each $0 < d \leq 1$, the border of each square $\{ (x_1,x_2) \in \mathbb{R}^2 : \text{max}( \lvert x_1 \rvert, \lvert x_2 \rvert) \leq d \}$ into the circle of radius $d$ centered at the origin; the manner in which these sets map into each other indicates why the function $\varphi$ is a bijection.  It is very easy to check its continuity as well, and I leave that task to the reader.

An inverse to this function is constructed in a similar way, so that each circle is mapped to the border of a square:

$\varphi^{-1}(y_1,y_2) = \begin{cases} 0 &\text{if } (y_1,y_2) = (0,0),\\ \displaystyle{\frac{\sqrt{y_1^2+y_2^2}}{\text{max}(\lvert y_1\rvert, \lvert y_2\rvert)}}(y_1,y_2) &\text{otherwise.}\end{cases}$

### An open interval and the real line

Let us find an homeomorphism from the unit interval $(-1,1)$ into the real line $\mathbb{R}$ based in an interesting construction called stereographic projection.  We start by mapping each $t \in (-1,1)$ into an angle $\theta \in (-\pi, \pi)$ by simple multiplication: $\theta = \pi t$.  This angle gives a single element in the unit circle $(\cos \pi t, \sin \pi t)$, except the point $(-1,0).$  The stereographic projection turns each point of the circle different than $(-1,0)$ into a unique point in the plane with coordinates $\big(1,\varphi(t) \big),$  where

$\varphi(t) = \displaystyle{ \frac{2\sin \pi t}{\cos \pi t +1} }.$

This is the function we are looking for.  Its continuity is easy to prove, and so are its one-to-one and onto properties.  In order to construct the inverse map, trace back from the real line to the vertical line $\{(1,s) : s \in \mathbb{R}\},$ from there to the unit disk by the inverse of stereographic projection through the point $(-1,0)$, and find the angle of the corresponding image.  Division by $\pi$ offers you a value in the unit interval $(-1,1).$  This value is the image of the inverse $\varphi^{-1}.$   I leave the construction of the analytical expression of this function to the reader as a nice exercise.