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.

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