Topological Homeomorphisms

This note is separate from Homeomorphisms from analysis. This note is primarily for MATH 190A, since we’re going to use it differently. Largely, they are the same.

Definition (Topological Homeomorphism)

A homeomorphism $(X,{\tau }_X)\to (Y,{\tau }_Y)$ is a map that is

• continuous
• bijective
• its inverse is continuous

We say $X\cong Y$ to mean $X$ is homeomorphic to $Y$.

Equivalently, we say two spaces, $(X,{\tau }_X)$ and $(Y,{\tau }_Y)$ are homeomorphic if $\exists$ mutually inverse continuous maps. Precisely,

$$(X,{\tau }_X)\stackrel{g}{\rightleftarrows }f(Y,{\tau }_Y)$$

where

$$f\circ g={\text{Id}}_{X}g\circ f={\text{Id}}_Y$$

Example 1:

We have $(-1,1)$ and $\mathbb{R}$ are homeomorphic.

Proof: We could use $f(x)=\tan (\pi x/2)$, with inverse being $g(y)=2{\tan }^{-1}(y)/\pi$.

Example 2:

$$f(x)=\frac{x}{1-x^2}$$

is homeomorphic with inverse

$$g(y)=\frac{2y}{1+\sqrt{1+4y^2}}$$

We assert both are continuous by Theorem (Continuity by Composition in a Topology).

Example 3 (Non-example):

Consider

$$\begin{array}{rl}f:[0,1)& \to S=\{x^2+y^2=1\}\subset {\mathbb{R}}^2\\ x& \mapsto (\cos 2\pi x,\sin 2\pi x)\end{array}$$

This map is a continuous bijection, but its inverse is not. In particular, the $f$ is not an open map.