Homeomorphisms on Topological Spaces

Recall from the Open and Closed Maps on Topological Spaces page that we said that if $X$ and $Y$ are topological spaces then a map $f : X \to Y$ is said to be an open map (or simply open) if for every open set $U$ of $X$ we have that $f(U)$ is open in $Y$.

Similarly, $f$ is said to be a closed map (or simply closed) if for every closed set $C$ of $X$ we have that $f(C)$ is closed in $Y$.

We will now look at a special type of map $f$ called a homeomorphism which will be the center of our focus for the next while.

Definition: Let $X$ and $Y$ be topological spaces. A bijective map $f : X \to Y$ is called a Homeomorphism if $f$ is both open and continuous. If a homeomorphism $f$ between $X$ and $Y$ exists then we say that $X$ and $Y$ are Homeomorphic written $X \cong Y$.

It is important to note that $f$ needs to be bijection.

If $f : X \to Y$ is a homeomorphism, then since $f$ is bijective and open we have that $f^{-1}$ is continuous. Therefore, a homeomorphism $f$ is such that $f$ and $f^{-1}$ are both continuous.

In the following proposition we state some of the most basic properties of topological spaces being homeomorphic which all arise from $\cong$ being an equivalence relation.

Proposition 1: Let $X$, $Y$, and $Z$ be topological spaces. a) If $X \cong X$ for all topological spaces $X$ (Reflexivity). b) $X \cong Y$ if and only if $Y \cong X$ (Symmetry). c) If $X \cong Y$ and $Y \cong Z$ then $X \cong Z$ (Transitivity).

There are many such examples of homeomorphisms. For example, take $X = (0, \infty)$ and $Y = \mathbb{R}$ where the topologies on $X$ and $Y$ are the usual topologies of open intervals. Then we claim that the following function $f : (0, \infty) \to \mathbb{R}$ defined for all $x \in (0, \infty)$ is a homeomorphism:

(1)

\begin{align} \quad f(x) = \ln x \end{align}

We first show that $f$ is a bijection by showing that $f$ is both an injection and a surjection. Let $x, y \in (0, \infty)$ and suppose that $f(x) = f(y)$. Then $\ln(x) = \ln(y)$. So $e^{\ln x} = e^{\ln y}$ which implies that $x = y$. So $f$ is indeed an injection. Now let $y = \ln x$. Then $x = e^y$. So, for all $y \in \mathbb{R}$ there exists an $x \in (0, \infty)$ such that:

(2)

\begin{align} \quad f(x) = f(e^y) = \ln(e^y) = y \end{align}

Hence $f$ is a surjection. Therefore $f$ is indeed a bijection.

Furthermore, it's not hard to see that $f$ is also continuous on all of $X = (0, \infty)$, and open, so $f$ is indeed a homeomorphism.