Topological Quotients

Recall from the Final Topologies page that if $X$ is a set and $\{ Y_i : i \in I \}$ is a collection of topological spaces then the final topology induced by $\{ f_i : i \in I \}$ on $X$ is the finest topology $\tau$ which makes the maps $f_i : Y_i \to X$ continuous for all $i \in I$.

We saw that if $\tau$ is the final topology induced by the collection of maps $\{ f_i : i \in I \}$ can be given by:

(1)

\begin{align} \quad \tau = \{ U \subseteq X : f^{-1}(U) \in \tau_i \: \mathrm{for \: all} \: i \in I \} \end{align}

That is, the final topology induced by $\{ f_i : i \in I \}$ is equal to the collection of subsets of $X$ whose inverse images are contained in every topology $\tau_i$ for $i \in I$.

We will now look at a special type of topology on a set known as a quotient topology but we will first need to define a special map known as a quotient map.

Definition: Let $(X, \tau)$ be a topological space and let $\sim$ be an equivalence relation on $X$. For each $x \in X$, denote the equivalence class of $x$ by $[x] = \{ y \in X : x \sim y \}$ and let $X \: / \sim$ denote the set of all of these equivalence classes. The Quotient Map with respect to the set $X$ and the equivalence relation $\sim$ is defined as the map $q : X \to X \: / \sim$ defined for all $x \in X$ by $q(x) = [x]$.

We are now ready to define a quotient topology on the set of equivalence classes $X \: / \sim$ where $\sim$ is an equivalence relation on $X$.

Definition: Let $(X, \tau)$ be a topological space and let $\sim$ be an equivalence relation on $X$. The Quotient Topology on $X \: / \sim$ is the final topology induced by the quotient map $q : X \to X \: / \sim$. Then the resulting Topological Quotient is the topological space of the set $X \: / \sim$ with the quotient topology.

Equivalently, the quotient topology on $X \: / \sim$ is the finest topology which maps the quotient map $q : X \to X \: / \sim$ continuous.

Proposition 1: Let $(X, \tau)$ be a topological space and $\sim$ be an equivalence relation on $X$. Furthermore, let $X \: / \sim$ have the quotient topology with respect to $\sim$. Then $U \subseteq X \: / \sim$ is open in $X \: / \sim$ if and only if $\displaystyle{\bigcup_{[x] \in U} [x]}$ is open in $X$.

• Proof: By definition, the quotient topology $\tau^*$ on $X \: / \sim$ is the final topology on the quotient map $q : X \to X \: / \sim$ and can be described as the following set:

(2)

\begin{align} \quad \tau^* = \{ U \subseteq X \: / \sim : f^{-1}(U) \in \tau \} \end{align}

• $\Rightarrow$ Suppose that $U \subseteq X \: / \sim$ is open in $X \: / \sim$. Then $q^{-1}(U)$ is open in $X$. But:

(3)

\begin{align} \quad q^{-1}(U) = \bigcup_{[x] \in U} [x] \end{align}

• So $\bigcup_{[x] \in U} [x]$ is open in $X$.

• $\Leftarrow$ Let $U \subseteq X \: / \sim$ and suppose that $\bigcup_{[x] \in U} [x]$ is open in $X$. Then $q \left ( \bigcup_{[x] \in U} [x] \right ) = U$. So $q^{-1}(U)$ is open and since the quotient topology is the finest topology making $q : X \to X \: / \sim$ continuous we have that $U$ is open in $X \: / \sim$. $\blacksquare$