# Math4201 Topology I (Lecture 17) ## Quotient topology How can we define topologies on the space obtained points in a topological space? ### Quotient map Let $(X,\mathcal{T})$ be a topological space. $X^*$ is a set and $q:X\to X^*$ is a surjective map. The quotient topology on $X^*$ is defined as follows: $$ \mathcal{T}^* = \{U\subseteq X^*\mid q^{-1}(U)\in \mathcal{T}\} $$ $U\subseteq X^*$ is open if and only if $q^{-1}(U)$ is open in $X$. In particular, $q$ is continuous map. #### Definition of quotient map $q:X\to X^*$ defined above is called a **quotient map**. #### Definition of quotient space $(X^*,\mathcal{T}^*)$ is called the **quotient space** of $X$ by $q$. ### Typical way of constructing a surjective map #### Equivalence relation $\sim$ is a subset of $X\times X$ satisfying: - reflexive: $\forall x\in X, x\sim x$ - symmetric: $\forall x,y\in X, x\sim y\implies y\sim x$ - transitive: $\forall x,y,z\in X, x\sim y\text{ and } y\sim z\implies x\sim z$ #### Equivalence classes Check equivalence relation. For $x\in X$, the equivalence class of $x$ is denoted as $[x]\coloneqq \{y\in X\mid y\sim x\}$. $X^*$ is the set of all equivalence classes on $X$. $q:X\to X^*$ is defined as $q(x)=[x]$ will be a surjective map.
Example of surjective maps and their quotient spaces Let $X=\mathbb{R}^2$ and $(s,t)\sim (s',t')$ if and only if $s-s'$ and $t-t'$ are both integers. This space as a topological space is homeomorphic to the torus. --- Let $X=\{(s,t)\in \mathbb{R}^2\mid s^2+t^2\leq 1\}$ and $(s,t)\sim (s',t')$ if and only if $s^2+t^2$ and $s'^2+t'^2$. with subspace topology as a subspace of $\mathbb{R}^2$. This space as a topological space is homeomorphic to the spherical shell $S^2$.
We will show that the quotient topology is a topology on $X^*$.
Proof We need to show that the quotient topology is a topology on $X^*$. 1. $\emptyset, X^*$ are open in $X^*$. $\emptyset, X^*$ are open in $X^*$ because $q^{-1}(\emptyset)=q^{-1}(X^*)=\emptyset$ and $q^{-1}(X^*)=X$ are open in $X$. 2. $\mathcal{T}^*$ is closed with respect to arbitrary unions. $$ q^{-1}(\bigcup_{\alpha \in I} U_\alpha)=\bigcup_{\alpha \in I} q^{-1}(U_\alpha) $$ 3. $\mathcal{T}^*$ is closed with respect to finite intersections. $$ q^{-1}(\bigcap_{\alpha \in I} U_\alpha)=\bigcap_{\alpha \in I} q^{-1}(U_\alpha) $$