> [!definition] > > Let $(X, \topo_X)$ be a [[Topological Space|topological space]] and $\sim$ be an [[Equivalence Relation|equivalence relation]] on $X$. Denote $\tilde X$ as the collection of [[Equivalence Class|equivalence classes]] and $\pi: X \to \tilde X$ with $x \mapsto \ol{x}$ as the map taking each element to its equivalence class. Then > $ > \tilde \topo = \bracs{U \subset \tilde X: \pi^{-1}(U) \in \topo_X} > $ > is a topology on $\tilde X$, known as the **quotient topology**[^1]. > > *Proof*. Let $\seqi{U} \subset \tilde \topo$, then > $ > \begin{align*} > \pi^{-1}\paren{\bigcup_{i \in I}U_i} = \bigcup_{i \in I}\pi^{-1}(U_i) \in \topo_X > \end{align*} > $ > therefore $\bigcup_{i \in I}U_i \in \tilde \topo$. Now let $U, V \in \tilde \topo$, then > $ > \pi^{-1}(U \cap V) = \pi^{-1}(U) \cap \pi^{-1}(V) \in \topo_X > $ > and $U \cap V \in \tilde \topo$. Lastly, $\pi^{-1}(\emptyset) = \emptyset \in \topo_X$ and $\pi^{-1}(\tilde X) = X \in \topo_X$. Therefore $\tilde \topo$ is a topology. > [!definition] > > Let $X$ and $Y$ be topological spaces, and $\pi: X \to Y$ be a surjective map. A set $E \subset X$ is **saturated** if there exists $F \subset Y$ such that $E = \pi^{-1}(F)$ is the complete preimage. > > The map $\pi$ is a **quotient map**[^2] if for any $U \subset Y$, $U$ is open if and only if $\pi^{-1}(U)$ is open in $X$. In other words, $\pi$ maps saturated open sets in $X$ to open sets in $Y$. > > *Proof*. Suppose that $\pi: X \to Y$ is a quotient map, then for any saturated set $U \subset X$, $\pi(U)$ is open in $Y$, and it maps saturated open sets to open sets. On the other hand, each preimage of an open set is a saturated open set, so $\pi$ is continuous. > > Now suppose that $\pi: X \to Y$ is continuous and maps saturated open sets in $X$ to open sets in $Y$. Let $U \subset Y$ be open, then $\pi^{-1}(U)$ is a saturated open set. For any saturated open set $U$, $\pi(U)$ is open, so $\pi$ is a quotient map. > [!theorem] > > Let $f: X \to Y$, then $f$ is **open** if it maps open sets to open sets, and **closed** if it maps closed sets to closed sets. If $f$ is surjective, continuous, and open/closed, then $f$ is a quotient map. > > *Proof*. If $f$ is open, then $f$ maps all saturated open sets to open sets. If $f$ is closed and $A \subset X$ is saturated and open, then $f(A)^c = f(A^c)$, so $f(A) = f(A^c)^c$ is open. > [!theorem] > > Let $\pi: X \to Y$ be a quotient map, and $A \subset X$ be a saturated set. Let $\tau: A \to \pi(A)$ be the restriction of $\pi$ to $A$, then > 1. If $A$ is open or closed, then $\tau$ is a quotient map. > 2. If $\pi$ is open or closed, then $\tau$ is a quotient map. > > *Proof*. First note that since $A$ is saturated, for any $B \subset \pi(A)$, $\pi^{-1}(B) \subset A$ and $\pi^{-1}(B) = \tau^{-1}(B)$. For any $B \subset X$, $\pi(A \cap B) = \pi(A) \cap \pi(B)$ as for any $y \in \pi(A) \cap \pi(B)$, there exists $b \in B$ such that $y = \pi(b)$. Because $A$ is saturated, $b \in A$ as well so $y \in \pi(A \cap B)$. > > Let $\widetilde{U} = \pi^{-1}(U) = \tau^{-1}(U) \subset A$ be a saturated (relatively) open set. > > Suppose that $A$ is open, then $\widetilde{U}$ is open in $X$. Since $\pi$ is a quotient map, $U$ is open in $Y$. Now suppose that $\pi$ is open, then there exists $V \subset X$ open such that $\widetilde{U} = V \cap A$. Since $\pi(V \cap A) = \pi(V) \cap \pi(A)$ and $\pi(V)$ is open, $\tau(\widetilde{U}) = \pi(V) \cap \pi(A)$ is relatively open in $\pi(A)$. > > Suppose that $A$ is closed, then $\widetilde{U}^c$ is closed in $X$. Since $\pi$ is a quotient map, $U^c$ is closed in $Y$, and $U$ is open in $Y$. Now suppose that $\pi$ is closed, then there exists $V \subset X$ open such that $\widetilde{U} = V \cap A$. Since $\pi(V^c \cap A) = \pi(V^c) \cap \pi(A)$ and $\pi(V^c)$ is closed, $\tau(\widetilde{U}) = \pi(V^c)^c \cap \pi(A)$ is relatively open in $\pi(A)$. > [!theorem] > > Let $\pi: X \to Y$ be a quotient map, and $g: X \to Z$ that is constant on each $\pi^{-1}(y)$, then $g$ induces a map $f: Y \to Z$ such that $g = f \circ \pi$. Then > 1. $f$ is continuous if and only if $g$ is continuous. > 2. $f$ is a quotient map if and only if $g$ is a quotient map. > > *Proof*. Let $U$ be an open set of $Z$, then $g^{-1}(U) = \pi^{-1}(f^{-1}(U))$. If $g$ is continuous, then $g^{-1}(U)$ is open, and so is $f^{-1}(U)$. > > If $g$ is a quotient map, then $f$ is surjective and continuous by the first statement. Let $U = f^{-1}(V)$ be a saturated open set. Since $\pi$ is continuous, $\pi^{-1}(U)$ is also a saturated open set. As $g$ is a quotient map, $V = g(\pi^{-1}(U)) = f(U)$ is open. Hence $f$ is a quotient map. > [!theorem] > > $\tilde X$ is [[Separation Axioms|T1]] if and only if every equivalence class is [[Closed Set|closed]] in $X$. > > *Proof*. Let $\ol{x} \in \tilde X$. If $\ol{x}$ is closed $\bracs{\ol{x}}^c$ is open and, $\pi^{-1}(\bracs{\ol{x}}^c)$ is [[Open Set|open]] in $X$. As $\pi^{-1}(\bracs{\ol{x}}^c) = \pi^{-1}(\bracs{\ol{x}})^c$ is open, $\pi^{-1}(\bracs{\ol x})$ is closed. Therefore the equivalence class of $x$ is closed. > > Suppose that $\ol{x}$ is closed in $X$, then $(\ol{x})^c$ is open and $\pi^{-1}(\bracs{\ol{x}}^c) = \pi^{-1}(\bracs{\ol{x}})^c$, meaning that $\bracs{\ol{x}}^c$ is open and $\ol{x}$ is closed. > [!theorem] "First Isomorphism Theorem" > > Let $g: X \to Z$ be a surjective continuous map. Let $x \sim y$ if $g(x) = g(y)$, and let $\tilde X$ be the quotient space equipped with the quotient topology, then > 1. The map $g$ induces a bijective continuous map $f: \tilde X \to Z$, which is a homeomorphism if and only if $g$ is a quotient map. > 2. If $Z$ is Hausdorff, then so is $\tilde X$. > > *Proof*. Suppose that $g$ is a quotient map, then $f$ is also a quotient map, as $f$ is bijective, all open sets are saturated open sets, so $f$ is a homeomorphism. > > Suppose that $Z$ is Hausdorff, then since $f$ is continuous and bijective, so is $\tilde X$. > [!theorem] > > Let $f: X \to Y$ be a quotient map. If $f$ is open, and $X$ is first/second countable, then $Y$ is also first/second countable. > > *Proof*. Since any neighbourhood base/base of $X$ gets mapped to a neighbourhood base/base of $Y$, the countability axioms are preserved. [^1]: Folland, chapter 4. [^2]: Munkres, chapter 2, section 22.