\section{Quotient Topologies}
\label{section:quotient-topology}

\begin{definition}[Saturated]
\label{definition:saturated}
    Let $X, Y$ be sets, $f: X \to Y$ be surjective, and $E \subset X$, then $E$ is \textbf{saturated} with respect to $f$ if $E = f^{-1}(f(E))$.
\end{definition}

\begin{definition}[Quotient Map]
\label{definition:quotient-map}
    Let $X, Y$ be topological spaces and $\pi: X \to Y$ be a surjective map, then the following are equivalent:
    \begin{enumerate}
        \item For any $U \subset Y$, $U$ is open if and only if $\pi^{-1}(U)$ is open.
        \item $\pi \in C(X; Y)$, and for any $U \subset X$ saturated and open, $\pi(U)$ is open.
    \end{enumerate}

    If the above holds, then $\pi$ is a \textbf{quotient map}.
\end{definition}
\begin{proof}
    $(1) \Rightarrow (2)$: Let $U \subset Y$ be open, then $\pi^{-1}(U)$ is open, so $f \in C(X; Y)$. If $V \subset X$ is saturated and open, then $\pi(V) = \pi(\pi^{-1}(\pi(V)))$ is open.

    $(2) \Rightarrow (1)$: Let $U \subset Y$ be open, then $\pi^{-1}(U)$ is open by continuity. If $V \subset Y$ and $\pi^{-1}(V)$ is open, then $V = \pi(\pi^{-1}(V))$ is open.
\end{proof}

\begin{definition}[Quotient Space]
\label{definition:quotient-topology}
    Let $X$ be a topological space, $\sim$ be an equivalence relation on $X$, then there exists $(\td X, \pi)$ such that:
    \begin{enumerate}
        \item $\td X$ is a topological space with ground set $X/\sim$.
        \item $\pi \in C(X; \td X)$.
        \item $\pi$ is constant on each equivalence class of $\sim$.
        \item[(U)] For any pair $(Y, f)$ satisfying (1), (2), and (3), there exists a unique $\td f \in C(\td X; Y)$ such that the following diagram commutes:
        \[
        \xymatrix{
        X \ar@{->}[d]_{\pi} \ar@{->}[rd]^{f} &  \\
        \td X \ar@{->}[r]_{\tilde f} & Y
        }
        \]

        \item $\pi$ is a quotient map.
    \end{enumerate}

    The space $(\td X, \pi)$ is the \textbf{quotient} of $X$ by $\sim$.
\end{definition}
\begin{proof}
    Let $\td X = X/\sim$. For each $U \subset \td X$, define $U$ to be open if and only if $\pi^{-1}(U) \subset X$ is open, then $(\td X, \pi)$ satisfies (1), (2), (3), and (5).

    (U): Since $f$ is constant on each equivalence class of $\sim$, there exists $\td f: \td X \to Y$ such that the diagram commutes. For any $U \subset Y$ open, $\td f^{-1}(U) = \pi(f^{-1}(U))$ is saturated with respect to $\pi$, so $\td f^{-1}(U)$ is open.
\end{proof}