\section{Quotient Spaces}
\label{section:lc-quotient}

\begin{definition}[Quotient Seminorm]
\label{definition:quotient-norm}
    Let $E$ be a vector space over $K \in \RC$, $\rho: E \to [0, \infty)$ be a seminorm, and $M \subset E$ be a vector subspace, then
    \[
    \rho_M: E/M \to [0, \infty) \quad x + M \mapsto \inf_{y \in x + M}\rho(y)
    \]

    is the \textbf{quotient} of $\rho$ by $M$.
\end{definition}
\begin{proof}
    Let $\lambda \in K$ with $\lambda \ne 0$, then $x \mapsto \lambda x$ is a bijection, and
    \begin{align*}
        \abs{\lambda}\rho_M(x) &= \abs{\lambda} \inf\bracs{\rho(y)|y \in x + M} = \inf\bracs{\abs{\lambda}\rho(y)|y \in x + M} \\
        &= \inf\bracs{\rho(\lambda y)|y \in x + M} =\inf\bracs{\rho(y)|y \in \lambda(x + M)} = \rho_M(\lambda(x + M))
    \end{align*}
    For any $x, x' \in E$, $y \in x + M$ and $y' \in x' + M$, $y + y' \in x + x' + M$, so
    \[
    \rho_M(x + x') \le \rho(y + y') \le \rho(y) + \rho(y')
    \]

    As this holds for all $y \in x + M$ and $y' \in x' + M$, $\rho_M(x + x') \le \rho_M(x) + \rho_M(x')$.
\end{proof}

\begin{definition}[Quotient Locally Convex Space]
\label{definition:lc-quotient}
    Let $E$ be a locally convex space over $K \in \RC$ and $M \subset E$ be a vector subspace, then there exists $(\td E, \pi)$ such that:
    \begin{enumerate}
        \item $\td E$ is a locally convex space over $K$.
        \item $\pi \in L(E; \td E)$.
        \item $\ker \pi \supset M$.
        \item[(U)] For any topological space $F$ and $f \in C(E;F)$ such that $f(x) = f(y)$ whenever $x - y \in M$, there exists a unique $\td f \in C(\td E; F)$ such that the following diagram commutes
        \[
        \xymatrix{
        E \ar@{->}[rd]^{f} \ar@{->}[d]_{\pi} &  \\
        \widetilde E \ar@{->}[r]_{\tilde f} & F
        }
        \]

        If $F$ is a TVS over $K$ and $f \in L(E; F)$, then $\td f \in L(E; F)$.
        \item If $\seqi{\rho}$ is a family of seminorms that induces the topology on $E$, then their quotients by $M$ induces the topology on $\td E$.
    \end{enumerate}
    The space $\td E = E/M$ is the \textbf{quotient} of $E$ by $M$.
\end{definition}
\begin{proof}
    By \autoref{definition:tvs-quotient}, (2), (3), (U) holds, and $\td E$ is a TVS over $K$.

    (1): Let $U \subset E$ be convex, then for any $x + M, y + M \in \pi(U)$ and $t \in [0, 1]$,
    \[
    (tx + M) + ((1 - t)y + M) = (tx + (1-t)y) + M \in U + M = \pi(U)
    \]

    so $\pi(U)$ is convex. Let $\fB = \bracs{U|U \in \cn_E(0) \text{ convex}}$, then $\bracs{\pi(U)|U \in \fB}$ is a fundamental system of neighbourhoods for the quotient topology on $E/M$. Therefore $E/M$ is locally convex.

    (5): By (U), each quotient seminorm is continuous on $\td E$, so the quotient topology contains the topology induced by the quotient seminorms. On the other hand, let $\pi(U) \in \cn_{\td E}(0)$, then there exists $J \subset I$ finite and $r > 0$ such that
    \[
    \bigcap_{j \in J}B_j(0, r) \subset U
    \]

    For each $j \in J$, let $\eta_j$ be the quotient of $\rho_j$ by $M$. Let $x + M \in E/M$ with $\eta_j(x) < r$ for all $j \in J$. For each $j \in J$, there exists $y_j \in x + M$ such that $\rho_j(y_j) < r$, so $y_j + M \in \pi(U)$. Therefore $x \in \pi(U)$ as well, and the quotient norms induce the quotient topology on $E/M$.
\end{proof}