% vim: tw=50
% 12/11/2022 11AM

\begin{flashcard}[QuotTopDefn]
    \begin{definition*}[Quotient Topology]
        Let $(X, \tau)$ be a topological space and
        $\sim$ an equivalence relation on $X$. Let $q
        : X \to X / \sim$ be the quotient map, i.e.
        $\forall x \in X$, $q(x) = [x]_\sim$. The
        \emph{quotient topology} on $X / \sim$ is
        \[ \rho = \cloze{\{G \subset X / \sim \mid q^{-1}(G)
        \in \tau\}} \]
    \end{definition*}
\end{flashcard}

\subsubsection*{Remarks}

\begin{enumerate}[(1)]
    \item $\rho$ is indeed a topology using
        $q^{-1}(\bigcup G) = \bigcup q^{-1}(G)$
        and $q^{-1}(G \cap H) = q^{-1}(G) \cap
        q^{-1}(H)$.
    \item $\rho$ is the largest topology on $X /
        \sim$ making the quotient map $q$
        continuous.
\end{enumerate}

\subsubsection*{Examples}

\begin{enumerate}[(1)]
    \item Take $\RR$ with the usual topology and
        $x \sim y$ if and only if $x - y \in \ZZ$.
        Then $\RR / \sim$ `is' $S'$, the unit
        circle (a subspace of $\RR^2$ with
        Euclidean topology). `is' means `is
        homeomorphic to'. (Proof later).
    \item As above but now $x \sim y$ if and only
        if $x - y \in \QQ$. What is quotient
        topology on $\RR / \sim$? Suppose $G
        \subset \RR / \sim$ is open, $G \neq
        \emptyset$. Then $q^{-1}(G) \subset \RR$
        is open and non-empty so contains some
        interval $(a, b) \subset q^{-1}(G)$ with
        $a \neq b$. Now take any $x \in \RR$. Then
        there exists $y \in (a, b)$ with $x - y
        \in \QQ$ i.e. $x \sim y$. Then $q(x) =
        [x]_\sim = [y]_\sim = G$. Hence $G = \RR /
        \sim$. \\
        So the quotient topology on $\RR / \sim$
        is the indiscrete topology.
\end{enumerate}
So the quotient topology on $\RR / \sim$ is the
indiscrete topology.
\begin{itemize}
    \item Quotients of metrizable spaces need not
        be metrizable.
    \item Quotients of Hausdorff spaces need not
        be Hausdorff.
\end{itemize}

\begin{hiddenflashcard}[quotient-of-metrizable-not-necessarily-metrizable]
Proof that if $X$ is a metrizable topological
space, and $Y$ is a quotient space of $X$, then
$Y$ is metrizable? \\
\cloze{
This statement is false. (As an example, can
consider $\RR / \sim$ where $x \sim y \iff x - y
\in \QQ$).
}
\end{hiddenflashcard}

\begin{hiddenflashcard}[quotient-of-Hausdorff-not-necessarily-Hausdorff]
Proof that if $X$ is a Hausdorff topological
space, and $Y$ is a quotient space of $X$, then
$Y$ is Hausdorff? \\
\cloze{
This statement is false. (As an example, can
consider $\RR / \sim$ where $x \sim y \iff x - y
\in \QQ$).
}
\end{hiddenflashcard}

\subsubsection*{Basics on equivalence relations
and quotients}

Suppose $X$ is a set and $\sim$ an equivalence
relation on $X$. We have $X / \sim = \{[x]_\sim
\mid x \in X\}$ and have quotient map $q : X \to X
/ \sim$, $x \mapsto [x]_\sim$. Clearly $q$ is
surjective. Suppose now $Y$ is also a set and $f :
X \to Y$. Assume $f$ \emph{respects} $\sim$, i.e.
$x \sim y \implies f(x) = f(y)$.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/31f24b44627d11ed.png}
\end{center}
Then there is a unique function $\ol{f} : X / \sim
\to Y$ such that $f = \ol{f} \circ q$. Indeed,
must have $\forall x \in X$,
\[ \ol{f}([x]_\sim) = \ol{f}(q(x)) = f(x) \]
As $f$ respects $\sim$, this is well-defined:
\[ [x]_\sim = [y]_\sim \implies x \sim y \implies
f(x) = f(y) \]

\begin{example*}
    Suppose $G$ is a group, $H$ another group,
    $\theta : G \to H$ is a homomorphism. Let $K =
    \Ker \theta$, and define $\sim$ on $G$ by $g
    \sim h \iff g^{-1} h \in K$. Then $G / K = G /
    \sim$ and
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/434eace8627d11ed.png}
    \end{center}
    Can check $\ol{\theta}$ is a homomorphism and
    injective so isomorphism onto $\theta(G)$.
    This is the first isomorphism theorem.
\end{example*}

\begin{proposition}
    Let $(X, \tau)$ be a topological space and
    $\sim$ an equivalence relation on $X$. Let
    $\rho$ be the quotient topology on $X / \sim$.
    Suppose $f : X \to Y$ is a continuous function
    respecting $\sim$, where $(Y, \sigma)$ is a
    topological space. Then there is a unique
    continuous function $\ol{f} : X / \sim \to Y$
    such that $f = \ol{f} \circ q$, where $q : X
    \to X / \sim$ is the quotient map.
\end{proposition}

\begin{proof}
    Define $\ol{f} : X / \sim \to Y$ by
    \[ \ol{f}([x]_\sim) = f(x) \]
    This is well-defined:
    \[ [x]_\sim = [y]_\sim \implies x \sim y
    \implies f(x) = f(y) \]
    Clearly $\ol{f} \circ q = f$. Let $G \in
    \sigma$. Then
    \[ q^{-1}(\ol{f}^{-1}(G)) = (\ol{F} \circ
    q)^{-1} (G) = f^{-1}(G) \in \tau \]
    as $f$ continuous. So by definition of
    quotient topology, $\ol{f}^{-1}(G) \in \rho$.
    Hence $\ol{f}$ continuous. Finally, if $f = h
    \circ q$ for some $h : X / \sim \to Y$ then
    $\forall x \in X$, $h([x]_\sim) = h(q(x)) =
    f(x) = \ol{f}([x]_\sim)$. So $h = \ol{f}$.
\end{proof}

\begin{remark*}
    This is what makes quotients useful. For
    example recall torus $T = \RR^2 / \sim$ for
    appropriate relation $\sim$.  Hopefully $T$ is
    homeomorphic to genuine torus as a subspace in
    $\RR^3$. $T$ is nasty. $\RR^2$ is nice. So
    always work `upstairs' in $\RR^2$ rather then
    `downstairs' in $T$. For example if you want
    to think about a continuous function on $T$ -
    instead think about an appropriate continuous
    function on $\RR^2$ respecting $\sim$.
\end{remark*}

\begin{example*}
    Recall we had $\RR$ with usual topology, $x
    \sim y$ if and only if $x - y \in \ZZ$ and $S'
    = \{x \in \RR^2 \mid \|x\| = 1\}$ with subspace
    topology inherited from Euclidean topology on
    $\RR^2$. We claimed $\RR / \sim$ is
    homeomorphic to $S'$. Define $f : \RR \to S'$
    by
    \[ f(x) = (\sin 2\pi x, \cos 2\pi x) \]
    Clearly $f$ is a continuous surjection, and it
    respects $\sim$. By proposition 40 there is a
    unique continuous $\ol{f} : \RR / \sim \to S'$
    with $\ol{f} \circ q = f$. Clearly $\ol{f}$ is
    a continuous bijection (for injectivity, note
    each $x \in S'$ is $f(a)$ for a unique $a \in
    [0, 1)$ and each $b \in \RR$ has $b \sim a$
    for a unique $a \in [0, 1)$) \\
    Now $\RR / \sim = q([0, 1])$ is a continuous
    image of a compact set so is compact. And $S'$
    is Hausdorff. Any continuous bijection from a
    compact space to a Hausdorff space is a
    homeomorphism. So done: $\ol{f}$ is a
    homeomorphism from $\RR / \sim$ to $S'$.
\end{example*}

\subsection{Connectedness}

Recall the \emph{Intermediate Value Theorem}: \\
If $f : [a, b] \to \RR$ is continuous, and without
loss of generality $f(a) < f(b)$, then
\[ [f(a), f(b)] \subset f([a, b]) \]
Moreover, if $x, d \in f([a, b])$ with $c < d$
then $[c, d] \subset f([a, b])$. Doesn't work more
generally, for example if we replace by $[a, b]$
by $[-1, 0) \cup (0, 1] = X$. Define: $X \to \RR$
by
\[ f(x) = \begin{cases}
    1 & x < 0 \\
    0 & x > 0
\end{cases} \]
Then $f$ is continuous on $X$, $0 \in f(X)$, $1
\in f(X)$ but for example $\half \not\in f(X)$ so
$[0, 1] \not\subset f(X)$. \\
What's gone wrong? $[-1, 0) \cup (0, 1]$ is
`disconnected'.

\begin{flashcard}[ConnectedTopDefn]
    \begin{definition*}
        A topological space $X$ is
        \emph{disconnected} if \cloze{there exist
        disjoint, non-empty open sets $U, V$ with
        $X = U \cup V$.} \\
        We say $X$ is
        \emph{connected} if \cloze{$X$ is not
        disconnected.}
    \end{definition*}
\end{flashcard}

\subsubsection*{Remarks}

\begin{enumerate}[(1)]
    \item Recall the $U \subset X$ is closed if
        and only if its complement is open. So
        \[ X \text{ disconnected} \iff \text{if
        there exist disjoint, non-empty closed
        sets $U, V$ with $X = U \cup V$.} \]
        \[ X \text{ connected} \iff \text{the only
        subsets of $X$ that are both open and
        closed are $\emptyset, X$} \]
\end{enumerate}
