% 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}