3.9 Quotient Topologies
Definition 3.9.1 (Saturated).label Let $X,Y$ be sets, $f:X\to Y$ surjective and $E\suf X$ then $E$ is saturated with respect to $f$ if $E=f\inv (f(E))$.
Definition 3.9.2 (Quotient Map).label Let $X$ be a topological space, $Y$ a set and $\pi:X\to Y$ a surjection then the following are equivalent:
- (1)
$Y$ has the final topology generated by $\pi$
- (2)
$\pi\in C(X;Y)$ and for any saturated open set $U\suf X$, $\pi(U)$ is open
- (U)
For each topological space $Z$ and function $f:Y\to Z$, $f\in C(Y;Z)\iff f\circ\pi\in C(X;Z)$
If the above holds, then $\pi$ is a quotient map.
Proof. Suppose (1) holds then (2) follows from Definition 3.1.8 and Definition 3.9.1. Suppose (2) holds then by continuity of $\pi$, $\pi\inv(V)$ is open whenever $V\suf Y$ is open. Observe $V=\pi(\pi\inv(V))$ for any $V\suf Y$ since $\pi$ is surjective hence $\pi\inv(V)=\pi\inv(\pi(\pi\inv(V)))$ so if $\pi\inv(V)$ is open then by assumption $\pi(\pi\inv(V))$ is open, showing (1). Suppose (1) holds then $f\circ \pi$ is continuous whenever $f$ is continuous. If $f\circ\pi$ is continuous then $f$ is continuous by Definition 3.1.8(U’), showing (U). Suppose (U) holds then Definition 3.1.8(4) shows (1).$\square$
Lemma 3.9.3.label Let $\pi:X\to Y$ be a surjective, continuous open map between topological spaces and $n\in\N^{+}$ then
- (1)
$\pi^{n}$ is an open surjetion, hence a quotient map
- (2)
For each topological space $Z$ and $f\in C(X^{n};Z)$ constant on $\pi$-fibers in each coordinate in the sense that for each $(x_{1},...,x_{n}).(x_{1}',...,x_{n}')\in X^{n}$
\begin{align*}\pi(x_{i})=\pi(x_{i}'),\fall i=1,...,n\implies f(x_{1},...,x_{n})=f(x_{1}',...,x_{n}')\end{align*}then there uniquely exists $\til{f}\in C(Y^{n};Z)$ such that $\til{f}\circ\pi^{n}=f$.
- (3)
Let $\mu\in C(X^{n};X)$ such that $\mu$ is compatible with the equivalence relation induced by $\pi$ in the sense that for each $(x_{1},...,x_{n}).(x_{1}',...,x_{n}')\in X^{n}$
\begin{align*}\pi(x_{i})=\pi(x_{i}'),\fall i=1,...,n\implies \pi(\mu(x_{1},...,x_{n}))=\pi(\mu(x_{1}',...,x_{n}'))\end{align*}then there uniquely exists $\til{\mu}\in C(Y^{n};Y)$ such that $\til{\mu}\circ\pi^{n}=\pi\circ \mu$.
Proof.
- (1)
The product of finitely many open maps is open and any open (or closed) surjection is a quotient map hence $\pi^{n}$ is a quotient map
- (2)
Define $\til{f}:Y^{n}\to Z$ by $\til{f}(\pi(x_{1}),...,\pi(x_{n}))\define f(x_{1},...,x_{n})$ then since $f$ is constant on $\pi$-fibers in each coordinate, $\til{f}$ is well-defined and unique. Continuity of $\til{f}$ comes from $\pi^{n}$ is a quotient map and $f$ is continuous.
- (3)
Apply (2) with $Z=Y$ and $f=\pi\circ\mu$.
$\square$
Definition 3.9.4 (Quotient Space).label Let $X$ be a topological space and $\sim$ an equivalence relation on $X$, then there exists $(\til{X},\pi)$ such that
- (1)
$\til{X}$ is a topological space with underlying set $X/\sim$
- (2)
$\pi\in C(X;\til{X})$
- (3)
$\pi$ is constant on each equivalence class of $\sim$
- (U)
For any pair $(Y,f)$ satisfying (1),(2), and (3), there exists a unique $\til{f}\in C(\til{X};Y)$ such that the following diagram commutes
\[\xymatrix{ X \ar@{->}[r]|-{f} \ar@{->}[d]|-{\pi} & Y \\ \tilde{X} \ar@{->}[ru]|-{\tilde{f}} & }\] - (5)
$\pi$ is a quotient map
The space $(\til{X},\pi)$ is the quotient of $X$ by $\sim$.
Proof. Let $\til{X}=X/\sim$ and $\pi$ maps $x$ to the equivalence class containing $x$. For each $U\suf \til{X}$, define $U$ to be open if and only if $\pi\inv(U)\suf X$ is open, then $(\til{X},\pi)$ satisfies (1),(2),(3), and (5). For (U), since $f$ is constatn on each equivalence class of $\sim$, $\til{f}:\til{X}\to Y$ exists. For any $U\suf Y$ open, $\til{f}\inv(U)=\pi(f\inv(U))$ is saturated with respect to $\pi$, so $\til{f}\inv(U)$ is open.$\square$
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