= Quotient topology
{wiki}
The <quotient topology> turns an instruction to glue points together into a precise <topological space>. Start with a <topological space> $(X,\tau_X)$ and an <equivalence relation> $\sim$. The <quotient set> $X/{\sim}$ consists of <equivalence classes>, and its <canonical projection of a quotient set> is
$$
q:X\longrightarrow X/{\sim},\qquad q(x)=[x].
$$
The <quotient topology> is defined by the rule using <preimages> of <open sets>:
$$
\tau_q=\{U\subseteq X/{\sim}:q^{-1}(U)\in\tau_X\}.
$$
Thus a <subset> of the <quotient space> is <open> if and only if its entire <preimage> is <open> in $X$. Each <fiber> $q^{-1}(\{[x]\})$ is the whole <equivalence class> $[x]$, so the <preimage> includes every point glued to any of its members.<Point-Set Topology by Romyar Sharifi>[§2.3, Definitions 2.3.1 and 2.3.7 and Lemma 2.3.6]{ref}
This is the finest <topology> making $q$ a <continuous map>: every <topology> with that property can contain only <subsets> whose <preimages> are <open>. It really is a <topology>, because <preimages> preserve the <empty set>, the whole <set>, arbitrary <unions>, and finite <intersections>. The gluing example below shows why checking a complete <preimage> matters, especially at the seam.
The <quotient topology> also makes <continuous maps> out of a glued space easy to construct. A <continuous map> $f:X\to Z$ descends to a unique <continuous map> $\bar f:X/{\sim}\to Z$ exactly when it is constant on every <fiber> of $q$, meaning $x\sim y\Rightarrow f(x)=f(y)$. Define $\bar f([x])=f(x)$; the <equivalence class representative> cannot affect the answer. For every <open set> $V\subseteq Z$,
$$
q^{-1}\bigl(\bar f^{-1}(V)\bigr)=f^{-1}(V).
$$
The right side is <open>, so the <quotient topology> makes $\bar f^{-1}(V)$ <open>. Conversely, if $f=\bar f\circ q$ for a <continuous map> $\bar f$, then <composition of continuous functions> makes $f$ <continuous>, and equal <equivalence classes> force equal values. This is the <universal property of the quotient topology>. <Topology and Groups/3.02 Quotient topology: continuous maps>{ref}
A <surjective> <continuous map> is not automatically a <quotient map>. On the same two-point <set> $D=\{a,b\}$, take the <identity map> from the <discrete topology> to the <indiscrete topology>. It is <continuous> and <surjective>, but $\{a\}$ is not <open> in the target even though its <preimage> is <open> in the domain. The <quotient topology> for this <identity map> would be the <discrete topology>. Thus <continuity> supplies only one direction of the rule using <preimages> of <open sets>; a <quotient map> requires both.
Back to article page
