The quotient topology · Topology

Intuition

Every construction so far has taken spaces apart: subspaces, products, components. Gluing goes the other way. Decide which points are to become the same, take the set of groups, and there is exactly one sensible topology on it: a set of groups is open when the points it is made of form an open set below. That is the quotient topology, and it is the finest topology making the gluing map continuous.

Fold a sheet of paper and press two edges together. The result is a new surface whose points are the pairs that were pressed together, and a region on it is open exactly when its unfolding was open on the sheet.

The gluing map $q$ sends each point to the group it belongs to. A subset of the new space is declared open exactly when its preimage is open below, so openness upstairs is what decides openness downstairs — and $q$ comes out continuous by construction.

One topology, decided from below

Let $\sim$ be an equivalence relation on $X$ , write $X/\!\sim$ for the set of classes and $q:X\to X/\!\sim$ for the map sending a point to its class. The quotient topology declares open exactly when is open in . A subset of is called saturated when it is a union of whole classes; the open sets downstairs are the images of the saturated open sets upstairs.

U\ \text{open in}\ X/\!\sim\ \iff q^{-1}(U)\ \text{open in}\ X

What the definition forces

$q$ is continuous and onto, and the quotient topology is the finest topology on the set of classes for which $q$ is continuous: any more open sets and one of the preimages would fail to be open.
A quotient map need not be open or closed. Gluing the ends of $[0,1]$ sends $[0,\tfrac12)$ to a set whose preimage is , which is not open.

Maps out of a quotient

f:X/\!\sim\ \to Y\ \text{continuous}\iff f\circ q\ \text{continuous}

One direction is free: a composition of continuous maps is continuous, and the gluing map is continuous. For the other, suppose the composition is continuous and take an open set in the target. Its preimage under the composition is open, and that preimage is exactly the preimage under the gluing map of the preimage under the map being tested. By the definition of the quotient topology, a subset of the glued space whose preimage is open is open, so the preimage under the map being tested is open, which is continuity.

Proof steps

f\ \text{continuous}\implies f\circ q\ \text{continuous}

A composition of continuous maps is continuous, and the gluing map is one.

V\ \text{open in}\ Y,\quad (f\circ q)^{-1}(V)\ \text{open in}\ X

Suppose the composition is continuous and take an open set in the target.

(f\circ q)^{-1}(V)=q^{-1}\big(f^{-1}(V)\big)

Preimages compose the other way round, which is the whole of the bookkeeping.

q^{-1}\big(f^{-1}(V)\big)\ \text{open}\implies f^{-1}(V)\ \text{open}

That is exactly what the quotient topology says about a subset of the glued space.

f\ \text{continuous}

Every open set of the target has open preimage, which is continuity.

Applications

Practice

Open Downstairs Means Open Upstairs

A set of classes is open exactly when the points making it up form an open set.

U\ \text{open}\iff q^{-1}(U)\ \text{open}

Try it

When is a subset $U$ of a quotient space open?