Quotient maps and final topology

Try

Imagine identifying several input points as one output point. A subset of the output should count as open exactly when all of its representatives form an open subset upstairs. This is the quotient-topology rule.

For a discrete six-point set partitioned into three classes, the quotient has three points and is discrete: every union of classes is open upstairs. Infinite or nondiscrete examples can behave very differently, so the finite case is a calculation rather than a general separation theorem.

Predict

Predict whether checking only one representative from each class is enough. It is not. The inverse image of an output subset contains whole fibers, and that complete set is what must be open.

Predict also whether every continuous surjection is a quotient map onto its stated target. An indiscrete target can make continuity artificially easy while omitting opens forced by the quotient rule.

Experiment

Compare the two implications.

Map property	Openness statement
Continuous	Target-open implies preimage-open
Quotient surjection	Target-open if and only if preimage-open
Open surjection	Source-open images are target-open
Closed surjection	Source-closed images are target-closed

A continuous open or continuous closed surjection is quotient, but quotient maps need not send arbitrary opens or closed sets to opens or closed sets.

Understand

Let q:X→Y be a surjection of sets, with X topological. Define the quotient topology on Y by V open exactly when q⁻¹(V) is open in X. Preimage identities verify the topology axioms. The topology is the finest one making q continuous: any target-open set under any topology making q continuous must have an open preimage and is therefore included in this quotient topology.

The mapping-out property is its main tool. A map h:Y→Z is continuous exactly when h∘q is continuous. Necessity follows by composition. Conversely, for every open W⊆Z, the preimage q⁻¹(h⁻¹(W)) is open. The quotient condition forces h⁻¹(W) open, proving continuity.

A continuous open surjection is quotient because if q⁻¹(V) is open, its image is open and equals V by surjectivity. For a continuous closed surjection, use the same argument on closed subsets and complements.

The identity from a two-point discrete space onto the same set with indiscrete topology is continuous and surjective but not quotient for that target topology. A singleton has open preimage while the target singleton is not open. Thus a continuous surjection only gives one direction of the quotient equivalence.

Initial constructions, such as products, control maps into a space through observations. Quotient constructions control maps out of a space through representative-independent functions. Confusing those directions can reverse the claimed finest or coarsest topology.

Apply

Take a map f on X that assigns the same value to every member of each fiber. Define h on quotient points by choosing a representative and applying f. Explain first why the value does not depend on the choice. Then, if f is continuous, use the mapping-out property to prove h continuous.

Repair “The quotient topology is coarser than every possible target topology.” Its defining extremal property is the finest target topology making this particular surjection continuous. Changing the map or its source changes that comparison.

Continue

The next lesson turns equivalence relations into concrete gluing constructions and studies saturated sets. Optional depth: prove that the composition of two quotient maps is quotient by applying the open-preimage equivalence twice. Do not assume the same is true for arbitrary restrictions without checking the needed hypotheses.