TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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Every quotient map $q : X \to Y$ induces a homeomorphism from $X$ modulo the relation " $q$ agrees" onto $Y$ , so up to homeomorphism the quotient maps out of $X$ are exactly the canonical projections

Statement

Let $q : X \to Y$ be a quotient map (The quotient topology of a surjection, quotient maps, saturated sets, and the quotient of a space by an equivalence relation with its canonical projection) and define a relation on $X$ by

$x \sim_{q} x^{'} : ⟺ q (x) = q (x^{'}) .$

Then $\sim_{q}$ is an equivalence relation, and the induced map

$\overset{ˉ}{q} : X / ⁣ \sim_{q} ⟶ Y, \overset{ˉ}{q} ([x]) : = q (x)$

is a homeomorphism (Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological), where $X / ⁣ \sim_{q}$ carries the quotient topology of its canonical projection $π$ (The quotient topology of a surjection, quotient maps, saturated sets, and the quotient of a space by an equivalence relation with its canonical projection). Moreover $\overset{ˉ}{q} \circ π = q$ .

So every quotient map out of $X$ is, up to a homeomorphism of its target, the canonical projection of $X$ onto one of its identification spaces: the target of a quotient map carries no information beyond the partition of $X$ into fibres.

Facts & Assumptions

Given: A quotient map $q : X \to Y$ , the relation $\sim_{q}$ above, the quotient set $Q : = X / ⁣ \sim_{q}$ with its canonical projection $π : X \to Q$ and the quotient topology of $π$ , and the map $\overset{ˉ}{q}$ of the statement.

[A1]

$q$ is a surjection and $V \subseteq Y$ is open exactly when $q^{- 1} [V]$ is open in $X$ ; $π$ is a surjection and $V \subseteq Q$ is open exactly when $π^{- 1} [V]$ is open in $X$ ; both $q$ and $π$ are continuous (The quotient topology of a surjection, quotient maps, saturated sets, and the quotient of a space by an equivalence relation with its canonical projection, Continuity of a map of topological spaces at a point and globally).

[A2]

Equality is reflexive, symmetric and transitive, and $[x] = {x^{'} : q (x^{'}) = q (x)}$ (The quotient topology of a surjection, quotient maps, saturated sets, and the quotient of a space by an equivalence relation with its canonical projection).

[L1]

For a quotient map $r$ and a continuous map $f$ constant on the fibres of $r$ , there is exactly one $\overset{ˉ}{f}$ with $\overset{ˉ}{f} \circ r = f$ , and it is continuous; and a map out of the target of $r$ is continuous exactly when its composite with $r$ is (For a quotient map $q : X \to Y$ , a map out of $Y$ is continuous iff its composite with $q$ is; a continuous map on $X$ constant on the fibres of $q$ factors uniquely through $q$ ; and a composite of quotient maps is a quotient map, claims 1 and 2).

[L2]

A homeomorphism is a continuous bijection with continuous inverse; a bijection has a unique two-sided inverse (Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological, Injection, surjection, bijection).

[L3]

Homeomorphy is an equivalence relation on spaces (A continuous bijection is a homeomorphism iff it is open iff it is closed, and homeomorphy is an equivalence relation on spaces, claim 2).

Proof

technique · direct

1.1

$\sim_{q}$ is an equivalence relation, being the relation " $q$ takes the same value", and equality is reflexive, symmetric and transitive.

1.2

$q$ is constant on the fibres of $π$ : if $π (x) = π (x^{'})$ then $x \sim_{q} x^{'}$ , that is $q (x) = q (x^{'})$ .

1.3

$π$ is constant on the fibres of $q$ : if $q (x) = q (x^{'})$ then $x \sim_{q} x^{'}$ , so $[x] = [x^{'}]$ and $π (x) = π (x^{'})$ .

2.1

By step 1.2 and [L1] applied to the quotient map $π$ and the continuous map $q$ , there is exactly one $\overset{ˉ}{q} : Q \to Y$ with $\overset{ˉ}{q} \circ π = q$ , and $\overset{ˉ}{q}$ is continuous; it satisfies $\overset{ˉ}{q} ([x]) = q (x)$ .

step 1.2A1L1

2.2

By step 1.3 and [L1] applied to the quotient map $q$ and the continuous map $π$ , there is exactly one $\overset{ˉ}{π} : Y \to Q$ with $\overset{ˉ}{π} \circ q = π$ , and $\overset{ˉ}{π}$ is continuous.

step 1.3A1L1

3.1

$\overset{ˉ}{q} \circ \overset{ˉ}{π} = {i d}_{Y}$ : composing with the surjection $q$ gives $\overset{ˉ}{q} \circ \overset{ˉ}{π} \circ q = \overset{ˉ}{q} \circ π = q = {i d}_{Y} \circ q$ , and a surjection may be cancelled on the right.

step 2.1step 2.2A1

3.2

$\overset{ˉ}{π} \circ \overset{ˉ}{q} = {i d}_{Q}$ : composing with the surjection $π$ gives $\overset{ˉ}{π} \circ \overset{ˉ}{q} \circ π = \overset{ˉ}{π} \circ q = π = {i d}_{Q} \circ π$ , and a surjection may be cancelled on the right.

step 2.1step 2.2A1

4.1

By steps 3.1 and 3.2 the maps $\overset{ˉ}{q}$ and $\overset{ˉ}{π}$ are mutually inverse bijections, and both are continuous by steps 2.1 and 2.2; so $\overset{ˉ}{q}$ is a homeomorphism with inverse $\overset{ˉ}{π}$ , and $\overset{ˉ}{q} \circ π = q$ by step 2.1. With step 1.1 this proves the theorem.

step 1.1step 2.1step 2.2step 3.1step 3.2L2L3∎

Remarks

This is the topological analogue of the first isomorphism theorem. A surjective homomorphism of groups factors through the quotient by its kernel and induces an isomorphism; a quotient map of spaces factors through the quotient by the partition into its fibres and induces a homeomorphism. In both cases the content is that the target is determined by the equivalence relation the map induces on the source.
The hypothesis that $q$ is a quotient map is not decorative. For a mere continuous surjection the induced $\overset{ˉ}{q}$ is still a continuous bijection, built by claim 2 of For a quotient map $q : X \to Y$ , a map out of $Y$ is continuous iff its composite with $q$ is; a continuous map on $X$ constant on the fibres of $q$ factors uniquely through $q$ ; and a composite of quotient maps is a quotient map applied to $π$ , but its inverse need not be continuous: that is exactly the failure recorded at level 8 in FALSE: every continuous bijection of topological spaces is a homeomorphism. What the quotient hypothesis buys is step 2.2, which manufactures the inverse as a continuous map.
A practical consequence, used on the companion page. To identify an explicitly described gluing $X / ⁣ \sim$ with a known space $Z$ , it suffices to produce a quotient map $X \to Z$ whose fibres are exactly the classes of $\sim$ ; the theorem then supplies the homeomorphism, and no map between equivalence classes ever has to be written down.

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Sources

Quotient space (topology) (Wikipedia) (standard reference, not scraped)
Isomorphism theorems (Wikipedia) (standard reference, not scraped)
J. Munkres, Topology, 2nd ed., §22 (standard reference, not scraped)

Every quotient map q:X→Yq:X→Y induces a homeomorphism from XX modulo the relation "qq agrees" onto YY, so up to homeomorphism the quotient maps out of XX are exactly the canonical projections