TheoremStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck 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

$x \sim_q x' \quad :\Longleftrightarrow \quad q(x) = q(x') .$

Then $\sim_q$ is an equivalence relation, and the induced map

$\bar q : X/\!\sim_q \;\longrightarrow\; Y, \qquad \bar q([x]) := q(x)$

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 $\pi$ (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 $\bar q \circ \pi = q$ .

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 $\pi : X \to Q$ and the quotient topology of $\pi$ , and the map $\bar q$ of the statement.

[A1]

$q$ is a surjection and $V \subseteq Y$ is open exactly when $q^{-1}[V]$ is open in $X$ ; $\pi$ is a surjection and $V \subseteq Q$ is open exactly when $\pi^{-1}[V]$ is open in $X$ ; both $q$ and $\pi$ 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 $\bar f$ with $\bar f \circ r = f$ , 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 $\pi$ : if $\pi(x) = \pi(x')$ then $x \sim_q x'$ , that is $q(x) = q(x')$ .

1.3

$\pi$ is constant on the fibres of $q$ : if $q(x) = q(x')$ then $x \sim_q x'$ , so $[x] = [x']$ and $\pi(x) = \pi(x')$ .

2.1

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

step 1.2A1L1

2.2

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

step 1.3A1L1

3.1

$\bar q \circ \bar\pi = \mathrm{id}_Y$ : composing with the surjection $q$ gives $\bar q \circ \bar\pi \circ q = \bar q \circ \pi = q = \mathrm{id}_Y \circ q$ , and a surjection may be cancelled on the right.

step 2.1step 2.2A1

3.2

$\bar\pi \circ \bar q = \mathrm{id}_Q$ : composing with the surjection $\pi$ gives $\bar\pi \circ \bar q \circ \pi = \bar\pi \circ q = \pi = \mathrm{id}_Q \circ \pi$ , 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 $\bar q$ and $\bar\pi$ are mutually inverse bijections, and both are continuous by steps 2.1 and 2.2; so $\bar q$ is a homeomorphism with inverse $\bar\pi$ , and $\bar q \circ \pi = q$ 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 $\bar q$ 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 $\pi$ , 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.

Depends on

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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 \to Yq:X→Y induces a homeomorphism from XXX modulo the relation "qqq agrees" onto YYY, so up to homeomorphism the quotient maps out of XXX are exactly the canonical projections