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

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

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

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). Then:

Characteristic property. For every space $W$ and every function $k : Y \to W$ , $k is continuous ⟺ k \circ q is continuous .$
Factorisation. Let $f : X \to W$ be continuous and constant on the fibres of $q$ , that is $q (x) = q (x^{'})$ implies $f (x) = f (x^{'})$ . Then there is exactly one function $\overset{ˉ}{f} : Y \to W$ with $\overset{ˉ}{f} \circ q = f$ , and it is continuous.
Composites. If $q : X \to Y$ and $p : Y \to Z$ are quotient maps then $p \circ q : X \to Z$ is a quotient map.

Facts & Assumptions

Given: A quotient map $q : X \to Y$ , a space $W$ , a function $k : Y \to W$ , a continuous $f : X \to W$ constant on the fibres of $q$ , and a further quotient map $p : Y \to Z$ .

[A1]

$q$ is a surjection and $V \subseteq Y$ is open exactly when $q^{- 1} [V]$ is open in $X$ ; the topology of $Y$ is the final topology of the one-element family $(q)$ (The quotient topology of a surjection, quotient maps, saturated sets, and the quotient of a space by an equivalence relation with its canonical projection, Injection, surjection, bijection).

[A2]

The same holds for $p : Y \to Z$ (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 final topology of a family $(g_{i})$ , a map $k$ out of the space is continuous exactly when every $k \circ g_{i}$ is (Characteristic properties: a map into a space with the initial topology is continuous iff every composite with the defining family is, a map out of a space with the final topology is continuous iff every composite with the defining family is, and the two topologies are respectively the coarsest and the finest making that family continuous, claim 4; Continuity of a map of topological spaces at a point and globally).

[L2]

A map of spaces is continuous exactly when preimages of open sets are open (For a map of spaces the following agree: continuity at every point, preimages of open sets open, preimages of closed sets closed, preimages of subbasic open sets open, and $f (\overline{A}) \subseteq \overline{f (A)}$ , clause (b)).

[L3]

Preimages compose: $(u \circ v)^{- 1} [T] = v^{- 1} [u^{- 1} [T]]$ ; a composite of surjections is a surjection (Injection, surjection, bijection).

[L4]

A topology on a set is a family of subsets of it (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

Proof

technique · direct

1.1

By [A1] the topology of $Y$ is a final topology of the one-element family $(q)$ , so [L1] gives claim 1 at once.

A1L1

1.2

Define $\overset{ˉ}{f} : = {(y, w) : there is x \in X with q (x) = y and f (x) = w}$ . It is total on $Y$ , since $q$ is surjective by [A1]; and it is single valued, since $q (x) = q (x^{'})$ implies $f (x) = f (x^{'})$ by hypothesis. So $\overset{ˉ}{f}$ is a function $Y \to W$ with $\overset{ˉ}{f} \circ q = f$ .

givenA1

1.3

Any $g : Y \to W$ with $g \circ q = f$ equals $\overset{ˉ}{f}$ : for $y \in Y$ pick $x$ with $q (x) = y$ , available by surjectivity, and then $g (y) = f (x) = \overset{ˉ}{f} (y)$ .

givenA1

1.4

$p \circ q$ is a surjection, being a composite of surjections.

A1A2L3

1.5

For $V \subseteq Z$ : $(p \circ q)^{- 1} [V] = q^{- 1} [p^{- 1} [V]]$ by [L3].

2.1

By step 1.2 the map $\overset{ˉ}{f}$ exists with $\overset{ˉ}{f} \circ q = f$ continuous, so $\overset{ˉ}{f}$ is continuous by step 1.1; with step 1.3 this is claim 2.

step 1.1step 1.2step 1.3

2.2

Let $V \subseteq Z$ . If $V$ is open in $Z$ then $p^{- 1} [V]$ is open in $Y$ by [L2] and [A2], hence $q^{- 1} [p^{- 1} [V]]$ is open in $X$ by [A1]; by step 1.5 that set is $(p \circ q)^{- 1} [V]$ .

step 1.5A1A2L2

2.3

Conversely, if $(p \circ q)^{- 1} [V]$ is open in $X$ , then $q^{- 1} [p^{- 1} [V]]$ is open in $X$ by step 1.5, so $p^{- 1} [V]$ is open in $Y$ by [A1], so $V$ is open in $Z$ by [A2].

step 1.5A1A2

3.1

By steps 1.4, 2.2 and 2.3 the map $p \circ q$ is a surjection for which $V$ is open in $Z$ exactly when $(p \circ q)^{- 1} [V]$ is open in $X$ ; that is claim 3. With steps 1.1 and 2.1 all three claims are proved.

step 1.1step 1.4step 2.1step 2.2step 2.3A1L4∎

Remarks

Claim 2 is how every quotient space in this library is identified. To produce a continuous map out of an identification space one never works with equivalence classes directly: one writes a continuous map on the original space, checks that it does not distinguish identified points, and quotes claim

Both examples of gluing on the companion page are exactly this move.

Uniqueness in claim 2 uses only surjectivity, and continuity of $\overset{ˉ}{f}$ uses only claim 1. Neither uses a choice principle: step 1.3 picks a preimage for a single $y$ inside a proof of an equation, which is an instance of existential instantiation and not a selection over an index set.
Claim 3 has no analogue for open maps or for closed maps in the direction one wants here. A composite of quotient maps is a quotient map, and that is what makes iterated identifications well behaved; whether a product of quotient maps is a quotient map is a different question, and it is not settled at this point in the reading order (see What the theory of these constructions still owes at this point in the reading order: preservation of quotient maps under products, separation beyond Hausdorff, and the invariants that tell the glued spaces apart).

Depends on

Used by

…and 12 more results.

Dependency tree · two levels

16 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

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

For a quotient map q:X→Yq:X→Y, a map out of YY is continuous iff its composite with qq is; a continuous map on XX constant on the fibres of qq factors uniquely through qq; and a composite of quotient maps is a quotient map