LemmaStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-04 (gpt-5.6-sol-codex-subscription)

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.

A continuous open surjection, a continuous closed surjection, and a continuous surjection admitting a continuous section are all quotient maps

Statement

Let $X$ and $Y$ be topological spaces and let $q : X \to Y$ be continuous (Continuity of a map of topological spaces at a point and globally). Each of the following three conditions makes $q$ 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).

$q$ is a surjection and an open map (Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological).
$q$ is a surjection and a closed map.
$q$ admits a continuous section: a continuous $s : Y \to X$ with $q \circ s = {i d}_{Y}$ . (Surjectivity of $q$ is then automatic and need not be assumed.)

Neither clause 1 nor clause 2 is necessary: a quotient map need be neither open nor closed. A witness that is a quotient map by clause 3 while failing clauses 1 and 2 is worked on the companion page, and is named in the remarks below.

Facts & Assumptions

Given: Topological spaces $X$ and $Y$ , a continuous map $q : X \to Y$ , a subset $V \subseteq Y$ , and, where the clause requires it, a continuous $s : Y \to X$ with $q \circ s = {i d}_{Y}$ .

[A1]

$q$ is a quotient map when it is a surjection and, for every $V \subseteq Y$ , $V$ is open in $Y$ exactly when $q^{- 1} [V]$ is open in $X$ (The quotient topology of a surjection, quotient maps, saturated sets, and the quotient of a space by an equivalence relation with its canonical projection).

[A2]

$q$ is an open map when images of open sets are open, and a closed map when images of closed sets are closed (Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological).

[L1]

A map is continuous exactly when preimages of open sets are open, and exactly when preimages of closed sets are closed (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 $f (\overline{A}) \subseteq \overline{f (A)}$ , clauses (b) and (c); Continuity of a map of topological spaces at a point and globally).

[L2]

If $q$ is surjective then $q [q^{- 1} [V]] = V$ for every $V \subseteq Y$ ; and $q^{- 1} [Y ∖ V] = X ∖ q^{- 1} [V]$ (Injection, surjection, bijection).

[L3]

A subset of a space is closed exactly when its complement is open (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

[L4]

$(u \circ v)^{- 1} [T] = v^{- 1} [u^{- 1} [T]]$ for composable functions (Injection, surjection, bijection).

Proof

technique · direct

1.1

If $V$ is open in $Y$ then $q^{- 1} [V]$ is open in $X$ , by continuity of $q$ and [L1]; this half of the quotient condition holds under all three hypotheses.

givenL1

1.2

Assume clause 3 and let $y \in Y$ ; then $y = q (s (y))$ , so $q$ is surjective.

given

1.3

Assume clause 1 and that $q^{- 1} [V]$ is open in $X$ . Then $q [q^{- 1} [V]] = V$ by [L2], and $q [q^{- 1} [V]]$ is open in $Y$ by [A2]; so $V$ is open.

givenA2L2

1.4

Assume clause 2 and that $q^{- 1} [V]$ is open in $X$ . Then $X ∖ q^{- 1} [V] = q^{- 1} [Y ∖ V]$ by [L2] and is closed by [L3], so $q [q^{- 1} [Y ∖ V]] = Y ∖ V$ is closed by [A2] and [L2]; hence $V$ is open by [L3].

givenA2L2L3

1.5

Assume clause 3 and that $q^{- 1} [V]$ is open in $X$ . Then $s^{- 1} [q^{- 1} [V]] = (q \circ s)^{- 1} [V] = V$ by [L4] and $q \circ s = {i d}_{Y}$ , and $s^{- 1} [q^{- 1} [V]]$ is open in $Y$ by continuity of $s$ and [L1]; so $V$ is open.

givenL1L4

2.1

Under clause 1 the map $q$ is a surjection by hypothesis and satisfies both halves of the quotient condition, by steps 1.1 and 1.3; so it is a quotient map by [A1].

step 1.1step 1.3A1

2.2

Under clause 2 the same holds by steps 1.1 and 1.4.

step 1.1step 1.4A1

2.3

Under clause 3 the map $q$ is a surjection by step 1.2 and satisfies both halves by steps 1.1 and 1.5.

step 1.1step 1.2step 1.5A1

3.1

Steps 2.1, 2.2 and 2.3 establish the three clauses.

step 2.1step 2.2step 2.3∎

Remarks

The three clauses are genuinely different, and neither clause 1 nor clause 2 reverses. The canonical projection of an identification space need be neither open nor closed, and the companion page's On $A = ([0, \infty) \times R) \cup (R \times {0})$ the first projection is a quotient map, by the section $x \mapsto (x, 0)$ , and is neither open nor closed ↗ is a quotient map that fails both conditions while satisfying clause 3. In the other direction, being an open map and being a closed map are unrelated notions, as Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological records with a two-point witness.
Clause 3 is the cheapest of the three in practice. Exhibiting a continuous right inverse is a one-line construction whenever one is available, and it requires no computation with images at all; clauses 1 and 2 require knowing what $q$ does to every open, respectively every closed, subset of $X$ .
Where openness usually comes from. For a quotient by an equivalence relation, $q$ is open exactly when the saturation of every open set is open (The quotient topology of a surjection, quotient maps, saturated sets, and the quotient of a space by an equivalence relation with its canonical projection), since $q [U]$ is open in the quotient exactly when $q^{- 1} [q [U]]$ is open in $X$ . That criterion is what the group-like quotients on the companion page verify, translation of an open set by a group element being a homeomorphism there.

Depends on

Used by

Dependency tree · two levels

17 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)
Open and closed maps (Wikipedia) (standard reference, not scraped)
Section (category theory) (Wikipedia) (standard reference, not scraped)