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

1.1givenL1

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.

1.2given

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

1.3givenA2L2

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.

1.4givenA2L2L3

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].

1.5givenL1L4

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.

2.1step 1.1step 1.3A1

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].

2.2step 1.1step 1.4A1

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

2.3step 1.1step 1.2step 1.5A1

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

3.1step 2.1step 2.2step 2.3∎

Steps 2.1, 2.2 and 2.3 establish the three clauses.

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.

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

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources