False statementConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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FALSE: every quotient map is an open map

Statement

False claim: every quotient map $q : X \to Y$ (The quotient topology of a surjection, quotient maps, saturated sets, and the quotient of a space by an equivalence relation with its canonical projection) is an open map, that is, carries open subsets of $X$ to open subsets of $Y$ (Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological).

The converse implication is the one that holds: a continuous open surjection is a quotient map (A continuous open surjection, a continuous closed surjection, and a continuous surjection admitting a continuous section are all quotient maps, clause 1). The claim above fails for the cheapest identification there is, collapsing a closed interval of $\mathbb{R}$ to a point. Take $X := \mathbb{R}$ with its usual topology (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y) = |x-y|$ is a metric, its open balls are the intervals $(x-r, x+r)$ , and it is unbounded, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not), $B := [0,1]$ (Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length), and let

$q : \mathbb{R} \to \mathbb{R}/B$

be the canonical projection of the quotient that identifies all of $B$ to one point and identifies nothing else (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 $q$ is a quotient map by construction, and $q[(-1,\ 1/2)]$ is not open.

Facts & Assumptions

Given: $\mathbb{R}$ with its usual topology; $B = [0,1]$ ; the equivalence relation on $\mathbb{R}$ whose classes are $B$ and the singletons $\{t\}$ for $t \notin B$ ; the quotient $\mathbb{R}/B$ with the quotient topology and its canonical projection $q$ ; and the set $U := (-1,\ 1/2)$ .

[A1]

$q$ is a surjection, the topology of $\mathbb{R}/B$ is the quotient topology of $q$ , and consequently $V \subseteq \mathbb{R}/B$ is open exactly when $q^{-1}[V]$ is open in $\mathbb{R}$ ; so $q$ is 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).

[A2]

$A \subseteq \mathbb{R}$ is saturated for $q$ exactly when $A \cap B$ is $\varnothing$ or $B$ , and $q^{-1}[q[A]]$ is the saturation of $A$ (The quotient topology of a surjection, quotient maps, saturated sets, and the quotient of a space by an equivalence relation with its canonical projection).

[A3]

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

[L1]

$(a,b) = \{t : a < t < b\}$ and $[0,1] = \{t : 0 \le t \le 1\}$ ; $U \subseteq \mathbb{R}$ is open in the usual topology exactly when every point of $U$ has a bounded open interval around it inside $U$ , and every bounded open interval is open (Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length, The absolute value makes $\mathbb{R}$ a metric space: $d(x,y) = |x-y|$ is a metric, its open balls are the intervals $(x-r, x+r)$ , and it is unbounded, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not).

[L2]

A continuous open surjection is a quotient map (A continuous open surjection, a continuous closed surjection, and a continuous surjection admitting a continuous section are all quotient maps, clause 1).

[L3]

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

Refutation

technique · direct

1.1

$U = (-1,\ 1/2)$ is open in $\mathbb{R}$ , being a bounded open interval.

1.2

$U \cap B = [0,\ 1/2)$ , which is neither $\varnothing$ , since it contains $0$ , nor $B$ , since $1 \in B$ and $1 \notin [0,1/2)$ ; so $U$ is not saturated.

A2L1

1.3

$(-1,\ 1]$ is not open in $\mathbb{R}$ : for every $r > 0$ the interval $(1-r,\ 1+r)$ contains $1 + r/2$ , which satisfies $1 + r/2 > 1$ and so lies outside $(-1,1]$ ; hence no bounded open interval around $1$ lies inside $(-1,1]$ .

2.1

$q^{-1}[q[U]] = U \cup B = (-1,\ 1]$ : the saturation of $U$ adds to $U$ exactly the class of each of its points, and the only non-singleton class meeting $U$ is $B$ itself, by step 1.2.

step 1.2A2L1

3.1

By step 2.1 and step 1.3 the set $q^{-1}[q[U]]$ is not open in $\mathbb{R}$ , so $q[U]$ is not open in $\mathbb{R}/B$ by [A1].

step 2.1step 1.3A1L3

4.1

By [A1] the map $q$ is a quotient map, and by step 1.1 and step 3.1 it carries the open set $U$ to a set that is not open; so $q$ is not an open map by [A3], and the claim is false.

step 1.1step 3.1A1A3L2∎

Remarks

The obstruction is saturation, and it is the general one. A quotient map 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), and collapsing a set with nonempty interior destroys that: an open set that meets $B$ without containing it acquires the whole of $B$ in its saturation, and $B$ has boundary points.
Closedness fails independently. The map above happens to be closed, since the saturation of a closed set $F$ is $F$ or $F \cup B$ , both closed; so this witness separates "quotient map" from "open map" only. A quotient map that is neither open nor closed needs a different construction, and one is worked on the companion page as On $A = ([0,\infty) \times \mathbb{R}) \cup (\mathbb{R} \times \{0\})$ the first projection is a quotient map, by the section $x \mapsto (x,0)$ , and is neither open nor closed ↗.
Why the converse direction is nevertheless useful. Most quotients that are identified with a known space in practice are open quotient maps, because their equivalence relation comes from translating by the elements of a group, and translation is a homeomorphism; both the circle and the torus on the companion page are of that kind.

Depends on

Used by

On A = ([0,∞) × ℝ) ∪ (ℝ × {0}) the first projection is a quotient map, by the section x ↦ (x,0), and is neither open nor closed Counterexample

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 78 results over 13 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Quotient space (topology) (Wikipedia) (standard reference, not scraped)
Open and closed maps (Wikipedia) (standard reference, not scraped)