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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

Statement refuted

Refuted: that a quotient map is an open map (FALSE: every quotient map is an open map), and in the same breath that a quotient map is a closed map.

Witness. In $\mathbb{R}^2$ with its usual topology (For $n \ge 1$ the product topology on $n$ copies of the usual topology of $\mathbb{R}$ is the metric topology of $d_\infty$ on $\mathbb{R}^n$ , and hence also of $d_1$ and $d_2$ , so $\mathbb{R}^n$ as a product and $\mathbb{R}^n$ as a metric space are one space) let

$A \;:=\; \big([0,\infty) \times \mathbb{R}\big) \cup \big(\mathbb{R} \times \{0\}\big)$

carry the subspace topology (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace), and let $q := \pi_0 \restriction A : A \to \mathbb{R}$ be the first projection. Then:

$q$ is a quotient map, because $s(x) := (x,0)$ is a continuous section: $s$ is continuous, takes values in $A$ , and $q \circ s = \mathrm{id}_{\mathbb{R}}$ (A continuous open surjection, a continuous closed surjection, and a continuous surjection admitting a continuous section are all quotient maps, clause 3).
$q$ is not open: $U := A \cap (\mathbb{R} \times (0,\infty)) = [0,\infty) \times (0,\infty)$ is open in $A$ and $q[U] = [0,\infty)$ is not open in $\mathbb{R}$ .
$q$ is not closed: $H^{+} := \{\, (x, 1/x) : x > 0 \,\}$ is closed in $\mathbb{R}^2$ , hence closed in $A$ , and $q[H^{+}] = (0,\infty)$ is not closed in $\mathbb{R}$ .

Facts & Assumptions

Given: $\mathbb{R}^2$ with the usual topology; the set $A$ above with the subspace topology; $q = \pi_0 \restriction A$ ; the map $s(x) = (x,0)$ ; the sets $U$ and $H^{+}$ of the statement.

[A1]

$\mathbb{R}^2$ carries the product topology, which is the metric topology of $d_\infty$ ; the boxes $V \times W$ with $V,W$ open in $\mathbb{R}$ form a basis; projections are continuous (For $n \ge 1$ the product topology on $n$ copies of the usual topology of $\mathbb{R}$ is the metric topology of $d_\infty$ on $\mathbb{R}^n$ , and hence also of $d_1$ and $d_2$ , so $\mathbb{R}^n$ as a product and $\mathbb{R}^n$ as a metric space are one space, The product set $\prod_{i \in I} X_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space, A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not).

[A2]

The open sets of $A$ are the traces $O \cap A$ with $O$ open in $\mathbb{R}^2$ , and the closed sets of $A$ are the traces of the closed sets (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace, For $A \subseteq S \subseteq X$ the closure of $A$ in $S$ is $\overline{A}^{X} \cap S$ , while the interior only contains $\operatorname{int}^{X}(A) \cap S$ , with equality when $S$ is open; and a dense subset of $X$ traces to a dense subset of every open $S$ ).

[L1]

A continuous surjection admitting a continuous section 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 3; The quotient topology of a surjection, quotient maps, saturated sets, and the quotient of a space by an equivalence relation with its canonical projection).

[L2]

A map into a binary product is continuous exactly when both components are; restrictions and corestrictions to subspaces of continuous maps are continuous; composites of continuous maps are continuous (A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice, Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace, Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous).

[L3]

The repaired hyperbola result proves that $H := \{(x,y) : xy = 1\}$ is closed in $\mathbb{R}^2$ as the preimage of the closed singleton $\{1\}$ under continuous multiplication (The hyperbola $\{(x,y) : xy = 1\}$ is closed in $\mathbb{R}^2$ and its image under the first projection is $\mathbb{R} \setminus \{0\}$ , which is not closed); an intersection of two closed sets is closed.

[L4]

$[a,\infty) = \{t : a \le t\}$ and $(a,\infty) = \{t : a < t\}$ ; a subset of $\mathbb{R}$ is open exactly when each of its points has a bounded open interval around it inside the set, and $(a,\infty)$ 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).

[L5]

For every real $\eta > 0$ there is a natural $m \ge 1$ with $1/m < \eta$ , and $0 < u \le v$ gives $0 < 1/v \le 1/u$ (For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ , Inverses of positives are positive, and reciprocation reverses order).

[L6]

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

Counterexample

technique · direct

1.1

$s$ is continuous into $\mathbb{R}^2$ , its components being the identity and the constant $0$ ; its values $(x,0)$ lie in $A$ , so its corestriction $s : \mathbb{R} \to A$ is continuous by [L2]. And $q(s(x)) = x$ , so $q \circ s = \mathrm{id}_{\mathbb{R}}$ .

A1L2

1.2

$U = A \cap (\mathbb{R} \times (0,\infty)) = [0,\infty) \times (0,\infty)$ : a point of $A$ with positive second coordinate cannot lie on $\mathbb{R} \times \{0\}$ , so it lies in $[0,\infty)\times\mathbb{R}$ with second coordinate positive. And $\mathbb{R}\times(0,\infty)$ is open in $\mathbb{R}^2$ by [A1] and [L4], so $U$ is open in $A$ by [A2].

A1A2L4

1.3

$H^{+} = H \cap \big([0,\infty)\times\mathbb{R}\big)$ is closed in $\mathbb{R}^2$ , being an intersection of two closed sets, the second being the complement of the open set $(-\infty,0)\times\mathbb{R}$ ; here $H^{+} \subseteq H$ has first coordinate positive, since $xy = 1$ forbids $x = 0$ . So $H^{+}$ is closed in $A$ by [A2], being $H^{+} \cap A$ with $H^{+} \subseteq A$ .

A1A2L3L4

1.4

$[0,\infty)$ is not open in $\mathbb{R}$ : for every $r > 0$ the interval $(-r,r)$ contains $-r/2 < 0$ , so no bounded open interval around $0$ lies inside $[0,\infty)$ .

1.5

$(0,\infty)$ is not closed in $\mathbb{R}$ : its complement $(-\infty,0]$ is not open, since for every $r > 0$ the interval $(-r,r)$ contains $r/2 > 0$ .

2.1

$q$ is continuous, being a restriction of the continuous $\pi_0$ ; and it is surjective, since $q(s(x)) = x$ for every $x$ .

step 1.1A1L2

2.2

$q[U] = [0,\infty)$ : for $x \ge 0$ the point $(x,1)$ lies in $U$ , and every point of $U$ has first coordinate in $[0,\infty)$ .

step 1.2L4

2.3

$q[H^{+}] = (0,\infty)$ : for $x > 0$ the point $(x,1/x)$ lies in $H^{+}$ , and every point of $H^{+}$ has positive first coordinate.

step 1.3L5

3.1

By steps 1.1 and 2.1 with [L1], $q$ is a quotient map. This is claim 1.

step 1.1step 2.1L1

3.2

By steps 1.2, 2.2 and 1.4 the map $q$ carries the open set $U$ to a set that is not open, so $q$ is not open by [L6]. This is claim 2.

step 1.2step 2.2step 1.4L6

3.3

By steps 1.3, 2.3 and 1.5 the map $q$ carries the closed set $H^{+}$ to a set that is not closed, so $q$ is not closed by [L6]. This is claim 3.

step 1.3step 2.3step 1.5L6

4.1

Steps 3.1, 3.2 and 3.3 give the three claims, so a quotient map need be neither open nor closed, which refutes the claim.

step 3.1step 3.2step 3.3∎

Remarks

What makes $q[U]$ fail to be open is the shape of $A$ near the negative axis. The set $U = A \cap (\mathbb{R} \times (0,\infty))$ is open in $A$ for the trivial reason that $\mathbb{R} \times (0,\infty)$ is open in $\mathbb{R}^2$ ; the shape of $A$ has nothing to do with that. What the shape does is fix the image. A point $(x,0)$ with $x < 0$ has $A$ -neighbourhoods lying entirely on the horizontal axis, because $A$ contains no point with $x < 0$ and $y \ne 0$ , so the axis to the left of the origin is, from inside $A$ , a half-line with nothing attached above or below it. Consequently $U$ reaches no first coordinate below $0$ and $q[U] = [0,\infty)$ , which contains the boundary point $0$ without containing any neighbourhood of it.
All three clauses of A continuous open surjection, a continuous closed surjection, and a continuous surjection admitting a continuous section are all quotient maps are needed. This map satisfies clause 3 and neither of the other two, so the section clause is not redundant; conversely the collapse map of FALSE: every quotient map is an open map is closed and not open, and the projection $q \times q$ of The square with opposite edges identified is homeomorphic to the product $(\mathbb{R}/\mathbb{Z}) \times (\mathbb{R}/\mathbb{Z})$ is open. The three clauses cover genuinely different situations.
The failure of closedness is the hyperbola again. $H^{+}$ is one branch of the closed set of The hyperbola $\{(x,y) : xy = 1\}$ is closed in $\mathbb{R}^2$ and its image under the first projection is $\mathbb{R} \setminus \{0\}$ , which is not closed, cut out by intersecting with a closed half-plane; the image loses the point $0$ for exactly the same reason as there.

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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)
A quotient map which is neither open nor closed (UC Riverside Math 205A notes) (standard reference, not scraped)

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