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

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 $d_{\mathrm{\infty }}$ 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

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

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 $q:={\pi }_0\upharpoonright A:A\to \mathbb{R}$ be the first projection. Then:

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

• What makes $q[U]$ fail to be open is the shape of $A$ near the negative axis. The set $U=A\cap (\mathbb{R}\times (0,\mathrm{\infty }))$ is open in $A$ for the trivial reason that $\mathbb{R}\times (0,\mathrm{\infty })$ 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 $y\mathrm{\ne }0$, 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 $q[U]=[0,\mathrm{\infty })$, 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 $(\mathbb{R}\mathrm{/}\mathbb{Z})\times (\mathbb{R}\mathrm{/}\mathbb{Z})$ 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.