I'm looking for an example of a quotient map $p : X \to Y$, with $X$ and $Y$ both locally compact Hausdorff spaces, such that there is a Hausdorff space $A$ for which $p \times 1_A : X \times A \to Y \times A$ is not a quotient map. Or, a proof that no such map exists.
I have tried a few common quotients of $\mathbb{R}$ and $[0,1]$ (for instance $\mathbb{R}/\mathbb{Z}$ where $\mathbb{Z}$ is identified to a point, or $\mathbb{R} \to S^1$, or $[0,1]/\lbrace \frac{1}{n}\rbrace$) and their products with $\mathbb{Q}$, but either the quotient wasn't locally compact Hausdorff or I couldn't figure out a saturated open set in $\mathbb{R} \times Q$ whose image is not open in the codomain.
More generally, it is enough for $A$ to be a sober space; this should give a counterexample to something with locales, for which soberness is sufficient. If anyone has suggestions or knows of an example, that would be greatly appreciated!
Let $Y=[0,1]$ be the unit interval and let $X$ be the topological sum of all convergent sequences in $Y$. Let $$p:X\rightarrow Y$$ be the obvious mapping which sends a sequence onto its image set in $Y$.
To be precise, let us say that a convergent sequence should consist of infinitely many distinct points and contain its limit. Thus $X$ is homeomorphic to a sum of copies of $\{1/n\mid n\in\mathbb{N}\}\cup\{0\}$. In particular, $X$ is locally compact metric and $Y$ is compact metric. The map $p$ is evidently continuous, and, since the continuity of maps out of $Y$ may be verified by their behaviour on convergent sequences, $p$ is a quotient mapping. In fact, it is not difficult to see that $p$ is hereditarily quotient (i.e. pseudo-open).
Now, let $Z$ have $[0,1]$ as its underlying set and topologise it thus: all points of $[0,1)$ are isolated, and $1$ has a neighbourhood system whose sets are complements of finite unions of convergent sequences, i.e. sets of the form $$\{1\}\cup([0,1]\setminus s_1\cup\dots\cup s_n)$$ where $s_1,\dots,s_n\subseteq [0,1]$ are convergent sequences.
Evidently $Z$ is a Hausdorff space, and, having a unique nonisolated point, $Z$ is furthermore paracompact and zero-dimensional.
Now, I claim that $p\times id_Z:X\times Z\rightarrow Y\times Z$ is not quotient. For consider the set $$D=\{(y,y)\in Y\times Z\mid y<1\}.$$ This is not closed, since clearly $(1,1)\in\overline D$. On the other hand, we can show that the inverse image $E=(p\times id_Z)^{-1}(D)\subseteq X\times Z$ is closed. To see this, take $(x,z)\not\in E$, so that $p(x)\neq z$.
$i)$ If $z<1$, then $p^{-1}([0,1]\setminus\{z\})\times\{z\}$ is a neighbourhood of $(x,z)$ in $X\times Z$ which is disjoint from $E$.
$ii)$ If $z=1$, then let $s$ be the sequence containing $x$. The set $s\times ([0,1]\setminus s\cup\{1\})$ is a neighbourhood of $(x,z)$ in $X\times Z$ which is disjoint from $E$.
In both cases we find that $(x,z)$ has a neighbourhood which does not meet $E$. Thus we conclude that $E$ is closed. It follows that $p\times id_Z$ is not quotient.
The example and construction come from the paper Bi-quotient maps and cartesian products of quotient maps by E. Michael. I tried to simplify it slightly, but didn't really succeed. This is partly because we have the following negative results:
Let $p:X\rightarrow Y$ be a quotient map between locally compact metric spaces. Then $p\times id_Z$ is a quotient map for every Hausdorff space $Z$ under any of the following conditions
Similarly, if $Z$ is compactly generated Hausdorff, then $p\times id_Z$ is quotient map for every quotient $p:X\rightarrow Y$ between locally compact Hausdorff spaces $X,Y$. This is well-known when $Z$ is locally compact (and doesn't need any assumptions on $X,Y$ in this case), but less well so if $Z$ is instead first-countable. Of course there are plenty of compactly generated spaces which are neither locally compact nor first-countable.