Disjoint arrows in the category of locales

Question

Call two arrows $f$ and $g$ disjoint if the pullback of $f$ by $g$ is the initial object. Here's my question: Does there exist a sublocale $j: J\to L$ which is not disjoint with any other (non-initial) sublocale of $L$, but which is disjoint with other non-initial arrows.

Answer 1

No, such thing cannot happen.

(Here was some blunder depending on the assumption that pullback of a nontrivial sublocale along a surjection is nontrivial, which, as Simon Henry points out in the comment below is not true.)

What is actually true is that if a sublocale of $L$ is not disjoint from any nontrivial sublocale then it coincides with $L$.

In fact even more is true: if a sublocale is not disjoint from any locally closed sublocale then it coincides with $L$.

Equivalent restatement in terms of nuclei on the frame of $L$ is this: if a nucleus $j$ has the property that $j\lor{\mathrm o}_a\lor{\mathrm c}_b=1$ implies ${\mathrm o}_a\lor{\mathrm c}_b=1$ for any $a,b\in L$ then $j$ is the identity map. Here ${\mathrm o}_a(x)=a\to x$ is an open nucleus and ${\mathrm c}_b(x)=b\lor x$ is a closed one, so the nucleus $[{\mathrm o}_a\lor{\mathrm c}_b](x)=a\to(b\lor x)$ is locally closed.

This follows from the well known fact that for any nucleus $j$ and any $a,b\in L$, $$ a\leqslant jb \iff {\mathrm c}_a\land{\mathrm o}_b\leqslant j. $$ Now each nucleus ${\mathrm c}_a\land{\mathrm o}_b$ is complemented, with $\neg({\mathrm c}_a\land{\mathrm o}_b)={\mathrm o}_a\lor{\mathrm c}_b$. But if $k$ is complemented then $k\leqslant j$ is equivalent to $j\lor\neg k=1$. Thus a $j$ with the above property will be such that $a\leqslant jb$ implies ${\mathrm o}_a\lor{\mathrm c}_b=1$ for any $a,b$, i. e. $a\to(b\lor x)=1$ for any $x$, i. e. $a\leqslant b$. Thus $jb=b$ for any $b$.