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.



Disjoint arrows in the category of locales
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 $b$.


