A locale is a complete lattice $L$ such that:
$\bullet$ $a\wedge\bigvee_{b\in B}b=\bigvee_{b\in B}(a\wedge b)$ for all $a\in L$ and $B\subseteq L$.
An $L$-set is a set $X$ together with a function $\delta:X\times X\rightarrow L$ such that:
$\bullet$ $\delta(x,y)=\delta(y,x)$ for all $x,y\in X$.
$\bullet$ $\delta(x,y)\wedge\delta(y,z)\leq\delta(x,z)$ for all $x,y,z\in X$.
An $L$-subset is a function $S:X\rightarrow L$ such that:
$\bullet$ $S(x)\leq\delta(x,x)$ for all $x\in X$.
$\bullet$ $S(x)\wedge\delta(x,y)\leq S(y)$ for all $x,y\in X$.
An $L$-singleton is an $L$-subset $S$ such that:
$\bullet$ $S(x)\wedge S(y)\leq\delta(x,y)$ for all $x,y\in X$.
Let $\mathrm{Sing}(X)$ be the set of all $L$-singletons of $X$.
Let $\sigma:X\rightarrow\mathrm{Sing}(X)$ be the function such that for every $x\in X$ then $\sigma_x$ is defined by:
$\sigma_x(t)=\delta(t,x)$.
A subset $C\subseteq X$ is $L$-compatible iff is satisfies:
$\bullet$ $\delta(x,x)\wedge\delta(y,y)\leq\delta(x,y)$ for all $x,y\in C$.
For a subset $C\subseteq X$ that is $L$-compatible, a glueing of $C$ is an element $z\in X$ such that:
$\bullet$ $\delta(x,z)=\delta(x,x)$ for all $C$.
$\bullet$ $\bigvee_{x\in C}\delta(x,x)=\delta(z,z)$.
I wanted to prove that $\sigma$ is surjective if and only if every subset $C\subseteq X$ that is $L$-compatible has at least one glueing $z\in X$.
I was able to prove that if $\sigma$ is surjective then every $L$-compatible subset has at least one glueing.
On the other hand, suppose that every $L$-compatible subset has at least one glueing. Then the empty set is $L$-compatible, so it has some glueing $o\in X$, so $\delta(o,o)=\bigvee_{x\in\emptyset}\delta(x,x)=0$, where $0$ is the minimum element of the locale $L$.
Let $S\in\mathrm{Sing}(X)$, then $0\leq S(o)\leq\delta(o,o)=0$, so $S(o)=\delta(o,o)=0$.
Moreover, considering the set $C=\{x\in X:S(x)=\delta(x,x)\}$, then $C$ is $L$-compatible, so it has a glueing $z\in X$.
Therefore $\delta(z,z)=\bigvee_{x\in C}\delta(x,x)=\bigvee_{x\in C}(\delta(x,x)\wedge\delta(x,x))= \bigvee_{x\in C}(S(x)\wedge\delta(x,z))\leq S(z)\leq\delta(z,z)$, so $S(z)=\delta(z,z)$.
Thus $\sigma_z(t)=\delta(z,t)=\delta(z,z)\wedge\delta(z,t)=S(z)\wedge\delta(z,t)\leq S(z)\wedge S(t)$.
Also $S(z)\wedge S(t)\leq\delta(z,t)=\sigma_z(t)$.
So $\sigma_z(t)=S(z)\wedge S(t)$.
However, I was not able to verify whether $S(z)\wedge S(t)=S(t)$ for all $t\in X$.
I have considered the set $C=\{x\in X:S(x)=\delta(x,x)\}$, that is always $L$-compatible, because I was able to prove that for all $S\in\mathrm{Sing}(X)$ and $z\in X$, if $\sigma_z=S$, then $z$ is necessarily a glueing of $C$. Indeed, for $x\in C$ we have $\delta(x,x)=S(x)=\sigma_z(x)=\delta(x,z)$; and for $x\in C$ then $\delta(x,z)\leq\delta(z,z)$; moreover $S(z)=\sigma_z(z)=\delta(z,z)$, so $z\in C$, so $\delta(z,z)=\bigvee_{x\in C}\delta(x,z)=\bigvee_{x\in C}\delta(x,x)$, so that $z$ is a glueing of $C$.