This question was inspired by [Can we build a continuous function from "fibers"/preimages defined over a topological base?](https://mathoverflow.net/questions/180282/can-we-build-a-continuous-function-from-fibers-preimages-defined-over-a-topolo)


Let $X,Y$ be sets and $L\subseteq \mathcal{P}(Y)$. Suppose $L$ has the following properties:


* $\emptyset, Y\in L$;
* $L$ is closed under arbitrary unions and intersections.


Let $F: L\to \mathcal{P}(X)$ be a function such that 


* $F(\emptyset) = \emptyset$ and $F(Y) = X$;
* $F$ commutes with arbitrary unions and intersections.


Is there a function $f: X \to Y$ such that $f^{-1}(U) = F(U)$ for all $U\in L$?



Constructing a function from preimages
This is true. Note that it suffices to show that for any $x\in X$, the set $U\_x:=\bigcap\limits\_{U\in L,x\in F(U)}{U} \cap \bigcap\limits\_{U\in L,x\notin F(U)}{(Y\setminus U)}$ is nonempty, since any map $f:X\rightarrow Y$ sending $x\in X$ to some element from $U\_x$ does the job.


To show the nonemptiness of $U\_x$, observe that $F(\bigcap\limits\_{U\in L,x\in F(U)}{U})=\bigcap\limits\_{U\in L,x\in F(U)}{F(U)}$ contains $x$ and that the whole intersection is empty if and only if we can write $\bigcap\limits\_{U\in L,x\in F(U)}{U}$ as a union of subsets of the form $(\bigcap\limits\_{U\in L,x\in F(U)}{U}) \cap V$ with $V\in L$ and $x\notin F(V)$. But since $F$ commutes with intersections and unions, this would imply $x\notin F(\bigcap\limits\_{U\in L,x\in F(U)}{U})$, a contradiction.


EDIT: I just realized something else. Note that we used $\mathrm{AC}$ in the above proof in order to construct $f$ from the family $(U\_x)\_{x\in X}$. Conversely, if working in $\mathrm{ZF}$, one assumes the existence of an $f$ for each such $F$, then for any family $(M\_i)\_{i\in I}$ of nonempty sets, w.l.o.g. pairwise disjoint, setting $X:=\{M\_i\mid i\in I\},Y:=\bigcup\limits\_{i\in I}{M\_i},L:=\{\bigcup\limits\_{i\in J}{M\_i}\mid J\subseteq I\}$ and $F(U):=\{M\in X\mid M\subseteq U\}$ for $U\in L$, we see that the corresponding $f$ is a choice function for $(M\_i)\_{i\in I}$. Hence this is yet another of the many equivalents of $\mathrm{AC}$.


