Is there a useful property for a concrete category $(\mathcal A, U : \mathcal A \to \mathcal X)$ such that in this case $U$ is fibre-complete iff $U$ is topological? (kind of like what "flat" is too "left exact). 


The motivation is the following: given a finitely complete category $\mathcal X$ we can define the concrete category of $\mathcal X$-objects equipped with relations / reflexive relations / preorders / congruences / etc. . These seem to be topological iff the functor $U : (X,R) \mapsto X$ is fibre-complete (unless I'm mistaken).



Generalization of topological functor which need not be fibre-complete
I assume that you mean "topological" in [this sense](https://ncatlab.org/nlab/show/topological+concrete+category), probably with the strict condition that $U(T)=X$ and $g=id$. In that case, an almost-answer is:



> 
> $U$ is a bifibration in preorders.
> 
> 
> 


A topological functor certainly has this property. Conversely, if $U$ is a fibre-complete fibration in preorders, then given any $U$-structured source $(f\_i : X \to U(S\_i))$, we can form the family of reindexings $f\_i^\* S\_i$ using the fibration structure and take their meet in the complete preorder $U^{-1}(X)$. The initiality of this lift follows from the universal properties of cartesian arrows and meets, together with the fact that reindexing functors $g^\*$ preserve meets, which follows from the fact that they have left adjoints because $U$ is also a bifibration.


Whether or not this is quite an answer depends on your definition of "fibre-complete" and size questions. In order for the above argument to work, we need the fibers to have possibly *large* meets; a complete *small* preorder automatically has large meets, but in general the fibers of a topological functor may be large. However if you define "fibre-complete" to mean the existence of *all* meets including large ones, then it works.


