Characterize a full functor as a morphism

Question

Suppose we have a category of categories, with the morphisms being functors between categories. Can we express the property that a functor is full purely in terms of its properties as a morphism?

P.S. I suppose we need the category of categories in question to be sufficiently rich.I wanted to say "consider the category of all categories", but I was afraid that Russel might get angry.

Answer 1

Yes, we can, supposed we are allowed to use a concrete functor in the answer.

Consider the obvious (not full) functor $J:[2]\longrightarrow [\to]$ from the $2$ element discrete category to the category with $2$ points and an arrow (besides the identities).

Then we have:

$F:\Bbb A\to\Bbb B$ is full $\ $ iff $\ $ it has the right lifting property w.r.t. $J$,

i.e. any commuting square
$\ \ \ [2] \,\to \Bbb A$
$\ \,J\! \downarrow \ \,\ \ \ \ \downarrow \!F$
$\ \ [\to] \to \Bbb B$

has a diagonal fill-in $[\to] \dashrightarrow \Bbb A$ which makes the diagram commutative.