TITLE: Characterize a full functor as a morphism
QUESTION [6 upvotes]: 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.

REPLY [6 votes]: 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.