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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.

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  • $\begingroup$ Epimorphisms need not be full: there is a category version of the fact that $\mathbb{Z} \to \mathbb{Q}$ is an epimorphism in the category of rings. $\endgroup$
    – Zhen Lin
    May 5 '14 at 18:34
  • $\begingroup$ Oh, thanks for the comment. I'll remove that paragraph then. $\endgroup$
    – Henrique
    May 5 '14 at 18:35
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    $\begingroup$ Certainly "consider the category of small categories" is fine. $\endgroup$ May 5 '14 at 20:25
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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.

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  • $\begingroup$ You are using the concrete function to define those two objects as "the only two categories that the concrete functor takes to the 2-set", correct? Then $J$ must be any given morphism between those categories? This is a great answer, but I would prefer a definition that doesn't take those objects as primitives. Is there a way to define them using their categorical properties? By the way, I'm particularly interested in the case where the categories are preorders, in case that makes it easier. $\endgroup$
    – Henrique
    May 5 '14 at 21:58
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    $\begingroup$ $J$ is a specific morphism in the category of small categories, it maps $[2]=\{A,B\}$ to $[\to]=\{A,B,f\}$ mapping $A\mapsto A,\ B\mapsto B$ (and the arrow $f:A\to B$ doesn't have preimage via $J$). The same works for preorders, as $J$ is already present there. $\endgroup$
    – Berci
    May 5 '14 at 23:09
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    $\begingroup$ @Henrique You can get this functor by various considerations: for instance, you can count the number of objects in $\mathbb{C}$ by looking at morphisms $\mathbb{1} \to \mathbb{C}$, where $\mathbb{1}$ is the terminal object; and in fact the left adjoint of the left adjoint of the functor represented by $\mathbb{1}$ sends $\mathbb{C}$ to its set of connected components, so we can also pick out the connected categories; finally, note that $\mathbb{2}$ is the "minimal" connected category with 2 objects, and that the morphism $2 \to \mathbb{2}$ in question is the unique one injective on objects. $\endgroup$
    – Zhen Lin
    May 5 '14 at 23:36

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