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Let $\mathbf{C}$ and $\mathbf{D}$ be two categories with classes of objects $\mathbf{C}_0$ and $\mathbf{D}_0$, respectively. Consider a function $f:\mathbf{C}_0\rightarrow \mathbf{D}_0$. When $f$ lifts to a functor $F:\mathbf{C}\rightarrow \mathbf{D}$? More precisely:

  • under which conditions there exists a functor $F:\mathbf{C}\rightarrow \mathbf{D}$ such that $F_0=f$?

Of course, if such a functor exists then it is generally highly non-unique. Indeed, the choice of an automorphism of $f(X)\in \mathbf{D}_0$ for every $X\in \mathbf{C}_0$ allows us to modify $F$ on morphisms.

Any help is welcome. If the question is stupid, I'm sorry.

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    $\begingroup$ Some necessary conditions: if there's an arrow $a\to b$ in $C$, there must also exist an arrow $f(a) \to f(b)$ in $D$. Isomorphic objects must be mapped to isomorphic objects. $\endgroup$ – Berci May 13 '20 at 18:02
  • $\begingroup$ Of course. Thank you for the comment. It would be very appreciate some sufficient conditions. $\endgroup$ – Math-Phys-Cat Group May 13 '20 at 18:04
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One easy case: if $\mathbf C = C(Q)$ is the free category on a quiver $Q$, then it is sufficient that $f$ extends to a quiver homomorphism from $Q$ to the underlying quiver of $\mathbf D$; i.e., that $\mathbf D(f(a), f(b))$ is non-empty whenever there is an edge from $a$ to $b$ in $Q$.

In this case, the condition mentioned in the comments (isomorphic objects must be mapped to isomorphic objects) is vacuously satisfied, since free categories have no non-identity isomorphisms.

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    $\begingroup$ Thank you for the answer. Actually, in this case I think that this condition in also necessary, or not? $\endgroup$ – Math-Phys-Cat Group May 13 '20 at 19:30
  • $\begingroup$ @math-phys-cat Yes, that's correct. $\endgroup$ – John Gowers May 14 '20 at 10:39

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