I was thinking it does. Here is my proof attempt (just the monic case):

Let $F: \mathcal{C} \to \mathcal{D}$ be a full and faithful functor, and let $f: A \to B$ be an arrow in $\mathcal{C}$ that is monic. Now consider $Ff: FA \to FB$. Take any two $g',h' \in \mathsf{Ar}(\mathcal{D})$ and suppose $Ff \circ g' = Ff \circ h'$. By fullness, $\exists g,h \in \mathsf{Ar}(\mathcal{C})$ with $g' = Fg, h' = Fh$. Hence we have $Ff \circ Fg = Ff \circ Fh$. By functoriality, $F(f \circ g) = F(f \circ h)$, and by faithfulness $f \circ g = f \circ h$. Then since $f$ is monic we have $g = h$, and hence $Fg = Fh$, i.e. $g' = h'$.

In my exercise, I am asked to provide a counterexample against this claim. Where am I going wrong?

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  • $\begingroup$ I'm pretty sure full means onto maps, at least according to the definition that I am using. $\endgroup$ – Stanley Aug 17 '17 at 15:25
  • $\begingroup$ Maybe you can study Yoneda lemma,there has counter example. $\endgroup$ – Jian Aug 17 '17 at 15:26
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    $\begingroup$ However, faithful functors reflect monics and epics. $\endgroup$ – Maxime Ramzi Aug 17 '17 at 15:44

The arrows $g'$ and $h'$ have some common codomain $Z$ in $\mathcal{D}$. If $Z$ is in the image of $F$, your argument works fine. But full functors need not be surjective on objects. And if $Z$ is not in the image of $F$, you certainly can't find $g$ and $h$ which map to $g'$ and $h'$!

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  • $\begingroup$ Thanks! That makes sense, I see the mistake now. $\endgroup$ – Stanley Aug 17 '17 at 15:38

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