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I have the suspicion that if $A$ is a subcategory of $B$, then the inclusion functor $A \rightarrow B$ is full. Is this right?

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I think the easiest possible counterexample is the group homomorphism $\{e\} \hookrightarrow \mathbb{Z}/2$ considered as a functor of one-object categories. – Aaron Mazel-Gee Oct 29 '12 at 17:50
up vote 2 down vote accepted

No. Let $A$ be the category of groups and $B$ be the category whose objects are groups but whose arrows are functions (not necessarily homomorphisms). Then there are set-theoretic maps (functions) between groups which are not group homomorphisms, hence the functor is not surjective on the $\operatorname{Hom}$ sets, which is what it means for a subcategory to be full.

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Thanks. I was not paying attention to the "between groups which are not group homomorphisms" part. – Kup Oct 29 '12 at 16:55
@Kup The category of groups is not a subcategory of the category of sets. So it's not a counterexample to your question. – Makoto Kato Oct 29 '12 at 17:13
@MakotoKato – Brett Frankel Oct 29 '12 at 17:14
@BrettFrankel Different groups can be defined on the same underlying set. – Makoto Kato Oct 29 '12 at 17:20
@MakotoKato Right you are. Silly me. – Brett Frankel Oct 29 '12 at 17:23

Let $B$ be a category. We define a category $A$ as follows. The class of objects of $A$ is the same as that of $B$. The morphisms of $A$ are monomorphisms of $B$. Then $A$ is a subcategoy of $B$. The inclusion functor $A \rightarrow B$ is not necessarily full. For example, if $B$ is the category of sets, $A \rightarrow B$ is not full.

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