# Full subcategory and inclusion functor

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

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.
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.