I have the suspicion that if $A$ is a subcategory of $B$, then the inclusion functor $A \rightarrow B$ is full. Is this right?
Full subcategory and inclusion functor
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category-theory
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3I think the easiest possible counterexample is the group homomorphism $\{e\} \hookrightarrow \mathbb{Z}/2$ considered as a functor of one-object categories. – 2012-10-29
2 Answers
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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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0Thanks. I was not paying attention to the "between groups which are not group homomorphisms" part. – 2012-10-29
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0@Kup The category of groups is not a subcategory of the category of sets. So it's not a counterexample to your question. – 2012-10-29
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0@MakotoKato http://en.wikipedia.org/wiki/Subcategory – 2012-10-29
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0@BrettFrankel Different groups can be defined on the same underlying set. – 2012-10-29
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0@MakotoKato Right you are. Silly me. – 2012-10-29
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0OK, it's a bit more artificial now, but that should take care of the issue. Thanks for pointing out the error. – 2012-10-29
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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.