1
$\begingroup$

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

  • 3
    I 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 2

2

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.

  • 0
    Thanks. I was not paying attention to the "between groups which are not group homomorphisms" part.2012-10-29
  • 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
  • 0
    @MakotoKato http://en.wikipedia.org/wiki/Subcategory2012-10-29
  • 0
    @BrettFrankel Different groups can be defined on the same underlying set.2012-10-29
  • 0
    @MakotoKato Right you are. Silly me.2012-10-29
  • 0
    OK, it's a bit more artificial now, but that should take care of the issue. Thanks for pointing out the error.2012-10-29
3

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.