# Injectivity of functors in terms of skeletons?

What happens if we define a functor $$F:C \to D$$ to be injective when it is injective on isomorphism classes, or equivalently when it gives an injective fucntion from the objects of the skeleton of $$C$$ to the skeleton of $$D$$?

Edit: To be more specific, how does this definition relate to a that of a fully faithful functor?

• Although I regularly ask myself questions of the form "What happens if I make this definition / construction / etc.", I think this question needs to be more specific. Do you want to compare this notion to other "injective-like" conditions on functors or something? Feb 11 at 16:19
• Your definition is the notion of "essentially injective functor", i.e. $F(A) \cong F(B) \implies A \cong B$, which is the analogue of essentially surjective functor for injective functions. Feb 11 at 16:30
• @Varkor: How does this relate to fullness and faithfullness? Feb 11 at 16:33

Call $$F : \mathbf C \to \mathbf D$$ essentially injective (mirroring the definition of essentially surjective functor) if $$F(A) \cong F(B)$$ implies that $$A \cong B$$. This matches your definition: a functor between skeletal categories is essentially surjective if and only if its object function is injective.
Recall that a functor is conservative if, whenever $$F(g) : F(A) \to F(B)$$ is an isomorphism, $$g : A \to B$$ is also an isomorphism.
These two notions are related: in particular, a conservative functor that is full is also essentially injective, because if $$F(g)$$ is an isomorphism, then so is $$g$$ and hence $$A \cong B$$. Fullness ensures that every morphism between $$F(A)$$ and $$F(B)$$ is in the image of $$F$$, so if an isomorphism exists, it will necessarily be reflected. (As Mike Shulman points out in the comments, it suffices for $$F$$ to be pseudomonic, which means that it is faithful, and full on isomorphisms.)