2
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ 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? $\endgroup$
    – Tim Campion
    Feb 11 at 16:19
  • 1
    $\begingroup$ 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. $\endgroup$
    – varkor
    Feb 11 at 16:30
  • $\begingroup$ @Varkor: How does this relate to fullness and faithfullness? $\endgroup$ Feb 11 at 16:33
4
$\begingroup$

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.)

Finally, every fully faithful functor is conservative, and hence essentially injective. It is easy to see that the converse does not hold: for instance, any functor from the category with two objects and a single isomorphism is essentially injective, but will rarely be fully faithful.

$\endgroup$
1
  • 4
    $\begingroup$ A weaker notion than fully-faithful which still implies conservativity and essential injectivity is being pseudomonic. $\endgroup$ Feb 11 at 18:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.