Let $Pres$ denote the 2-category of locally presentable categories (categories that are accessible/cocomplete), cocontinuous functors, and natural transformations. Let $Cat$ denote the 2-category of (small) categories, functors, and natural transformations. There is an inclusion functor $i:Pres\hookrightarrow Cat$. Does this admit a left/right adjoint?

  • 2
    $\begingroup$ The presheaf category is a left adjoint, that is $\mathrm{Fun}^L(P(C),D)\cong \mathrm{Fun}(C,D)$. $\endgroup$ Aug 26, 2016 at 13:15
  • $\begingroup$ @DenisNardin I think there may be some size issues in trying to apply that here... $\endgroup$
    – David Roberts
    Aug 26, 2016 at 13:23
  • 1
    $\begingroup$ @DavidRoberts I think that the size issues appear already. when you try to include presentable categories inside small categories (that is in the definition of the functor $i$). $\endgroup$ Aug 26, 2016 at 13:26
  • 3
    $\begingroup$ Yeah, this isn't making much sense: small locally presentable categories must be preorders: a small-cocomplete small category is a preorder. $\endgroup$
    – Todd Trimble
    Aug 26, 2016 at 13:44
  • $\begingroup$ What happens if the smallness condition is dropped? $\endgroup$
    – user84563
    Aug 26, 2016 at 17:26

1 Answer 1


Size conditions are not such a big deal when it comes to formulating the question, at least. We just let Cat be the non-locally-small 2-category of locally-small categories (perhaps we require that the object sets be no bigger than the universe, it turns out not to matter). The inclusion $\mathsf{Pres}^L \to \mathsf{Cat}$ preserves reasonable 2-limits, but not 2-colimits, see here. So it might have a left biadjoint but it certainly does not have a right biadjoint.

But we can show that size issues do prevent the left biadjoint from existing. Let $C$ be a discrete category with object set $C_0$ of the size of the universe. Then if $FC$ were a reflection of $C$ in $\mathsf{Pres}^L$, we would have $\mathsf{Pres}^L(FC,D) = \mathsf{Cat}(C,D) = D^{C_0}$ for any locally presentable $D$. If $D = \mathsf{Set}$, say (the category of small sets), it's easy to see that $D^{C_0}$ has the cardinality of the powerset of the universe. But the hom-category $\mathsf{Pres}^L(E,D)$ between any two locally presentable categories is no bigger than the size of the universe (this is because any cocontinuous functor $E \to D$ is the left Kan extension of a functor defined on the small subcategory of $\lambda$-presentable objects for some $\lambda$; there are at most universe-many such functors, and taking the union over $\lambda$ still yields universe-many). So no such $FC$ can exist.

Along the lines of Dennis Nardin's comment, though, the inclusion of locally class-presentable categories into $\mathsf{Cat}$ has a left biadjoint given by sending $C$ to the free cocompletion of $C$ (which is the full category of presheaves $PC$ on $C$ when $C$ is small, but a proper subcategory of $PC$ when $C$ is large).

  • $\begingroup$ Are locally class-presentable categories complete? Because the small presheaves cocompletion need not be complete. $\endgroup$
    – Todd Trimble
    Aug 26, 2016 at 22:01
  • $\begingroup$ Oh, you're right, I think they're complete by definition. I should say "cocomplete class-accessible categories". $\endgroup$
    – Tim Campion
    Aug 26, 2016 at 22:17
  • $\begingroup$ I'm still maybe confused: a small-cocomplete accessible category is automatically small-complete, so the question is whether something like that carries over to this class-accessible setting (which I have not studied). $\endgroup$
    – Todd Trimble
    Aug 26, 2016 at 23:15
  • $\begingroup$ I think it's not automatic in the class accessible case, so chorny and rosicky simply define a locally class presentable category to be a class accessible category which is both complete and cocomplete. $\endgroup$
    – Tim Campion
    Aug 26, 2016 at 23:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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