$\newcommand\Psh{\mathit{Psh}}\newcommand\Pr{\mathit{Pr}}$Let $\Psh$ be the category of presheaf categories and cocontinuous functors which preserve tiny objects. There is a functor $(-)^\ast : \Psh \to \Psh$ sending $\Psh(C) \mapsto \Psh(C^\text{op})$. This functor is an involution in the sense that $(\Psh(C)^\ast)^\ast = \Psh(C)$. Note that there is a non-full inclusion $i : \Psh \to \Pr^L$ into the category of presentable categories and cocontinuous functors.

Question: Is there an involution $(-)^\star : \Pr^L \to \Pr^L$ such that $i(\Psh(C)^\ast) = (i(\Psh(C)))^\star$?

(I have freely mixed and matched terminology here from 1-categories and $\infty$-categories. The above question is really two questions: one in the 1-categorical case and another in the $\infty$-categorical case. Please ask if it's unclear what I'm saying!)


  • Of course, the opposite category to a presentable category is rarely presentable. Note that the putative involution I'm asking about would not be obtained by taking the opposite category.

  • The duality involution on $\Psh$ is related to dualizability with respect to the Lurie tensor product. I'm pretty sure I've been told that the only dualizable objects in $\Pr^L$ are the retracts of presheaf categories. But I don't think that rules out an involution of the form I'm asking about. (I'm a bit confused on this point too, because the involution I'm asking about would anyway be covariant rather than contravariant like the one related to dualizability.)

  • $\begingroup$ Isn't $Psh$ equivalent to $\mathbf{Cat}$ under the requirement that functors preserve tiny objects? $\endgroup$
    – varkor
    May 3 at 13:38
  • $\begingroup$ @varkor Yes, $Psh$ is equivalent to the category of small, idempotent-complete categories and all functors. $\endgroup$
    – Tim Campion
    May 3 at 13:38
  • 1
    $\begingroup$ In the 1-categorical case, I think this is the "reflexivity" question posed here by Brandenburg, Chirvasitu, and Johnson-Freyd? They show there's a pretty broad class of categories where you get reflexivity even when you don't have dualizability. They don't give any counterexamples that aren't reflexive as far as I can tell. $\endgroup$ May 3 at 16:09
  • $\begingroup$ Oh, key point I missed, they're working in the k-linear setting which it looks like you aren't. $\endgroup$ May 3 at 16:55
  • $\begingroup$ @NoahSnyder : there's also the question of variance: taking duals is contravariant (although Tim already pointed this out). $\endgroup$ May 3 at 18:52

2 Answers 2


The answer is no, even if you restrict to the full subcategory of $Pr^L$ spanned by the $Psh(C)$'s. I'll answer in the $1$-categorical case but : a- the $\infty$-categorical case follows because presentable $1$-categories are presentable $\infty$-categories and b- even if it didn't strictly follow, one easily convinces oneself that the same method works.

Indeed, your involution provides, for any $Set\to Psh(C)$, a functor $Set = i(Set) \to i(Psh(C)) = Psh(C^{op})$, i.e., for any presheaf $F$ on $C$, a canonical presheaf on $C^{op}$.

Specifically, for every $F: C^{op} \to Set$ it gives you some $\iota F : C\to Set$ in a way compatible with left Kan extension along small functors $C\to D$. Note that on representables, it sends $\hom(-,x)$ to $\hom(x,-)$.

Now I claim that $\iota$ can be extended to a functor.

Namely, say I have a natural transformation of presheaves of $C$, $F\to G$, viewed as $\Delta^1 \to Psh(C)$, then I can extend it to $Psh(\Delta^1) \to Psh(C)$ and the two inclusions $\Delta^{\{i\}}\to \Delta^1$ show that applying my involution $i$ and restricting along $\Delta^1\to Psh(\Delta^1)$ gives me a transformation $\iota G\to \iota F$ (there is an inversion of direction because of $\Delta^1$ vs $(\Delta^1)^{op}$.

Furthermore, by looking at $\Delta^2$, it is easy to see that this really makes $\iota$ into a functor. In particular $\iota : Psh(C)\to Psh(C^{op})^{op}$ is a functor which restricts to the identity along the Yoneda embeddings.

Because $i$ is an involution and not only a functor, you can do the same thing in the opposite direction, and the fact that it's an involution shows that the composite $Psh(C)\to Psh(C^{op})^{op} \to Psh(C)$ is the identity, and same of course in the other direction. In particular, $Psh(C)\simeq Psh(C^{op})^{op}$, which is impossible.


$\newcommand\Pr{\mathit{Pr}}\newcommand\Pres{\mathit{Pres}}$Not an answer, but too long for a comment. Gabriel–Ulmer duality lends some intuition here. For simplicity I consider the finitely presentable case, though it should be clear everything generalises to the non-finite setting. There is a biequivalence between the 2-category $\Pr^L$ and the 2-category $\mathbf{Rex}$ of small finitely cocomplete categories and finitely cocontinuous functors. Since the duality involution on $\mathit{Psh}$ is inherited from that on $\mathbf{Cat}$, it is not a significant generalisation to rephrase the question as follows:

Is there a duality involution on $\mathbf{Rex}$ such that the following square commutes up to pseudonatural equivalence (where the 2-functor $\mathbf{Cat} \to \mathbf{Rex}$ is the free cocompletion under finite colimits)?

$\require{AMScd}\begin{CD} \mathbf{Rex}^{\text{co}} @>{(-)^{\star}}>> \mathbf{Rex} \\ @AAA @AAA \\ \mathbf{Cat}^{\text{co}} @>>{(-)^{\text{op}}}> \mathbf{Cat}. \end{CD}$

In this form, it seems unlikely there will be such a duality involution (certainly taking the opposite category does not work). However, if instead of considering the finitely cocomplete categories, we consider those categories with finite colimits and finite limits, a similar square does commute. By applying Gabriel–Ulmer duality again, we should get a duality involution on the sub-2-category of $\Pres^L$ whose objects have finitely complete subcategories of finitely presentable objects, and whose functors preserve finite limits, which satisfies the (pseudo)commutativity condition you are looking for.


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