The setting for this question is the (2,1) category Topoi.

  • 0-cells in Topoi are Grothendieck topoi.
  • 1-cells are geometric morphisms and have the direction of the right adjoint.
  • 2-cells are invertible natural transformation (between the left adjoints).

Johnstone (and Lurie) proves that this 2-category has pseudo colimits that can be computed in Cat. To be precise, given a diagram in Topoi its colimit is the limit of the corresponding diagram in Cat associated to the inverse images.

Q. Does the representable functor $\mathsf{pt}:= \text{Topoi(Set, )} $ preserve directed colimits?

After some naive optimism, I have the feeling that the answer is no, but I suspect that it might be known.

  • A candidate counterexample would be a topos $\mathcal{E}$ whose subtopoi lattice is not infinitary distributive, in fact a point is an atom in this coHeyting algebra.
  • Another idea is to build a topos with with some points as a directed colimit of topoi without points, but I am not sure that this is possible at all.
  • If a counterexample exist, I expect that it's possible to exhibit a localic counterexample.
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    $\begingroup$ Is your functor $\mathrm{pt}$ valued in sets or in categories (which I feel is the most natural choice)? $\endgroup$ – Denis Nardin Feb 4 at 10:35
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    $\begingroup$ It takes value in categories! Thanks for the question! $\endgroup$ – Ivan Di Liberti Feb 4 at 10:39
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    $\begingroup$ I believe I can see that $Set$ is $\kappa$-presentable in the category of locally $\kappa$-presentable topoi, but I don't think it's any more presentable than that. However, I don't have a counterexample... $\endgroup$ – Tim Campion Feb 4 at 14:52

If $\mathcal{C}$ is a small category with finite limits then geometric morphisms from ${\rm Set}$ to the presheaf topos ${\rm PSh}(\mathcal{C})$ are in bijection with left exact functors $\mathcal{C} \to {\rm Set}$, or, equivalently, with objects in the pro-category ${\rm Pro}({\cal C})$. More explicitly, if ${\bf X} = \{X_i\}_{i \in {\cal I}}$ is a pro-object in ${\cal C}$ then the corresponding point ${\bf X}^*:{\rm PSh}(\mathcal{C}) \to {\rm Set}$ sends a presheaf $F :\mathcal{C}^{\rm op} \to {\rm Set}$ to ${\rm colim}_{i \in {\cal I}} F(X_i)$. If ${\cal C}_1,{\cal C}_2$ are two categories with finite limits and $f: {\cal C}_1 \to {\cal C}_2$ is a functor (which does not necessarily preserve finite limits) then we have a geometric morphism $f_*: {\rm PSh}(\mathcal{C}_1) \leftrightarrows {\rm PSh}(\mathcal{C}_2): f^*$ given by restriction and right Kan extension. We can then check that the functor on points ${\rm Pro}({\cal C}_1) \to {\rm Pro}({\cal C}_2)$ induced by $f_*$ sends $\{X_i\}_{i \in {\cal I}}$ to $\{f(X_i)\}_{i \in {\cal I}}$.

If we now take a sequence $$ {\cal C_1} \to {\cal C_2} \to ... \to {\cal C_n} \to ... \to $$ of categories with finite limits (where the functors ${\cal C_i} \to {\cal C_{i+1}}$ are not assumed to preserve finite limits) then $$ {\rm colim}^{\rm Topoi}_i {\rm PSh}({\cal C}_i) \simeq {\rm lim}^{\rm Cat}_i {\rm PSh}({\cal C}_i) \simeq {\rm PSh}({\rm colim}_i{\cal C_i}), $$ but in general the map ${\rm colim}_i{\rm Pro}({\cal C_i}) \to {\rm Pro}({\rm colim}_i{\cal C_i})$ is not an equivalence. For example, it is often not essentially surjective: take ${\cal C_n} = [n]$ with each consecutive map $[n] \to [n+1]$ being the inclusion as $[n] \cong \{1,...,n+1\} \subseteq [n+1]$ (these categories have finite limits because they are posets in which every finite subset has a minimal element).

  • $\begingroup$ Thanks for your answer! I completely believe it, because I dot expect the Ind-completion to be an accessible KZ-doctrine over Cat, but is there an evident argument that proves $\text{colim} \text{Pro}([n]) \not \cong \text{Pro}([\omega])$? $\endgroup$ – Ivan Di Liberti Feb 7 at 10:01
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    $\begingroup$ @IvanDiLiberti, Note that for every $n$ we have ${\rm Pro}([n]) \simeq [n]$, but there is a non-constant object in ${\rm Pro}([\omega])$, namely, the inverse system given by the identity diagram $\omega \to \omega$ (and which corresponds to the terminal left exact functor sending everything to the point). $\endgroup$ – Yonatan Harpaz Feb 7 at 10:52
  • $\begingroup$ Thanks! Accepted. $\endgroup$ – Ivan Di Liberti Feb 7 at 10:53

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