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Let $\mathit{Pr}^L$ be the $\infty$-category of presentable $\infty$-categories and continuous functors in some universe. Is it presentable itself a larger universe?

Zippy
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2 Answers2

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Let us fix a universe and use the words "large" and "small" with respect to that universe. Presentable $\infty$-categories are typically large $\infty$-categories (since, as the previous answer mentioned, small $\infty$-categories are rarely presentable). One might then expect that $Pr^L$ would be a "very large" $\infty$-category (like the $\infty$-category $\widehat{Cat}_\infty$ of possibly large $\infty$-categories). In that case $Pr^L$ would turn from being very large to just large upon increasing the universe, and it would be natural to ask if it then becomes presentable. However, the $\infty$-category $Pr^L$ is actually not "very large", but just large. This is because presentable $\infty$-categories, though being large, are actually determined by a small amount of data. To formally prove this one might consider, for example, for each cardinal $\kappa$, the subcategory $Pr^L_{\kappa} \subseteq Pr^L$ consisting of $\kappa$-compactly generated presentable $\infty$-categories and functors preserving $\kappa$-compact objects between them (see section 5.5.7 of higher topos theory). Proposition 5.5.7.10 loc. cit. shows that when $\kappa > \omega$ the $\infty$-category $Pr^L_{\kappa}$ is equivalent to the $\infty$-category of small $\infty$-categories admitting $\kappa$-small colimits, and hence $Pr^L_{\kappa}$ is large (but not very large). Consequently, $Pr^L$ is a large colimit of large $\infty$-categories, and hence large (but again not very large). It follows that if we increase the universe $Pr^L$ will become small, and it will not be so natural to ask if it is presentable. On the other hand, since $Pr^L$ is just large you might ask if it is presentable without increasing the universe. In principle the answer would have to be no, because then $Pr^L$ would contain itself, and we know that such stories do not end well (although I admit I do not have a direct proof in mind, and would like to see one). Morally, $Pr^L$ should not be a presentable $\infty$-category, but some $(\infty,2)$-version of the notion, similarly to how the $\infty$-category of $\infty$-topoi should be something like an $(\infty,2)$-topos (but not an $(\infty,1)$-topos). Unfortunately, I am not aware of these ideas being made precise anywhere.

Edit: a formal argument why $Pr^L$ is not presentable (in the current universe) is that it is not locally small, see comments below.

Yonatan Harpaz
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  • I disagree: in principle the answer should be yes. Indeed, $Pr^L$ contains all limits and colimits (the strict version of this statement is due to Greg Bird in his unpublished '76 thesis, if my memory is correct), and feels like it is generated under colimits by some basic building blocks. The reason Cantor's paradox does not apply is that we are in the homotopical world, not the set-theoretic world. Indeed, there are good homotopical models (although now I am speaking outside my expertise) in which all functors, and in particular the "power set" functor, have fixed points. These worlds ... – Theo Johnson-Freyd Jul 04 '15 at 20:16
  • ... are well-studied in some areas of computer science, because they allow for coinductive, rather than inductive, reasoning. A more down-to-earth reason not to be afraid of Cantor's paradox is the well-known theorem that the homotopy category of topological spaces is not concretizable (because every non-empty object has a proper class, not a set, of subobjects). – Theo Johnson-Freyd Jul 04 '15 at 20:19
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    Even "very large" categories become small when you pass to a larger universe. For instance, if $V_\kappa$ is a universe, then the collection of all "large" categories in $V_\kappa$ has cardinality only $2^\kappa$, which is far smaller than any inaccessible larger than $\kappa$. – Eric Wofsey Jul 04 '15 at 20:24
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    @EricWofsey, you're right, but after you made this enlargement it will not be interesting anymore to ask if the original $\infty$-category is presentable. Instead, you should stop when it is just large, and ask if it is presentable then. – Yonatan Harpaz Jul 04 '15 at 20:32
  • Sure. I just wanted to correct your assertion that a very large category might still be large in a larger universe. – Eric Wofsey Jul 04 '15 at 20:34
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    I think there is just a terminology gap here. To avoid confusion, one should not (as I did) use the language of "large" and "small" and the language of universes at the same time. By a large $\infty$-category I just meant a small $\infty$-category of the next universe. Consequently, the $\infty$-category of possibly large $\infty$-categories is what would be the $\infty$-category of small $\infty$-categories in the next universe, and hence not small in the next universe. These kind of objects is what I called "very large". – Yonatan Harpaz Jul 04 '15 at 20:52
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    It just occurred to me that $Pr^L$ is not presentable simply because it is not locally small (the space of functors between two presentable $\infty$-categories is the maximal $\infty$-groupoid of a presentable $\infty$-category, and as such is large). However, when considered as an $(\infty,2)$-category, the mapping categories in $Pr^L$ between every two objects are presentable, and hence controlled by a small amount of data. This might be considered as evidence that $Pr^L$ should be something like $(\infty,2)$-presentable. – Yonatan Harpaz Jul 04 '15 at 21:00
  • @YonatanHarpaz I would rather think of the $(\infty, 2)$-category of $(\infty, 1)$-categories as the archetypical $(\infty, 2)$-topos, just as the 1-category of 0-categories is the archetypical 1-topos. – Zhen Lin Jul 04 '15 at 23:42
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    @EricWofsey, Let $\lambda > \tau > \kappa$ be strongly inaccessible cardinals and let $V_{\kappa}, V_{\tau}, V_{\lambda}$ be the associated universes. If $V_{\kappa}$ is my universe then I call elements in $V_{\kappa}$ "small sets", elements of $V_{\tau}$ "large sets" and elements of $V_{\lambda}$ "very large sets". If I change my universe to $\tau$ it means I now refer to sets in $V_{\tau}$ as "small" and to sets in $V_{\lambda}$ as "large" -> – Yonatan Harpaz Jul 05 '15 at 20:46
  • -> Now the category of large categories is very large (its object set cannot belong to $V_{\tau}$, but it does belong to $V_{\lambda}$), and will become just large (but not small), when I change my universe to $V_{\tau}$. – Yonatan Harpaz Jul 05 '15 at 20:46
  • @YonatanHarpaz: Ah, OK, sorry, I was misreading your definitions. – Eric Wofsey Jul 05 '15 at 21:38
  • @YonatanHarpaz Let $C$ and $D$ be presentable $\infty$-categories. There exists some regular (small) cardinal $\chi$ such that $C\cong Ind_{\chi}C_{\chi}$ and $D\cong Ind_{\chi}D_{\chi}$, where $C_{\chi}$ and $D_{\chi}$ are essentially small and admit $\chi$-colimits. Then the set (0-simplicies) of colimit preserving functors $C\to D$ is bijective to the set of $\chi$-colimit preserving functors $C_{\chi}\to D_{\chi}$, which is small. Is there something wrong with this argument? – Ilan Barnea Jul 10 '15 at 02:07
  • @IlanBarnea, colimit preserving functors $C \longrightarrow D$ are in bijection with the collection all functors $C_{\chi} \longrightarrow D$, which is big. This also made me realize that I need to correct the definition of $Pr^L_{\kappa}$ in the answser above. – Yonatan Harpaz Jul 10 '15 at 17:55
  • @YonatanHarpaz In your answer above you define $Pr^L_{\kappa}$ to be the full subcategory of $Pr^L$ spanned by $\kappa$-compactly generated presentable $\infty$-categories. Then by Proposition 5.5.7.10 in HTT we know that if $\kappa > \omega$ the $\infty$-category $Pr^L_{\kappa}$ is equivalent to the $\infty$-category of small $\infty$-categories admitting $\kappa$-small colimits. This last category is locally small so it looks to me that $Pr^L$ is also locally small – Ilan Barnea Jul 10 '15 at 18:44
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    This definition is not correct, it should be the (not full) subcategory spanned by $\kappa$-compactly generated presentable $\infty$-categories and functors which preserve $\kappa$-small objects. This is the definition to which Proposition 5.5.7.10 applies to and that is indeed a (non-full) subcategory which is small. The argument appearing in the answer above still applies to show that $Pr^L$ is just large, and not very large. – Yonatan Harpaz Jul 10 '15 at 18:58
  • Thanks for making the edit. This is all much clearer now to me at least. Previously it seemed like the two answers were at odds, but no longer. – David White Jul 11 '15 at 11:52
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When you pass to a larger universe, all categories that were in your old universe become small. A small category which is not a poset cannot be closed under colimits (think about taking coproducts with index sets larger than your category), and so cannot be presentable. All this applies equally well to $(\infty,1)$-categories, and ought to apply to any other reasonable generalized kind of category.

Eric Wofsey
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