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Recently I've received an email from Sori Lee about an earlier question I had asked, and we ended up with a number of questions about 2-limits and 2-bilimits which I couldn't quite answer, and decided to ask here.

Throughout this question let XXX = lax, oplax, strong/pseudo, or strict.

The first question that came up was the following:

What is an example of an XXX bilimit that isn't an XXX $2$-limit.

The first potential example that came to my mind would be the following: the $2$-limit of the empty diagram in the $2$-category $\mathsf{Cats}$ of categories, functors, and natural transformations would be the punctual category $\mathsf{pt}$ having a single object and a single (identity) morphism, whereas I think any contractible groupoid would serve as the $2$-bilimit of that diagram.

Question 1. Is this correct? More generally, if we have a diagram $D$ in a bicategory $\mathcal{C}$ with $2$-limit $\mathsf{2lim}(D)$ and an object $L$ of $\mathcal{C}$ equivalent to $\mathsf{2lim}(D)$, must $L$ be a 2-bilimit of $D$? If not, is there a useful criterion for when this is the case?

Question 2. Are there any examples (especially naturally occurring) of bicategories that are co/complete with respect to all XXX $2$-bilimits, but not with respect to all XXX $2$-limits?

Question 3. Lax/oplax/pseudo $W$-weighted $2$-limits can be seen as special cases of strict $Q(W)$-weighted $2$-limits where $Q(W)$ is a lax/oplax/pseudo morphism classifier. Is the same true if we replace "$2$-limits" with "$2$-bilimits"?

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    $\begingroup$ I'm not quite sure what you want a strict weighted bilimit to be, actually. It classifies strictly 2-natural cones (cylinders)? But it doesn't really even make sense to look at strictly 2-natural cones over anything but a strict 2-functor, right? So to have any hope for something like that to be true you'd have to first make sure you could strictify the diagram you wanted a limit of. $\endgroup$ Commented Oct 6, 2023 at 3:08

2 Answers 2

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Q1. Yes, every object that is equivalent to a bicategorical limit is again a bicategorical limit. This is easy to deduce from the definition. It can also be deduced from the corresponding fact about representable pseudofunctors.

Q2. The 2-category of monoidal categories has a bicategorical initial object $\{1\}$, but no $2$-categorical initial object. Otherwise, $\{1\}$ would be $2$-categorical initial, i.e. the category of monoidal functors $\{1\} \to \mathcal{C}$ would be isomorphic to the terminal category $\star$ for all $\mathcal{C}$. But this is not the case:

It is well-known that the lax monoidal functors $\{1\} \to \mathcal{C}$ are monoid objects $(M,m,e)$ in $\mathcal{C}$. The functor is strong monoidal iff $e : 1 \to M$ and $m : M \otimes M \to M$ are isomorphisms. It follows that the category of strong monoidal functors is isomorphic to the category of isomorphisms $1 \to M$. This is clearly equivalent to $\star$, but not isomorphic to $\star$, since there are many isomorphisms $1 \to M$.

Essentially, the notion of isomorphism of categories is often too strict.

A more interesting example: In the $2$-category of cocomplete symmetric monoidal categories (this includes the condition that $\otimes$ is cocontinuous in each variable), $(\mathbf{Set},\times)$ is bicategorical initial, but there is no $2$-categorical initial object. If $R$ is a commutative ring, then $(\mathbf{Mod}_R,\otimes)$ is bicategorical (and not $2$-categorical) initial in the $2$-category of cocomplete symmetric monoidal $R$-linear categories.

If $X,Y$ are two quasi-compact quasi-separated $R$-schemes, then $\mathbf{Qcoh}(X \times_R Y)$ is the bicategorical coproduct of $\mathbf{Qcoh}(X)$ and $\mathbf{Qcoh}(Y)$ in the $2$-category of cocomplete symmetric monoidal $R$-linear categories (this is the main result of Localizations of tensor categories and fiber products of schemes), but there will not be any $2$-categorical coproduct.

Another simple example: consider the $2$-category of categories with coproducts (or small categories with finite coproducts, this also works) and functors preserving coproducts. Here, the coproduct of a family $(\mathcal{C}_i)_{i \in I}$ is the subcategory of $\prod_{i \in I} \mathcal{C}_i$ consisting of those objects $(x_i)_{i \in I}$ such that almost all $x_i$ are initial. (This is a categorification of the construction of coproducts of abelian groups.) In fact, if $\mathcal{D}$ is a category with coproducts, then there is an equivalence of categories (the $\sqcup$ subscript indicates coproduct-preserving functors) $$\textstyle\hom_{\sqcup}(\coprod_{i \in I} \mathcal{C}_i,\mathcal{D}) \simeq \prod_{i \in I} \hom_{\sqcup}(\mathcal{C}_i,\mathcal{D}).$$ But there is no isomorphism, so again there is no $2$-categorical coproduct.

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    $\begingroup$ PS: I prefer the term "bicategorical limit" instead of "bilimit", because it is more explicit and cannot be confused with other notions where "bi" stands for "twosided". $\endgroup$ Commented Oct 5, 2023 at 10:02
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    $\begingroup$ Nice clear counterexamples — and perhaps worth adding that your example for Q2 is a very typical kind of example: in 2-categories of “categories with structure”, whose maps preserve that structure not-necessarily-on-the-nose, the colimits will generally be only bicategorical not strictly 2-categorical. $\endgroup$ Commented Oct 5, 2023 at 11:29
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    $\begingroup$ Very interesting; am I mistaken, or is your example a non-circular $2$-categorical characterization of the category of sets with Cartesian products? I say non-circular because I’ve seen it argued that characterizations like ‘the terminal Grothendieck topos’ are circular because the definition of a Grothendieck topos involves the category of sets, but that doesn’t seem to be the case here. (maybe in the cocompleteness condition?) $\endgroup$
    – Alec Rhea
    Commented Oct 5, 2023 at 16:10
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    $\begingroup$ @MartinBrandenburg Wow, I was expecting there might be only artificial examples for Q2, but the examples you mentioned are *so* nice. Thank you for the answer, Martin! :) $\endgroup$
    – Emily
    Commented Oct 5, 2023 at 20:08
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    $\begingroup$ I hope that someone else can answer Q3. I saw that you accepted my answer, even though I didn't answer Q3. Maybe ask a separate question? $\endgroup$ Commented Oct 5, 2023 at 23:07
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I wanted to flesh out Martin's nice concrete examples to Question 2 a bit with a more general perspective. As with the 2-category of monoidal categories and strong or lax monoidal functors, basically every 2-category of structured categories and pseudo morphisms will be bicategorically complete and cocomplete but not 2-categorically so. This includes 2-categories like the 2-category of categories admitting $J$-shaped limits or colimits and functors preserving the same as well as various relatives of monoidal categories, and quite generally, the 2-categories of algebras and pseudo-morphisms for any 2-monad $T$ on 2-categories such as those of categories or of multicategories.

These 2-categories, as well as strict 2-initial or 2-terminal objects, lack strict (co)equalizers, pullbacks, and pushouts. For a simple example, let $\star$ be the terminal category and $I$ the category freely generated by an isomorphism. These are both trivial examples of categories with finite products, and both functors $\star\rightrightarrows I$ preserve these products (for instance, because they are equivalences of categories.) But not only does this diagram have no equalizer, there is no object whatsoever in the 2-category of categories with finite products that strictly equalizes these two morphisms! This behavior is quite typical; note that however $\star$ is the equalizer of this diagram in the bicategorical sense.

A good keyword for learning more on this topic is "PIE": a 2-category admitting the specific set Products, Inserters, and Equifiers of weighted 2-limits is always bicategorically complete.

As for Question 3, I commented on the main post that I'm not quite sure that a strict weighted bicategorical limit is a well-defined notion. However certainly it's true that pseudo- lax- and oplax-limits, in the bicategorical sense, can also be given in terms of weighted limits (themselves defined in the pseudo sense.) I think I would approach this by observing that not only does every bicategory have a strictification, i.e. an equivalent 2-category, but it has a continuously equivalent 2-category, namely its image under the bicategorical Yoneda embedding, which is a complete 2-category for which the Yoneda embedding preserves whatever limits exist. So you can convert, say, a bicategorical pseudolimit to a weighted one by converting the corresponding 2-categorical pseudolimit to a weighted one under the Yoneda embedding.

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  • $\begingroup$ 2-categories whose 1-morphisms are lax morphisms between structured categories usually aren't bicategorically complete or cocomplete. For example, given a monoidal category $C$ and a monoid $m$ in $C$, the co-slice category $m/C$ doesn't admit in general a monoidal structure making it the (bicategorical) lax limit of the arrow $m \colon 1 \to C$ in the 2-category of monoidal categories and lax monoidal functors. $\endgroup$ Commented Oct 6, 2023 at 4:33
  • $\begingroup$ @AlexanderCampbell Oh, interesting, I guess I had never actually thought this through beyond the pseudo case. Yes, I suppose you more or less need your monoid operation to be invertible to get a monoidal coslice…At least in return for this public mistake I get to take away some natural examples of non-representable multicategories I hadn’t noticed before! $\endgroup$ Commented Oct 6, 2023 at 5:00
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    $\begingroup$ @KevinArlin Thank you so much for the answer, Kevin! This perspective on completeness with respect to bilimits vs. 2-limits is so nice! On Q3, it's been ages since I've done anything with bilimits actually, I think I'll try reading about them again and see if I can rephrase it into a question that makes sense. Thank you so much again! $\endgroup$
    – Emily
    Commented Oct 6, 2023 at 23:33
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    $\begingroup$ @Emily Thanks for the kind words. $\endgroup$ Commented Oct 7, 2023 at 2:32

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