## Lax Functors and Equivalence of Bicategories?

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Lax functors of bicategories were introduced at the very inception of bicategories, and I'm trying to get a better feel for them. They are the same as ordinary 2-functors, but you only require the existence of a coherence morphism, not an isomorphism. The basic example I'm looking at are when you have a lax functor from the singleton bicategory to a bicategory B. These are just object b in B with a monad T on B.

My Question: If I have an equivalence of bicategories A ~ A', do I a get equivalent bicategories of lax functors Fun(A, B) and Fun(A', B)? If not, is there any relation between these two categories?

So let me be more precise on the terminology I'm using. I want to look at lax functors from A to B. These are more general then strong/pseudo and much more general then just strict functors. For a lax functor we just have a map like this:

$F(x) F(g) \to F(fg)$

for a strong or pseudo functor this map is an isomorphism, and for strict functor it is an identity. I don't care about strict functors.

I'm guessing that these form a bicategory Fun(A,B), with the 1-morphism being some sort of lax natural transformation, etc, but I don't really know about this. Are there several reasonable possibilities?

When I said equivalence between A and A' what I meant was I had a strong functor F:A --> A' and a strong functor G the other way, and then equivalences (not isomorphisms) FG = 1, and GF = 1. This seems like the most reasonable weak notion of equivalence to me, but maybe I am naive.

I haven't thought about equivalences using lax functors. Would they automatically be strong? What I really want to understand is what sort of functoriality the lax functor bicategories Fun(A, B) have?

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You're asking about lax functors and not about pseudofunctors, right? Just making sure. – Harry Gindi Dec 3 2009 at 14:23

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Chris, could you make a couple of things in the question more explicit? First, when you say "equivalence of bicategories", do you intend the functors F and G going back and forth to be weak/pseudo, or just lax? And do you intend FG to be isomorphic to the identity, or just equivalent? (Standard terminology, which I'm no fan of, would choose the first answer to both questions.) Finally, are the 1-morphisms in Fun(A, B) the weak/pseudo natural transformations, or the lax ones? – Tom Leinster Dec 3 2009 at 16:50
As far as the 1-morphisms in Fun(A,B) are concerned, I would be happy to know what happens in either case, i.e. where we use only weak/pseudo transformations or when we use the lax ones. – Chris Schommer-Pries Dec 3 2009 at 18:07
Thanks for clarification. I agree, this is very much the most reasonable notion of equivalence. It's just that in the (questionable) classical terminology, this is called "biequivalence", with "equivalence" reserved for when FG is actually isomorphic to the identity. – Tom Leinster Dec 3 2009 at 18:56

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Thank you for asking this question! I no longer feel guilty about not believing in lax functors. – Reid Barton Dec 3 2009 at 19:49
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## 2 Answers

First of all, for any two bicategories A and B, there is a bicategory $Fun_{x,y}(A,B)$ where x can denote either strong, lax, or oplax functors, and y can denote either strong, lax, or oplax transformations. There's no problem defining and composing lax and oplax transformations between lax or oplax functors, and the lax/oplax-ness doesn't even have to match up. It's also true that two x-functors are equivalent in one of these bicategories iff they're equivalent in any other one. That is, any lax or oplax transformation that is an equivalence is actually strong/pseudo.

Where you run into problems is when you try to compose the functors. You can compose two x-functors and get another x-functor, but in general you can't whisker a y-transformation with an x-functor unless x = strong, no matter what y is, and moreover if y isn't strong, then the interchange law fails. Thus you only get a tricategory with homs $Fun_{x,y}(A,B)$ if x=y=strong. (In particular, I think this means that there isn't a good notion of "equivalence of bicategories" involving lax functors.)

For a fixed strong functor $F\colon A\to A'$, you can compose and whisker with it to get a functor $Fun_{x,y}(A',B) \to Fun_{x,y}(A,B)$ for any x and y. However, the same is not true for transformations $F\to F'$, and the answer to your question is (perhaps surprisingly) no! The two bicategories are not equivalent.

Consider, for instance, A the terminal bicategory (one object, one 1-morphism, one 2-morphism) and A' the free-living isomorphism, considered as a bicategory with only identity 2-cells. The obvious functor $A' \to A$ is an equivalence. However, a lax functor from A to B is a monad in B, and a lax functor from A' to B consists of two monads and a pair of suitably related "bimodules". If some lax functor out of A' is equivalent to one induced by composition from A (remember that "equivalence" doesn't depend on the type of transfomation), then in particular the two monads would be equivalent in B, and hence so would their underlying objects. But any adjunction in B whose unit is an isomorphism gives rise to a lax functor out of A', if we take the monads to be identity 1-morphisms, the bimodules to be the left and right adjoint, and the bimodule structure maps to be the counit and the inverse of the unit. And of course can have adjunctions between inequivalent objects.

By the way, I think your meaning of "equivalence" for bicategories is becoming more standard. In traditional literature this sort of equivalence was called a "biequivalence," because for strict 2-categories there are stricter sorts of equivalence, where you require either the functors to be strict, or the two composites to be isomorphic to identities rather than merely equivalent to them, or both. These stricter notions don't really make much sense for bicategories, though. For instance, in a general bicategory, even identity 1-morphisms are not isomorphisms, so if "equivalence" were to demand that FG be isomorphic to the identity, a general bicategory wouldn't even be equivalent to itself!

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I was surprised (and delighted) to discover that the answer is no. Here is an example; in fact it's a special case of the example Mike referred to, but worked out in more detail.

Let S be a set and cd(S) the codiscrete category on S (Hom(x, y) = • for every x and y in S). Let BSet denote the one-object bicategory corresponding to the monoidal category (Set, ×). As remarked here on the nlab, a lax functor cd(S) → BSet is the same data as a category with object set S. However, there is a catch: the morphisms (whether lax or strong natural transformations) are not just functors between categories fixing the objects. Instead, if C and D are categories corresponding to two lax functors cd(S) → BSet, then a morphism (X, F) from C to D consists of

• for each object s of S, a set X(s),
• for each pair of objects s and t of S, a map F(s, t) : C(s, t) × X(t) → X(s) × D(s, t),
• such that obvious identity and composition laws hold.

Depending on which notion of natural transformation you want to take, the map F(s, t) might be an isomorphism, or it might go the other way. I'll consider all three possibilities simultaneously.

We can compose 1-morphisms: the composition (Z, H) of (X, F) and (Y, G) has Z(s) = X(s) × Y(s), and H formed from F and G in an obvious way. The identity morphism has X(s) = • and F(s, t) = id.

We also have 2-morphisms from (X, F) to (Y, G), which are families of maps X(s) → Y(s) making some diagrams involving X, F, Y, G commute.

This doesn't yet seem to be any familiar 2-category. Let's extract the maximal 2-groupoid by taking only invertible 1- and 2-morphisms. (That's certainly an equivalence-invariant thing to do.) The invertible 2-morphisms are the ones where the maps X(s) → Y(s) are isomorphisms. For (X, F) to be invertible, we first need for every s, an object Y(s) and an isomorphism X(s) × Y(s) = •. That can only happen if X(s) = • for every s. (This is the only property of (Set, ×) I really care about: it has no invertible objects besides •.) Now F is just an ordinary functor C → D which is the identity on objects, and for (X, F) to be invertible F needs to be an isomorphism. When (X, F) and (Y, G) are invertible, so X(s) = Y(s) = •, there is at most one 2-morphism between them, exactly when F and G are equal as maps C(s, t) → D(s, t). In conclusion, the maximal 2-groupoid contained in the 2-category of lax functors cd(S) → BSet is the 1-groupoid of categories with object set S and isomorphisms fixing the objects. Note that it didn't matter for the argument which notion of natural transformation I chose.

Now as S varies over nonempty sets, these resulting groupoids are nonequivalent categories. However, surely the 2-categories cd(S) are all equivalent under any reasonable definition, since they equivalent 1-categories.

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