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Stream: theory: category theory

Topic: the 2-category of adjoints

Tim Campion (Jul 24 2022 at 16:30):

Let $C$ be a 2-category. Then there is a wide, locally-full sub-2-category $Adj(C) \subseteq C$ , whose 1-morphisms are those 1-morphisms of $C$ which have a right adjoint in $C$ .

The construction $C \mapsto Adj(C)$ is an endofunctor of $2Cat$ , and the inclusion $Adj(C) \to C$ makes this into a co-pointed endofunctor. I think it's even co-well-pointed. However, I don't think that this co-well-pointed endofunctor is a comonad.

This is a little weird. On the one hand I tend to think of $C \mapsto Adj(C)$ as an "intrinsically interesting construction". On the other hand, I tend to think of endofunctors which are not co/monads as "technical tools" --surely any endofunctor worth its salt will have the decency to arise from some more fundamental adjunction, and thus be a co/monad, right? So I guess I'm wondering

Is there some other construction $C \mapsto foo(C)$ , related to $C \mapsto Adj(C)$ , which is more fundamental (and which perhaps is a co/monad or arises from some adjunction)? Whenever I see $C \mapsto Adj(C)$ , should I really be thinking about $C \mapsto foo(C)$ ?
If not, then does $C \mapsto Adj(C)$ have a bit more structure than being a co-well-pointed endofunctor, which might make me feel a bit more "at ease" with this construction?

Tim Campion (Jul 24 2022 at 16:33):

For instance, one construction which might be relevant is the construction which turns $C$ into the double category whose horizontal part is $C$ and whose vertical part is $Adj(C)$ (with the "obvious" squares). I'm not sure if this construction has better formal properties than $C \mapsto Adj(C)$ .

Mike Shulman (Jul 24 2022 at 17:01):

Well, I must say I disagree with your opinion about endofunctors. I think the world is littered with endofunctors that are not monads or comonads.

Mike Shulman (Jul 24 2022 at 17:09):

But it is possible to decompose $\rm Adj$ (or something equivalent to it) into a composite of other functors in a way that might be illuminating. For instance, consider the composite ${\rm Comp}\circ H \circ \rm Sq$ where:

$\rm Sq : 2Cat \to Dbl$ constructs the double category of squares (= quintets) in a 2-category.
$H : \rm Dbl \to Dbl$ constructs the horizontal opposite of a double category.
$\rm Comp : Dbl\to 2Cat$ constructs the 2-category of [[companion pairs]] in a double category.

The composite ${\rm Comp}\circ H$ constructs the 2-category of [[conjunctions]] in a double category, and a conjunction in ${\rm Sq}(C)$ is precisely an adjunction in $C$ . So this composite constructs the 2-category of adjunctions in a 2-category. This is biequivalent to your ${\rm Adj}(C)$ , but includes the whole adjunction as data, rather than its mere existence as a property.

In addition, we have an adjunction $\rm Sq \dashv Comp$ , so the composite $\rm Comp \circ Sq$ is a monad (though it is biequivalent to the identity). So this modified version of $\rm Adj$ is a "twist" ( $H$ ) away from being a monad.

John Baez (Jul 24 2022 at 17:13):

Nice!

Tim Campion (Jul 24 2022 at 17:25):

Oh that is very nice!

Tim Campion (Jul 24 2022 at 17:26):

@Mike Shulman I've had trouble finding an actual reference for the adjunction $Sq \dashv Comp$ -- is this in the literature somewhere?

Tim Campion (Jul 24 2022 at 17:27):

While I'm at it -- in the $\infty$ -categorical setting, Gaitsgory and Rozenblyum conjectured that $Sq$ is fully faithful, but I don't know a reference for this in the strict setting even. Do you?

Tim Campion (Jul 24 2022 at 17:29):

(I also don't know a reference for the left adjoint of $Sq$ , which is related to the Gray tensor product, but this is taking me increasingly further afield!)

Mike Shulman (Jul 24 2022 at 17:29):

No, I don't know a reference for that adjunction.

Mike Shulman (Jul 24 2022 at 17:31):

We discussed the putative full-faithfulness of $\rm Sq$ here. I don't think I have anything to add to that.

Tim Campion (Jul 24 2022 at 17:32):

Oh you're right -- we've talked about this before!

Tim Campion (Jul 24 2022 at 17:33):

Right -- I think my takeaway from that discussion was that if $Sq$ is fully faithful, then it's only when you take the domain of $Sq$ to be (Rezk-complete) $(\infty,2)$ -categories (or maybe $(m,2)$ -categories for large enough $m$ ).

David Kern (Aug 09 2022 at 14:51):

Tim Campion said:

Mike Shulman I've had trouble finding an actual reference for the adjunction $Sq \dashv Comp$ -- is this in the literature somewhere?

There is Theorem 1.7 in Grandis–Paré's Adjoints for double categories.