4
$\begingroup$

Let $F : A \to B$ be a functor, and suppose that the right Kan extension $T = Ran_F F : B \to B$ exists. Then $T$ is a monad, the codensity monad of $F$. Moreover, unless I'm mistaken there is a canonical factorization of $F$ through the category of algebras of $T$: $A \xrightarrow{\tilde F} Alg(T) \xrightarrow{U^T} B$ (at least up to some canonical 2-cell?).

I want to think of $\tilde F : B \to Alg(T)$ as a "jazzed-up" version of $F$, which remembers everything that $F$ remembers in a "maximally structured" way. I'd like to express this as some kind of universal property of $\tilde F$.

Question 1: Is the factorization $F = U^T \circ \tilde F$ part of some kind of factorization system whose right half is the monadic functors? If so, what is the left half (i.e. what are the characteristic properties of the functor $\tilde F$)?

Question 2: If we don't have a factorization system, then is there still something to be said about the passage $F \mapsto \tilde F$? Is it at least left adjoint to something (as a functor from $Cat_{A/}$ to some subcategory of $Cat_{A/}$, perhaps)?

If it helps to assume that $A,B,F$ have nice properties, that's fine by me. I'm also interested to understand the dual situation of the the density comonad.

Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ Nunes's Semantic Factorization and Descent may be related to what you are looking for (in particular, the "semantic factorisation" described within). $\endgroup$
    – varkor
    Feb 10, 2022 at 18:46

1 Answer 1

Reset to default
2
$\begingroup$

There is something in $Cat/B$ : $Alg(T) \to B$ is the universal monadic right adjoint through which $F$ factors.

There's probably a cleaner way to see this, but here is how I think about it :

It is easy to see that a factorisation of $F :A \to B$ through a monadic functor $B^M \to B$ is the same as an action of the monad $M$ on $F$.

Now, codensity monad are the same as "endomorphisms monad" (in the sense of endomorphism objects in a monoidal category acting on a category), so such an action is the same as a morphism of monad $M \to End(F)=T$ which in turn corresponds to a forgetfull functor $B^{End(F)} \to B^M$ over $B$.

Putting together, you get that a functor $A \to B^M$ over $B$, is the same as functor $B^{End(F)} \to B^M$ over $B$.

I guess an advantages of this point of view is that it is easier to generalizes $(\infty,1)$-categories as the endomorphism monad point of view is how Lurie deals with monadicity in Higher Algebra...

Improve this answer
$\endgroup$
7
  • $\begingroup$ Great, thanks! So I guess that just leaves the question of which functors are left orthogonal to monadic functors... $\endgroup$
    – Tim Campion ♦
    Feb 10, 2022 at 17:55
  • $\begingroup$ @TimCampion I'm pretty sure there is a mistake in what I'm going to say but... If I assume all categories involved are locally presentable and all functors accessible then monadic right adjoint functor are stable under pullback, so the argument above is enough to show that one has a unique factorization system. The left class is then exactly the functors whose factorization gives $A \to B = B$, that is the functor whose codensity monad is the identity, that is the codense functor ? That's seem strange though... $\endgroup$
    – Simon Henry
    Feb 10, 2022 at 18:32
  • 1
    $\begingroup$ I think I got confused -- even with presentability assumptions, monadic functors are not closed under composition, right? This leads me to think that the best one can hope for is an "over $B$" statement. Another point leading me to think this is that the functors you describe would be exactly the codense functors, but I'm pretty sure that $\tilde F$ need not be codense -- the factorization functor is not idempotent if we allow $B$ to vary. $\endgroup$
    – Tim Campion ♦
    Feb 10, 2022 at 18:36
  • $\begingroup$ No I think I'm right on that part. Using Beck's criterion you can see that a pullback of a monadic right adjoint functor is monadic when it is a right adjoint. So in situation where the special adjoint functor theorem applies, pullback of monadic functor are monadic ? $\endgroup$
    – Simon Henry
    Feb 10, 2022 at 18:48
  • 1
    $\begingroup$ Probably you're right about stability under pullbacks -- what I'm saying is that in order to get a factorization system, that's not all you need to check. One of the other things you need to check is stability under composition, which I think fails. For instance, if $C$ is a small category and $K$ is a reflective subcategory of presheaves on $C$, then $K \to Psh(C) \to Psh(Ob C)$ is a composite of monadic functors which is typically not monadic. $\endgroup$
    – Tim Campion ♦
    Feb 10, 2022 at 18:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.