# Constrained factorization of monads

Suppose $A\dashv B\colon \cal H\to K$ are adjoint functors; if the comonad $AB$ has a right adjoint $T\colon\cal K\to K$ then $T$ is a monad in a natural way.

I would like to tell when "$T$ can be factored through $B$", i.e. $T=CB$ for some $B\dashv C$, $C\colon \cal H\to K$.

This is sort of a converse to the general statement that when you have $A\dashv B\dashv C$ then the comonad $AB$ has as right adjoint the monad $CB$.

Since my interest in this question comes from cartesian closed categories (I can provide details, see below), I tried to skim the Elephant looking for a clue, but I didn't find the right keyword... can you help me?

Edit: This is where I found the problem: consider a category with pullbacks $\cal C$ and the functor ${\cal C}/U\to {\cal C}/V$ obtained by an arrow $\varphi\colon V\to U$; it is well known that if $\cal C$ is a topos, then there is an adjunction $$(\Sigma_\varphi \dashv \varphi^*\dashv \Pi_\varphi) \colon {\cal C}/V\underset{\xrightarrow[\Pi_\varphi]{}}{\overset{\xrightarrow{\Sigma_\varphi}}{\longleftarrow}} {\cal C}/U$$ and since the functor $\Sigma_\varphi\circ \varphi^*$ (a comonad) coincides with "pulling back with $\varphi$" (without changing base), the right adjoint $\Pi_\varphi\circ \varphi^*$ must be the internal hom in ${\cal C}/U$, which is hence cartesian closed.

Suppose now that you don't know $\cal C$ is a topos, but suppose that you know that $\Sigma_\varphi\circ \varphi^*$ has a right adjoint $T$, forced to be a monad. If you are able to factor $T$ along $\varphi^*$ as $T=P_\varphi\circ \varphi^*$ then there is a unique choice for $P_\varphi$. If this factorization is possible for a particular $\varphi$, then ${\cal C}/U$ is cartesian closed.

Skimming through the nlab page about LCCC I see (Prop. 2) that whenever every slice category ${\cal C}/U$ is a cartesian closed category (i.e. $\Sigma_\varphi \varphi^*\dashv T$ for any $\varphi$), then for every morphism $\varphi\colon X\to Y$ the dependent product $\varphi^*\dashv \Pi_f$ exists. This is the "constrained factorization" I wondered, since this is telling you that anytime $\Sigma_\varphi\circ \varphi^*$ admits a right adjoint, this right adjoint factors through $\varphi^*$ giving your "dependent product".

So we come my question: I wanted to generalize this situation to other contexts.

• I don't understand your question. What does it mean "$T$ can be factored through $B$", i.e. $T=CB$ for some $B \dashv C$? What does "some" mean here? If $B$ is given, then there are not so many right adjoints $C$... Nov 27, 2013 at 22:59
• "Some" = it is unique up to natural iso; nothing more. :) Nov 28, 2013 at 11:25

## 1 Answer

If $A\dashv B\colon \cal H\to K$, then the condition $AB$ has right adjoint $T$ is clearly equivalent to the condition that:

$$\hom(A(B(-)), =) \approx \hom(B(-), B(=))$$

is representable. You may actually think of $\hom(B(-), B(=))$ as of the canonical "promonad" (a monad in the bicategory of profunctors) associated to $B$. Such a promonad is representable if and only if $B$ (as a functor) has an absolute right lifting along itself (this is a form of op-density condition).

Therefore, up to the fact that $B$ has left adjoint, your question may be rephrased as: "when does a functor, which has an absolute right lifting along itself, have a right adjoint?". I doubt there are general conditions (unless trivial) to answer this question.

• Thank you for your answer. I like your use of promonads! I edited the OP to make clearer my intent (which yesterday seemed quite foggy even to me). Nov 28, 2013 at 11:24
• @tetrapharmakon, the case of a codomain fibration is very special --- there are a lot of different concepts that coincide for a codomain fibration. I don't see how you could generalise this situation in the proposed direction. However, I'm willing to believe, that it may be possible to generalise the case of a codomain fibration in the following way: if a fibration has products along canonical morphisms between cartesian products in the base category, and is fibrewise cartesian closed with finite limits, then under some mild conditions, it has all fibred products. Nov 28, 2013 at 14:36