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

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.