It's there a way to take a composite monad and a monad map to create a map of the composite?

Question

Let us suppose you start with two monads, $M_S = \langle S , \eta_S, \mu_S \rangle$ and $M_T = \langle T , \eta_T, \mu_T \rangle$ and suppose you have a distributive law, $\lambda: ST \rightarrow TS$ to allow them to combine into a new monad as such $\langle ST , \eta_{ST}, \mu_{ST} \rangle$. Now suppose you have some monad maps defined by natural transformations $F_L: S \rightarrow Q$ and $F_R: T \rightarrow Q'$. Is there a straightforward way to take these monad maps and get new composite monads?

Like this?:

$M_L = \langle QT, \eta_{QT}, \mu_{QT} \rangle$

$M_R = \langle TQ', \eta_{TQ'}, \mu_{TQ'} \rangle$

Answer 1

So, having such a morphism doesn't help to make $QT$ or $SQ'$ into monads. To convince you of this, take $S$ to be the identity monad. For any monad $T$ and $Q$, you always have a (unique) distributive law $ST \to TS$ and a unique morphism of monad $S \to Q$, but when it comes to make $QT$ into a monad this is just two completely general monads, so you'll need a distributive law to get a monad structure. So in general, you can't do anything.

There is still something interesting to be said though.

First a related question: $f:S \to Q$ induce a natural transformation $ST \to QT$ and given distributive laws $\lambda : ST \to TS$ and $\lambda' : QT \to TQ$ one can wonder whether this natural transformation is a morphism of monads.

This has a simple answer: One can show that this happen exactly if the square with $\lambda$ and $\lambda'$ horizontally and $fT$ and $Tf$ vertically commutes.

Ok, now coming back to your problem, the reason why it doesn't work in general is because the map $S \to Q$ gives you no control on $Q$ beyond the "image" of this map. So, a general thing to do is to add the assumption that the maps $S \to Q$ is such that for each object $X$, the map $S(X)$ to $Q(X)$ is an epimorphism (so that $Q$ is a quotient of $S$ in some sense, to make sure that $Q$ is nicely controlled by $S$) then the map $\lambda'$, if it satisfies the condition above, become fully determined by $\lambda$ in a precise sense: The map $ST \overset{\lambda}{\to} TS \overset{Tf}{\to} TQ $ can have at most one factorization into $ST \overset{fT}{\to} QT$, and if such a factorization exists then the map $QT \to TQ$ filling the square is $\lambda'$

If I'm not mistaken - assuming just that $\lambda$ is a distributive law and $f$ is morphism of monads which is levelwise an epimorphism, then if such a factoization $\lambda'$ as above exists (again - if it exists it is unique) one can show that it is a distributive law, hence making $QT$ into a monad and the morphism $ST \to QT$ into a monad morphism.