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The Eilenberg–Moore construction is essentially a 2-functor from the “2-category of monads on locally small categories” to the “2-category of locally small categories”, modulo certain size issues. In the reverse direction, it is known that any functor $\Phi : \mathcal{C}^\mathbb{S} \to \mathcal{D}^\mathbb{T}$ such that $U^\mathbb{T} \Phi = F U^\mathbb{S}$ for some functor $F : \mathcal{C} \to \mathcal{D}$, where $U^\mathbb{S} : \mathcal{C}^\mathbb{S} \to \mathcal{C}$ and $U^\mathbb{T} : \mathcal{D}^\mathbb{T} \to \mathcal{D}$ are the respective forgetful functors, must come from a unique morphism of monads $\mathbb{S} \to \mathbb{T}$. (This is basically a stronger version of Theorem 6.3 in Toposes, triples and theories.) It is also not hard to come up with functors $\mathcal{C}^\mathbb{S} \to \mathcal{D}^\mathbb{T}$ that do not arise in this fashion.
But what about the 2-cells? Is every natural transformation between functors induced from a morphism of monads also induced by a 2-cell between the monad morphisms? Clearly, the answer is no – any natural transformation that doesn’t factor through the forgetful functor $U^\mathbb{S}$ can’t be induced by a 2-cell between monad morphisms. But I haven’t been able to come up with a proof or counterexample when I assume that the natural transformation does factor through $U^\mathbb{S}$.
Is every natural transformation between functors induced from a morphism of monads also induced by a 2-cell between the monad morphisms? Clearly, the answer is no – any natural transformation that doesn’t factor through the forgetful functor $U^\mathbb{S}$ can’t be induced by a 2-cell between monad morphisms. But I haven’t been able to come up with a proof or counterexample when I assume that the natural transformation does factor through $U^\mathbb{S}$.
Good question – the short answer is ’yes’.
Just to lay all the cards on the table, I assume that by a monad morphism from a monad $(C, S, m_S: S S \to S, u_S: 1_C \to S)$ to a monad $(D, T, m_T: T T \to T, u_T: 1_D \to T)$, you mean a pair $(F: C \to D, \phi_F: T F \to F S)$ where the natural transformation $\phi_F$ satisfies an obvious compatibility with the monad multiplications and the monad units (taking the shape of a pentagon and a unit, much as in the case of a distributive law, which is actually a special case). (I point this out because if you don’t think about it too hard, one could guess the opposite direction for $\phi_T$!)
By a 2-cell from a monad morphism $(F, \phi_F)$ to a monad morphism $(G, \phi_G)$, you must mean a transformation $\eta: F \to G$ satisfying an obvious compatibility with the $\phi$’s.
Okay, suppose functors $F, G: C \to D$ have lifts to functors $\Phi, \Psi: C^S \to D^T$; then, as you say, $F$ and $G$ become endowed with appropriate transformations $\phi_F$, $\phi_G$, making them monad morphisms. Now suppose we have a transformation $\theta: \Phi \to \Psi$ which “descends” to a transformation $\eta: F \to G$, in the sense that $\eta U_S = U_T \theta$, where $U_S$, $U_T$ are the forgetful functors from Eilenberg-Moore categories. We want to show $\eta$ is a transformation between the monad morphisms.
Here is the critical diagram you need to stare at:
$\array{ T F & \stackrel{T u_T F}{\to} & T T F & \stackrel{T\phi_F}{\to} & T F S & \stackrel{\phi_F S}{\to} & F S S & \stackrel{F m_S}{\to} & F S \\ ^\mathllap{T \eta} \downarrow & & & & \downarrow^\mathrlap{T\eta S} & & & & \downarrow^\mathrlap{\eta S} \\ T G & \underset{T u_T G}{\to} & T T G & \underset{T\phi_G}{\to} & T G S & \underset{\phi_G S}{\to} & G S S & \underset{G m_S}{\to} & G S }$It’s not too hard to see, using the fact that $(F, \phi_F)$ is a monad morphism and one of the unit laws for a monad, that the long top horizontal composite is $\phi_F$; similarly, the long bottom horizontal composite is $\phi_G$. So we’re done if we show the entire diagram commutes. Now the right-hand rectangle commutes essentially because $\eta$ lifts to a transformation $\theta: \Phi \to \Psi$. The left rectangle commutes because it’s $T$ applied to a naturality square for $\eta$ in disguise – one has to rewrite $\phi_F \circ u_T F = F u_S$ and $\phi_G \circ u_T G = G u_S$, again using the fact that $(F, \phi_F)$ and $(G, \phi_G)$ are monad morphisms, to remove the disguise.
Perfect, thanks! I got stuck because I subdivided that diagram too much…
Yup! The same thing happened to me :-)
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