Is a monad functor also known as a monad map?

Question

Suppose I have a monad $M_S = \langle S , \eta_S, \mu_S \rangle$. I want to map this monad to another monad, $M_Q = \langle Q , \eta_Q, \mu_Q \rangle$. What is the minimum I have to define to give this monad map? Can it be defined just with a natural transformation $F: S \rightarrow Q$. Is there a freedom to just map one or both of the natural transformations to new natural transformations?

I am looking at Street 72 and in it he defines a monad functor.

Is this the same as a monad map? If so, it looks like you need to define a natural transformation $\phi: S \rightarrow Q$, since $U$ in their notation is identity if we stay on the same base category. Is the monad functor the same as a monad map? I mean, are they saying that a monad functor takes a monad to another monad always? Why don't we have the freedom to change the natural transformations to give a new monad? Why don't we have to check/change the natural transformations?

Answer 1

$\require{AMScd}$I have thought a bit about this recently, because (surprisingly) there is no source that explains the matter in a clear comprehensive fashion: as far as I remember Street doesn't make this simple remark in his paper.

He defines a 2-category $Mnd(Cat)$ (in fact, $Mnd(K)$ for every 2-category $K$) and right after, an obvious functor $Und:Mnd(Cat)\to Cat$ "projecting on the first component", i.e. sending a monad $(C,t)$ to the category $C$, a morphism $(U,\phi) : (C,t)\to (D,s)$ to $U : C\to D$, and similarly on 2-cells.

Now: this definition of a monad morphism is "correct", and modeled on the idea that a monoid is its own regular representation, so monad morphisms shall be equivariant maps (intertwiners: you see why a morphism in the sense of Street is such thing); but what about the good old (and true) motto that a monad is a monoid in the category of endofunctors? Shouldn't then morphisms of monads be monoid homomorphisms, i.e. natural transformations $\alpha : t\Rightarrow s$ such that the usual algebra morphism diagrams $$\begin{CD} tt @>\alpha*\alpha>> ss @.@. 1 @>\eta^t>> t\\ @V\mu^tVV@VV\mu^sV @.@| @VV\alpha V\\ t @>>\alpha> s @.@. 1 @>>\eta^s> s \end{CD}$$ commute?

Well, yes!

So is there a way to relate the two definitions?

Well, yes [whew.jpeg meme]! And the first definition is "correct" because it allows to recover the second: look again at the functor $Und:Mnd(Cat)\to Cat$: its "fiber" over $C$, i.e. the category having objects all the monads $(C,t)$ on $C$ and morphisms the monad maps such that $Und(U,\phi)=id_C$ (so, $U$ is the identity functor, and $\phi : t\Rightarrow s$) is made exactly of the $\phi$'s (specializing Street's diagrams at the beginning of p151) that are "monoid homomorphisms".