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There is a notion of morphism from a monad $T:\mathscr C\to \mathscr C$ to another one $T':\mathscr C'\to \mathscr C'$. It arose here on MO e. g. in "Functors between monads": what are these really called? or Distributive law between Kleisli triples. Such morphisms have been considered by Pumplün, Street and probably others, they are functors $F:\mathscr C\to\mathscr C'$ together with a transformation $T'F\to FT$ with some coherence conditions generalizing those for a distributive law. The idea is to have an induced functor between Eilenberg-Moore algebras.

While this is certainly "the" correct notion, I've recently encountered a situation when it is not. With several colleagues we are studying varieties like meet-semilattices-with-a-nucleus; "correct" morphisms $(M,j)\to(M',j')$ must be $f:M\to M'$ which preserve meets and satisfy $j'f=fj$, not just $j'f\leqslant fj$ as it would be if one would view $j$ and $j'$ as monads and take monad morphisms in the above sense.

The natural "categorified" version of such morphisms would then be natural isomorphisms $T'F\cong FT$; that is, a $\textit{pseudo}$ version, while Pumplün-Street notion would then be the $\textit{lax}$ version.

Question: can one distinguish such "strict"/"pseudo" morphisms of monads among more general ones by some abstract-nonsensical properties? Could be something like having functors both between Eilenberg-Moore and Kleisli categories, compatible in certain way. Somehow the precise formulation escapes me.

Is anybody aware of some work on these?

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I don't know of very much work about these specifically, or a characterization of them in terms of how they act on Eilenberg-Moore or Kleisli categories. But there is a precise sense in which they are the "pseudo" (or "strong") to Street's "lax" morphisms. Namely, there is a 2-monad $M$ on $Cat$ whose algebras are categories equipped with a monad, and these are the pseudo and lax $M$-morphisms respectively.

By the way, I've found uses for this kind of monad morphism too; see for instance Definition 11.5 and Lemma 11.10 of this paper.

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  • $\begingroup$ Sorry I definitely see it is relevant but cannot get to the actual connection. What would be these strict/strong/pseudo guys in your context? Cartesian morphisms in some associated fibration? $\endgroup$ – მამუკა ჯიბლაძე Jul 26 '17 at 3:51
  • $\begingroup$ @მამუკა ჯიბლაძე what do you mean by "in your context"? Are you asking about my paper with Peter that I linked in the second paragraph? I was just saying that there we used pseudo monad morphisms (under the name "strong monad morphisms") at the cited places. $\endgroup$ – Mike Shulman Jul 26 '17 at 13:54
  • $\begingroup$ Yes I meant this, and also that I do not see well enough why these strong monad morphisms appear in your context. The little that I was able to realize is that these fibred monads that you obtain arise when you have a "uniform mechanism" to produce monads on many categories at once. Like, for example, the free monoid monad exists on many (all?) categories with countable coproducts and finite products. Clearly any functor preserving these coproducts and products comes equipped with such a strong morphism between free monoid monads. But your setup seems to be much more general than this. $\endgroup$ – მამუკა ჯიბლაძე Jul 26 '17 at 15:50
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    $\begingroup$ @მამუკა ჯიბლაძე To construct a free monoid monad using countable coproducts, you need the finite products to preserve the countable coproducts in each variable. The transfinite construction of free algebras (ncatlab.org/nlab/show/transfinite+construction+of+free+algebras) is indeed more general, but your intuition is right: whenever a monad is constructed in a particular way, then any functor that preserves the constructions that go into it will also be a strong monad morphism. $\endgroup$ – Mike Shulman Jul 28 '17 at 5:14
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    $\begingroup$ I don't know where it is written down, but it has as easy description. A monad on a category $C$ is the same as a monoid in the strict monoidal category $[C,C]$ of endofunctors. The augmented simplex category $\Delta_a$ is the free strict monoidal category on a monoid. Thus, a monad on $C$ is the same as a strict monoidal functor $\Delta_a \to [C,C]$, or equivalently an action on $C$ of the monoid $\Delta_a$ in $\mathrm{Cat}$. Thus, $M(C) = \Delta_a \times C$. $\endgroup$ – Mike Shulman Jul 28 '17 at 5:52

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