Let $M, N$ be monads of rank $\lambda$, where $\lambda$ is a regular cardinal (I'm primarily interested in the case of finitary monads). Is there a known characterization of functors $\mathrm{Alg}~N \to \mathrm{Alg}~M$ (of course, I mean the Eilenberg-Moore category) coming from monad morphisms $M \to N$ (that is, a morphism of monoids in the monoidal category $\mathrm{End}(\mathrm{Set})$)?
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asked Jun 8, 2023 at 20:51
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The functor $U_{({-})} : \mathrm{Mnd}(\mathrm{Set})^\circ \to \mathrm{CAT}/\mathrm{Set}$, sending each monad to the forgetful functor from its category of algebras, is fully faithful (Theorem 3 of Frei's Some remarks on triples). Therefore, a functor between categories of algebras for two monads (regardless of rank) corresponds to a monad morphism if and only if the functor commutes with the forgetful functors.
Given such a functor $F : \mathrm{Alg}(N) \to \mathrm{Alg}(M)$, we define a monad morphism $M \to N$ as follows. For each set $x$, we have a free $N$-algebra $Nx$, which is sent by $F$ to an $M$-algebra on $Nx$. Concretely, such an algebra comprises a function $MN x \to N x$ satisfying two laws. Given this function, we precompose by the unit $M\eta_x : Mx \to MNx$ of $N$, producing a function $Mx \to Nx$. This defines a natural transformation $M \Rightarrow N$, and it can be checked that this is a monad morphism.
In fact, the corresponding statement is true for monads on any category, not just $\mathrm{Set}$.
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1$\begingroup$ Thanks a lot! Moreover, all such functors automatically have rank $\lambda$ (because their composition with a forgetful functor (which has rank $\lambda$ and is conservative) is a functor of rank $\lambda$) and preserve all limits (for a similar reason), hence are right adjoint functors. $\endgroup$ Commented Jun 8, 2023 at 21:39
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2$\begingroup$ The fact is that I expected at least such functor properties as right adjoint, rank and wanted to know the full list. But it turned out that everything is so simple. Thanks again! $\endgroup$ Commented Jun 8, 2023 at 21:41
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$\begingroup$ I have now found that the same is true for an arbitrary category $C$ ($\mathrm{Set}$ is not essential here). See Francis Borceux - Handbook of categorical algebra II, Prop 4.5.9. $\endgroup$ Commented Jun 8, 2023 at 22:35
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2$\begingroup$ @ArshakAivazian: yes, I should have mentioned that this doesn't just work for Set (Frei works with monads on an arbitrary base category, for instance). Glad I could help :) $\endgroup$– varkorCommented Jun 9, 2023 at 5:33