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In order to test the monadicity of a functor, there is a precise monadicity theorem (PM) as well as a crude monadicity theorem (CM), see the nlab. In CM, the forgetful functor should create reflexive coequalizers.

What is a nice and specific example which shows that CM is only sufficient, but not necessary; i.e. what is a monad which doesn't preserve reflexive coequalizers? I couldn't find such an example.

Besides, is the condition in CM satisfied in almost all practical / non-pathological examples? For example it holds for algebraic monads on $\mathsf{Set}$. What is the intuitive difference between PM and CM? The background for this rather soft question is the following: I would like to prove a certain theorem about certain monads, and in one step it would be very convenient if the monad preserved reflexive coequalizers. Now I wonder if there is any problem when I just add this assumption to the theorem.

Given a monad defined by generators and relations (on a category different from $\mathsf{Set}$), how can I test practically if the forgetful functor creates reflexive coequalizers?

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    $\begingroup$ This question was asked and answered several years ago, but I thought I should comment anyway: observe that the conditions in the crude monadicity theorem are composable. That is, a pair of "crudely monadic functors" compose to another "crudely monadic functor". However, there are a few classical examples showing that monadic functors don't compose to monadic functors in general, so at least one of the functors involved in such examples must fail to be monadic. This perhaps raises the question of whether there exists a maximal class of composable monadic functors in a given context. $\endgroup$
    – Morgan Rogers
    Commented Mar 13, 2019 at 14:08

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For $X$ an infinite set, the monad $(-)^X$ (induced from the comonoid structure on $X$ with respect to cartesian product) does not preserve reflexive coequalizers. See page 538 of this paper by Adámek, Koubek, and Velebil. Correspondingly, the forgetful functor for the category of algebras does not preserve reflexive coequalizers (hence also cannot reflect/create them since the forgetful functor reflects isomorphisms).

Any finitary (i.e., filtered-colimit preserving) monad on $Set$ does preserve reflexive coequalizers, because finitary powers $(-)^n$ do. This implies that the forgetful functor will reflect reflexive coequalizers. So the practical difference shows up in finitary vs. infinitary monads.

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    $\begingroup$ Thank you for this instructive example. The coequalizer of $f^I,g^I$ in that example is the set of equivalence class of "numberings" of $I$, where two such numberings are equivalent iff their difference is bounded. Thus if $I$ is finite, there is only one class, but if $I$ is infinite, there will be many classes. // Instead of $(-)^X$ we could also take the power set monad? $\endgroup$
    – Martin Brandenburg
    Commented Jan 5, 2013 at 13:45
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    $\begingroup$ Yes, good question. I'm not sure off-hand for the covariant power set monad, whose algebras are sup-lattices, but I'll think about it. However, for the double power-set monad where the monadic functor can be identified with $P: \mathrm{Set}^{op} \to \mathrm{Set}$, it's true that this monad preserves reflexive coequalizers (and in fact it's the crude monadicity theorem which is typically used here, as in Mac Lane-Moerdijk, IV.5 theorem 2 (page 182 in my edition)). So you're right that some infinitary monads do preserve reflexive coequalizers. $\endgroup$
    – Todd Trimble
    Commented Jan 5, 2013 at 15:53
  • $\begingroup$ The same example works for the power set monad. $\endgroup$
    – Martin Brandenburg
    Commented Jan 12, 2013 at 16:19
  • $\begingroup$ I'm glad you worked that out, Martin, because I had forgotten my promise to think about it. :-( $\endgroup$
    – Todd Trimble
    Commented Jan 12, 2013 at 17:06

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