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](http://ncatlab.org/nlab/show/monadicity+theorem). In CM, the forgetful functor should create [reflexive coequalizers](http://ncatlab.org/nlab/show/reflexive+coequalizer).


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?



The crude monadicity theorem
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](http://www.emis.de/journals/CMUC/pdf/cmuc0003/adamvele.pdf) 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. 


