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I was recently reading free monad, and in the section where it is shown that an algebraically free monad is free, it seems to me there is a slight level slip: on the one hand we have 1-categories of (pointed) endofunctors and of monads on , and on the other we have a 2-category (insofar as an algebraically free monad said to involve an equivalence, not an isomorphism).
I think this must be one of those (maybe rare?) situations where we want to be “evil” and demand an isomorphism of categories instead of equivalence, in order to make the argument go through that an algebraically free monad is free. (Kelly in his long paper on transfinite construction of free algebras also mentions isomorphisms.) Agree?
If so, then a small remark (on the occasional “necessity of evil”) should also go in monadicity theorem as well.
Hmmm. There’s probably some kind of trick one can use to rectify the equivalence to an isomorphism, since the forgetful functor from the category of algebras for an endofunctor or monad is an amnestic isofibration. But yes, on the face of it, it does seem as if this situation calls for an isomorphism of categories over in the strictest possible sense.
I think Zhen is right: the equivalence must be an equivalence over in , and all the objects of involved are 2-categorically discrete, in that any natural transformation into them is an identity. Thus, any equivalence between them is automatically an isomorphism. And yes, this should be mentioned; thanks for noticing it!
Thanks Zhen, and thanks Mike! I was a little slow on the uptake because I’m actually on vacation, but it makes sense now. I inserted a remark at free monad taking into account amnestic isofibrations (I see now they were already mentioned at monadic functor, but I added a similar remark at monadicity theorem).
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