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Jul 15, 2022 at 19:02 history edited Glorfindel CC BY-SA 4.0
broken link fixed, cf. https://meta.mathoverflow.net/q/5301/70594
Dec 22, 2009 at 19:50 comment added Tom Leinster Greg: yes, the general proof that monadic functors create limits, when specialized to this case, is indeed the same as your proof. (I don't think there could be two substantially different proofs of your claim.) One might say that the inclusion of abelian groups into groups is monadic "because" the subcategory is obtained by imposing an equation in the sense of universal algebra, and monadic functors are all about universal algebra. For instance, every forgetful functor mentioned in part 1 of my answer to mathoverflow.net/questions/5786 is monadic.
Dec 22, 2009 at 17:46 comment added Greg Kuperberg @Tom: I suppose that monadic functors capture and generalizes the ideas in the proof that I sketch? @unknown: You're welcome. If I may make a request too, can you register with your name?
Dec 22, 2009 at 14:29 vote accept user2734
Dec 22, 2009 at 12:36 comment added Tom Leinster Greg's claim also follows from a well-known categorical fact: that if A is a reflective subcategory of a category C then the inclusion of A into C is monadic. (For example, the inclusion of abelian groups into groups is monadic.) Monadic functors create limits, and in particular finite products, giving Greg's Claim.
Dec 22, 2009 at 3:58 history answered Greg Kuperberg CC BY-SA 2.5