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Dec 5, 2016 at 7:33 comment added Tim Campion I think if you just write down the definitions it's obviously isomorphic. In both $\mathbf{Endo}$ and $\mathbf{Set}^T$, every object has a uniquely-defined carrier object in $\mathbf{Set}$, and you can write down the bijection between $T$-algebra structures on $A$ and endomorphisms of $A$, which underlies the comparison functor (or rather -- since the comparison functor is well-defined only up to isomorphism -- the obvious choice of comparison functor). I don't see any problems unless maybe if you're trying to work in a weak metatheory or something...
Dec 4, 2016 at 18:36 comment added David Roberts I edited slightly to help the question be more obvious.
Dec 4, 2016 at 18:36 history edited David Roberts CC BY-SA 3.0
Added an actual question mark
Dec 4, 2016 at 17:52 review First posts
Dec 4, 2016 at 18:04
Dec 4, 2016 at 17:48 history asked P. Corazza CC BY-SA 3.0