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- Thread starter rasmhop
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Hurkyl

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Well, the easiest answer is to use foundations that include large cardinals. Hom(A,-) and F are just large sets, and so Nat is a large set of large sets.

Second-order ZFC should do the trick too. Hom(A,-) and F are first-order classes, so Nat(Hom(A,-), F) would be a second-order class.

I think you could manage the same trick in first-order NBG, since Hom(A,-) and F are classes (in the sense of being objects), and Nat(Hom(A,-),F) would be a class (in the sense of being a logical predicate).

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For a fixed object [itex]B[/itex] in category C both hom(A,B) and FB are sets so the class of functions from hom(A,B) to FB is a set. Hence if [itex]\eta[/itex] is a natural transformation from hom(A,-) to F, then [itex]\eta_B[/itex] is a set and thus [itex]\{\eta_B | B \in ob(C)\}[/itex] is a class in the sense of NBG. Hence we can define Nat(hom(A,-),F) in terms of a formula [itex]\varphi(x,p_1,\ldots,p_n)[/itex] where [itex]p_1,\ldots,p_n[/itex] are free variables and x is a class. This is sufficient to allow us to set up the bijection from Nat(hom(A,-),F) to FA in the usual sense and that shows that Nat(hom(A,-),F) is actually a set.

It just seems odd to me that such an argument is omitted, and I still wonder whether there is some easier way to do this that doesn't resort to using logical predicates.

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