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*To*: categories@mta.ca (Categories)*Subject*: categories: Re: question on "model functor"*From*: "Dr. P.T. Johnstone" <P.T.Johnstone@dpmms.cam.ac.uk>*Date*: Fri, 15 Sep 2000 18:08:14 +0100 (BST)*In-Reply-To*: <39C22090.5802016D@informatik.uni-bremen.de> from "Lutz Schroeder" at Sep 15, 2000 03:13:52 PM*Sender*: cat-dist@mta.ca

> The following question looks so natural that somebody's > bound to have looked into it: > > Does the functor > > Cat^op ---> CAT > > A |--> [A,Set] > > reflect isomorphisms (more generally: limits)? The question is not very well posed: since Cat and CAT are 2-categories, one ought to be asking about whether it reflects equivalences. For this, the answer is negative (and well known): a functor A --> B induces an equivalence [B,Set] --> [A,Set] iff it induces an equivalence between the idempotent-completions of A and B. So the inclusion of any non-idempotent- complete category in its idempotent-completion provides a counterexample. However, if you insist on asking about isomorphisms rather than equivalences, the answer is yes. It's easy to see that if F: A --> B fails to be surjective (resp. injective) on objects then the induced functor [F,Set] fails to be injective (resp. surjective); so if [F,Set] is an isomorphism then F must be bijective on objects, and this combined with inducing an equivalence of idempotent-completions is enough to make it an isomorphism. But this is not a very meaningful result. Provided you assume a sufficiently powerful form of the axiom of choice, [A,Set] and {B,Set] will be isomorphic whenever they are equivalent (since each has a proper class of objects in each isomorphism class, except for the initial object which is unique in its isomorphism class). The isomorphism will not, of course, be induced by a functor from A to B; but it will be naturally isomorphic to a functor that is (at least provided B is idempotent-complete). Peter Johnstone

**References**:**categories: question on "model functor"***From:*Lutz Schroeder <lschrode@informatik.uni-bremen.de>

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