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From: "Dr. P.T. Johnstone"
Newsgroups: gmane.science.mathematics.categories
Subject: Re: question on "model functor"
Date: Fri, 15 Sep 2000 18:08:14 +0100 (BST)
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In-Reply-To: <39C22090.5802016D@informatik.uni-bremen.de> from "Lutz Schroeder" at Sep 15, 2000 03:13:52 PM
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> 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