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)? related questions (i.e. ones that would solve the problem or at least help solving it) are -- Are epimorphisms in Cat those functors F:A-->B for which F[A] generates B by composites and inverses of isomorphisms (in B)? -- is the source of all covariant hom-functors an extremal mono-source in CAT? The answers are probably in the literature - but where? Thanks a lot, Lutz Schroeder -- ----------------------------------------------------------------------------- Lutz Schroeder Phone +49-421-218-4683 Dept. of Computer Science Fax +49-421-218-3550 University of Bremen lschrode@informatik.uni-bremen.de P.O.Box 330440, D-28334 Bremen http://www.informatik.uni-bremen.de/~lschrode -----------------------------------------------------------------------------
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
Lutz Schroeder asked: Does the functor Cat^op ---> CAT A |--> [A,Set] reflect isomorphisms (more generally: limits)? Peter Johnston's answer to the first question is the right one, but one could expand. There's no doubt that his argument works well for small categories (which, I take it, is what was intended by the notation), there's a little worry about it working for _locally_ small categories, and there's no hope at all that it works for arbitrary categories, even if we restrict to those categories with small numbers of objects. (As for the second -- parenthetical -- question: see the bottom.) For the fun of it, consider the one-object category whose maps are named by pairs <X,f> where X is a finite set and f is a permutation thereon. I'll say that <X,f> and <Y,g> name the same map if the supports of f and g are both contained in the intersection of X and Y, and further, when f and g are restricted to that intersection, they become equal. Given the names of any two maps, one can always choose names with the same first coordinates and that's enough to tell you how to compose them. There's a functor from this big one-object category to the tiny one- object category usually called the group, Z_2, to wit, the signature functor. This functor is carried by [-,Set] to an isomorphism in CAT. Also for the fun of it, here's a proof for locally small categories that doesn't require considering things like small projectives in super-big categories. It's routine to reduce to the case where T may be assumed the be the inclusion functor of a lluf subcategory (one that contains all identity maps) *A* into a category *B*. The next step is to show that *A* is a retract of *B*. For any object A, let H^A denote the covariant set-valved functor on *A* represented by A. Assuming that [T,Set] is an isomorphism, let G^A be the unique extension of H^A to *B*. For any f:A -> X in *A* and g:X -> Y in *B* (G^A'g)'(f) is a map in *A* from A to Y. Note that this is not defined as a composition of maps (indeed, G^A'g is not defined as a map in any category other than the category of sets) but as the application of a function G^A'g on an element f in G^A'X, hence I will avoid using catenation for other than composition by inserting prime-marks (for application). The functoriality says that for h:Y -> Z we have 1: (G^A'h)'((G^A'g)'f) = (G^A'(hg))'f (and, of course, (G^A'1)'(f) = f). Now let j:B -> A be a map in *A*. The natural transformation H^j extends uniquely to a transformation G^j. But since the two Gs do the same thing to objects as the two Hs we know that G^j and H^j do the same thing to those objects. We have, therefore, 2: ((G^A'g)'(f))(j) = (G^B'g)'(fj). Construct U:*B* --> *A* by defining U(g) = (G^X'g)'1 for g:X -> Y. For h:Y -> Z in *A* we have (using 1) U(hg) = (G^X'(hg))'1 = (G^X'h)'((G^X'g)'1) = (G^X'h)'(Ug) and (using 2) the later is equal to (G'X'h)'(1Ug) = (G^Y'h)'1)(Ug) = (Uh)(Ug). Consider the two endo-functors on *B*, the identity functor and the idempotent *B* --> *A* --> *B*. The hypothesis that [T,Set] is an isomorphism says that for any set-valued F on *B* it is the case that F is equal to *B* --> *A* --> *B* --> Set. But the set-valued functors (indeed, just the representables) are collectively faithful, and that forces *B* --> *A* --> *B* to be the identity functor, which, in turn, forces each of *B* --> *A* and *A* --> *B* to be identity functors. The later functor is the given T. Finally, as for the question about reflecting limits: if I may quote Cats and Alligators: "It seems to be a general principle that almost any property of interest is reflected by [isomorphism-reflecting embeddings] that preserve it" (1.33) In particular there's an easy argument for the case of limits in a complete category because a cone on a diagram fails to be a limit precisely when the induced map from the limit fails to be an isomorphism.
participants (3)
-
Dr. P.T. Johnstone -
Lutz Schroeder -
Peter Freyd