Equivalences and psuedo-equivalences (two items)
The answer to this question must be in the literature somewhere but I haven't been able to track it down or figure it out for myself. Let A and B be bicategories. By a functor from A to B I shall mean a morphism of underlying reflexive globular sets that preserves all operations and identities on the nose and satisfies the standard coherence conditions. A psuedo-functor is a morphism that preserves operations and identities only up to 1 cells that are equivalences in B ( a 1-cell f is an equivalence if there is a 1-cell g such that fg and gf are both defined and isomorphic to the respective identitity 1-cells). (These 1-cells must of course satisfy additional, standard coherence conditions.) An equivalence from A to B is a functor that is essentially surjective on 0-cells (every 0-cell of B is equivalent in B to a 0-cell in the image of F) and which induces an equivalence of 1-categories between A( x,y ) and B ( Fx, Fy ) for all 0-cells x, y in A. A psuedo-equivalence from A to B is a psuedo-functor that has these same properties. Question 1 : If F is a psuedo-equivalence from A to B does there exist an equivalence G from A to B? (references/counterexamples?) Question 2: Same as question 1 but with A and B strict 2-categories. Qestion 3: If the answer to question 1 is yes, then can G be chosen to be equivalent to F in the bicategory whose 0-cells are psuedo-functors from A to B? I shall be very grateful for any guidance the list can provide on these questions. Carl Futia ------------------------------------------------------ SECOND ITEM: Subject: Correction to my last query I have to apologize for a silly error in my last request for help. The problem lies with the definition of psuedo functor. I should have said that a psuedo functor preserves operations up to a 2-cell isomorphism, not a 1-cell equivalence. Sorry! Carl Futia
Carl Futia asks the following questions: (1) If F:A-->B is a biequivalence of bicategories, is it equivalent to a strict homomorphism? (2) As in (1), but suppose that A and B are 2-categories. The answer to both questions is no. Here's an example. Let A be the 2-element group {0,1}, seen as a 2-category with one object, two arrows, and no non-trivial 2-cells. Let B be the 2-category with one object, with the integers as arrows (and composition given by addition) and with a unique, invertible 2-cell between arrows m and n if m-n is even, and no other 2-cells. The only strict homormorphism (i.e. 2-functor) from A to B sends both arrows of A to the identity arrow 0 of B. There is, however, a biequivalence F:A-->B, sending 0 to 0 and 1 to 1. Note also that there is an obvious 2-functor G:B-->A which is a biequivalence. So the example also illustrates that for a 2-functor which is a biequivalence it may not be possible to choose an ``inverse biequivalence'' which is a 2-functor. This example appeared in: Stephen Lack, A Quillen model structure for 2-categories, K-Theory 26:171-205, 2002 as Example 3.1 on page 178. In that context G:B-->A is actually a trivial fibration, so the fact that there is no 2-functor F with GF=1 also shows that the 2-category A is not cofibrant. Those interested in the rest of the paper should also look at the sequel: Stephen Lack, A Quillen model structure for bicategories, available from http://www.maths.usyd.edu.au/u/stevel/papers/qmcbicat.html which corrects an error in the model structure definition given in the earlier paper, and also extends the model structure to bicategories, giving a Quillen equivalence between the two model categories. Steve Lack.
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