Equivalences and psuedo-equivalences (two items)

Equivalences and psuedo-equivalences (two items)

[Topos8]

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

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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

re: Equivalences and psuedo-equivalences (two items)

[Steve Lack]

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.