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.