Jean B?nabou wrote in part:
(i) If F: A -> B and G: B -> C are full and faithful essentially surjective functors, so is GF. How do you compose your equivalences?
Good question! In this case, we can compose by pullback. But when I wrote "the obvious thing to do", maybe I should have written "one obvious thing to try". After all, it might not work; in this case, it does. If it didn't work, another obvious thing to try would be zigzags. In this case, this gives an equivalent 2-groupoid. So in principle, one could do either, but spans are simpler. Generalising from the 2-groupoid of categories to the 2-category of them, we can use spans A <- X -> B where only A <- X needs to be ff eso. This is what Michael Makkai did, and it allows one to avoid AC while retaining the usual results about the 2-category of categories. (Pace Tom Leinster's recent comment under this thread, Makkai actually required A <- X to be strictly surjective on objects, but again this does not matter; the resulting 2-category is equivalent.) Beyond this, let me just say that I agree with Marta's answers. --Toby [For admin and other information see: http://www.mta.ca/~cat-dist/ ]