Sigh! Peter is, of course, correct. It actually occurred to me a couple days after I wrote my reply that Ab[C^op] is not the opposite of Ab[C] but I never looked into it. Michael On Sat, 16 Dec 2000, Peter Freyd wrote:
There was a little trap waiting for Mike and me to fall into. We were seeking conditions on a category C that force Ab[C] to be abelian and we came up with the same condition. I wrote, "Note that the conclusion (abelian) is self-dual but the condition (effective regular) is not." The trouble is: the definition of Ab[C] is not self-dual. I suppose my assertion is true as it stands but the implicit message is wrong. Mike had no such luck; he made it explicit: "[I]t is sufficient that either C or C^op be exact." (Yes, my "effective regular" and Mike's "exact" are equivalent -- it follows just from regularity that the pullback of a cover against itself is a pushout.)
It took me a while to find a counterexample and I'm not happy with the one I found. Adjoin to the equational theory of abelian groups a new constant and no further axioms. Let C be the category of finite models. As is the case for the finite models of any equational theory, C is effective regular (and Ab[C] is isomorphic to the category of finite abelian groups). But Ab[C^op] is not abelian. It's empty. (C^op doesn't have a terminator.)