whoops

whoops

[Peter Freyd]

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.)

Re: whoops

[Michael Barr]

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.)
> 
>