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* Query about Ab[C]
@ 2000-12-14  4:19 Bill Rowan
  2000-12-14 16:25 ` Dr. P.T. Johnstone
  2000-12-14 17:39 ` Michael Barr
  0 siblings, 2 replies; 4+ messages in thread
From: Bill Rowan @ 2000-12-14  4:19 UTC (permalink / raw)
  To: categories


Hi all,

Ab[C] is just my notation for the category of abelian group objects in
the category C.  I was wondering if there is a simple characterization of
those categories C for which Ab[C] is abelian.

Bill Rowan



^ permalink raw reply	[flat|nested] 4+ messages in thread

* Re: Query about Ab[C]
  2000-12-14  4:19 Query about Ab[C] Bill Rowan
@ 2000-12-14 16:25 ` Dr. P.T. Johnstone
  2000-12-14 17:39 ` Michael Barr
  1 sibling, 0 replies; 4+ messages in thread
From: Dr. P.T. Johnstone @ 2000-12-14 16:25 UTC (permalink / raw)
  To: categories

> 
> 
> Hi all,
> 
> Ab[C] is just my notation for the category of abelian group objects in
> the category C.  I was wondering if there is a simple characterization of
> those categories C for which Ab[C] is abelian.
> 
> Bill Rowan

You can't hope to characterize them: knowing properties of Ab[C] can't
tell you everything about C. For example, if C has a strict terminal object,
then Ab[C] is abelian (because it's degenerate), but that gives you no
information about what else C might contain. If you're looking for a
sufficient condition on C, a canonical one is "Barr-exact" (= effective
regular, in Freyd's terminology): Ab[C] inherits Barr-exactness from C,
and abelian is equivalent to Barr-exact plus additive. Conversely, every
abelian category A is isomorphic to Ab[C] for a suitable Barr-exact C,
namely C = A.

Peter Johnstone




^ permalink raw reply	[flat|nested] 4+ messages in thread

* Re: Query about Ab[C]
  2000-12-14  4:19 Query about Ab[C] Bill Rowan
  2000-12-14 16:25 ` Dr. P.T. Johnstone
@ 2000-12-14 17:39 ` Michael Barr
  1 sibling, 0 replies; 4+ messages in thread
From: Michael Barr @ 2000-12-14 17:39 UTC (permalink / raw)
  To: categories

I suspect that it is too much to hope for a characterization.  But it is
sufficient that either C or C^op be exact.  For C to be exact, it is
required that it have finite limits, coequalizers of equivalence
relations, that regular epis be stable under pullback and that every
equivalence relation is the kernel pair of its coequalizer.  

On Wed, 13 Dec 2000, Bill Rowan wrote:

> 
> Hi all,
> 
> Ab[C] is just my notation for the category of abelian group objects in
> the category C.  I was wondering if there is a simple characterization of
> those categories C for which Ab[C] is abelian.
> 
> Bill Rowan
> 




^ permalink raw reply	[flat|nested] 4+ messages in thread

* Re: Query about Ab[C]
@ 2000-12-14 14:44 Peter Freyd
  0 siblings, 0 replies; 4+ messages in thread
From: Peter Freyd @ 2000-12-14 14:44 UTC (permalink / raw)
  To: categories

Bill Rowan asks if there is a simple characterization of those
categories  C  for which  Ab[C]  is abelian. I doubt if there can be
a useful necessary and sufficient condition.

A sufficient condition can be found on page 91 of Cats and Alligators,
to wit, that the category be effective regular (where "effective"
means that every equivalence relation is effective,i.e. it appears as
a pullback of a map against itself). Note that the conclusion
(abelian) is self-dual but the condition (effective regular) is not.

I'm pessimistic about a useful necessary and sufficient condition
because of the following: Let  C  be a category with cartesian squares
(needed to define abelian-group-object) such that  Ab[C]  is abelian. 
Let  C' be a full subcategory closed under cartesian squaring that 
contains the image of the forgetful functor from  Ab[C]  back to  C.
Then  Ab[C'] =  Ab[C]. An example of the sort of pathological 
categories to be found among such  C' is the category of all groups in
which the commutator subgroup is a product of a finite number of
simple groups each of which was described prior to 30 June 1973.



^ permalink raw reply	[flat|nested] 4+ messages in thread

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2000-12-14  4:19 Query about Ab[C] Bill Rowan
2000-12-14 16:25 ` Dr. P.T. Johnstone
2000-12-14 17:39 ` Michael Barr
2000-12-14 14:44 Peter Freyd

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