Chu(Ab,circle) is abelian

Chu(Ab,circle) is abelian

[Michael Barr]

The category of topological abelian groups is not abelian.  The reason is
not hard to explain.  In topological spaces, points are primordial and a
monomorphism of points need not be a monomorphism of the attached frames
(that is a surjection of the open set lattices).  When it is, then
ignoring the usual separation axiom, it is a subspace and thereby a
kernel.  In frames, the reverse is true.  Now the open sets are all and
you ignore the points.  In Chu(Ab,circle), the points and open sets are
treated with equal respect.  Now a monomorphism must be injective on the
points and surjective on the opens and is obviously a kernel.  The dual is
also true.  

At another conceptual level, abelianess is defined by certain exactness
condition, which concerns canonical arrows, usually between limits and
colimits.  A category is pointed if the canonical map 0 --> 1 is an
isomorphism and there is then a canonical arrow A + B --> A x B and when
that is an isomorphism, it is additive.  There is a map from the domain of
any monomoprhism to the kernel of its cokernel....  The conditions are
self dual and a limit is computed in a Chu category as the limit of its
first component and colimit of its second and all the required
isomoprhisms remain isomorphisms.  

Of course, this is just as true of Chu(A,_|_) whenever A is abelian (and,
of course, closed monoidal).  The contrast with the case of topological
and that of localic abelian groups is striking.

I realized all this as a result of listening to Peter Freyd's lecture in
Saint John, NB last week.  He was trying to discover the initial abelian
category with one object.  It is self dual, but not this one since this
contains no non-zero bijective (that is objects that are simultaneously
injective and projective) while Freyd's category has enough of them.
Still it might be interesting.  And to anticipate Vaughan's question, no
chu(Ab,circle) is not abelian, roughly for the same reasons as topological
and localic abelian groups.

Re: Chu(Ab,circle) is abelian

[Vaughan Pratt]

>And to anticipate Vaughan's question, no
>chu(Ab,circle) is not abelian, roughly for the same reasons as topological
 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
>and localic abelian groups.

Actually I was going to ask about Ab.  I thought it was abelian groups
but (maybe I'm just the last to be told) the context suggests Ab =
topological abelian groups.  Is there life in discrete circles?

(To reconcile the subject line with the underlined line you need to know
Mike's usage: the large print giveth and the small print taketh away.
Chu is to chu as preordered sets are to posets, or topological spaces
to T_0 spaces.)

Vaughan

Re: Chu(Ab,circle) is abelian

[Michael Barr]

Ab is discrete abelian groups.  Thus the circle has, for these purposes,
the discrete topology.  Since you are mapping discrete groups to it, the
topology is irrelevant.  But the small chu category is much more like
hausdorff topological abelian groups (in fact, is equivalent to two full
subcategories of them) and is not abelian for similar reasons.  One take
on this is that separation conditions are more or less incompatible with
effective equivalence relations (= monics are kernels in the additive
case).  And extensionality is incompatible with the dual.

On Fri, 19 Jun 1998, Vaughan Pratt wrote:

> 
> >And to anticipate Vaughan's question, no
> >chu(Ab,circle) is not abelian, roughly for the same reasons as topological
>  ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
> >and localic abelian groups.
> 
> Actually I was going to ask about Ab.  I thought it was abelian groups
> but (maybe I'm just the last to be told) the context suggests Ab =
> topological abelian groups.  Is there life in discrete circles?
> 
> (To reconcile the subject line with the underlined line you need to know
> Mike's usage: the large print giveth and the small print taketh away.
> Chu is to chu as preordered sets are to posets, or topological spaces
> to T_0 spaces.)
> 
> Vaughan
>