One thing is clear: your ideal abelian category is not abelian. Furthermore, you don't get to choose your sub- and quotient objects; they are imposed by the category. Moreover, although a weaker topology (or an abelian group with a weaker topology, which is what I assume is meant) is certainly a subobject, it is not regular, which every subobject in an abelian category must be. In fact, the only abelian categories of topological abelian groups I am aware of are the discrete groups and the dual category of compact groups. For me, the ideal category of topological abelian groups is SP(LCA), the subobjects of products of locally compact abelian groups. It is not abelian, but it is *-autonomous. It is equivalent to the category of weakly topologized abelian groups or SP(R/Z), subobjects of powers of the circle. On Mon, 30 Apr 2001, Bill Rowan wrote:

I am attempting to construct the ideal abelian category within which live complete, hausdorff abelian topological groups. The idea is that the quotients of such a group, in the abelian category, would be completions of the group with respect to topologies coarser than the given one. The subobjects would be those topologies. Of course, having a topology as an object in the abelian category means we have to have objects in the category other than abelian groups.

Of course I want to know if this has been done before. Also, what other ideas are there about the ideal abelian category containing these groups? Mac Lane felt that compactly-generated spaces formed the ideal base category for topological algebra. I seem to be using the category of complete, hausdorff uniform spaces as a base category. I wrote a paper on (universal) algebras with a compatible uniformity, and got some nice results about the congruence (actually, uniformity) lattices. But, admittedly, algebras with compatible uniformities have drawbacks as a foundation for topological algebra because even something like the complex numbers cannot be formalized as such, the multiplication not being uniformly continuous.

Bill Rowan