A question about extensive categories

A question about extensive categories

[Jiri Adamek]

Dear colleagues,

Does anyone know whether every extensive and locally finitely presentable
category fulfils the following condition:

For every omega op-chain of coproduct injections i_n: A_n+1 -> A_n
with all A_n finitely presentable some i_n is an isomorphism.

We need this for investigating iterative monads in such categories,
and we have not managed to prove it, nor to find a counterexample.

Thanks,
Jiri Adamek, Stefan Milius and Jiri Velebil

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Re: A question about extensive categories

[reinhard.boerger]

Dear colleagues

on 29 Nov 2006 at 15:58 Jiri Adamek wrote:

> Does anyone know whether every extensive and locally finitely
> presentable
> category fulfils the following condition:
>
> For every omega op-chain of coproduct injections i_n: A_n+1 -> A_n
> with all A_n finitely presentable some i_n is an isomorphism.

Here is the answer:

Consider the category of pairs of sets X,Y together with an isomorphism s:XxY->Y, with
inverse Y->XxY, Y|->(py,qy). Then X is locally finitely presentable, because it is a two-sorted
variety with operations s,p,q and equations s(py,qy)=y=qs(x,y), ps(x,y)=x. One might think of
the elements of Y as versions of sequences in X; an element y can be considered as a
version of the sequence (pq^n(y))_n, where n runs over all natural numbers (including 0).
Then p maps evey "sequence" to its initial (0-th) member, q means ommitting the initial
member, an s(x,y) is obtained from y by adding the new initial member x. Then it is quite
easy to see that the category is also extensive.

The free algebra A=(N,S) on one element of Y can then be described as follows: N is the set
of natural numbers and s is the set of all sequences (x(n))_n in N with the property that the
exisit a natural number m and an integer k with x(n)=n+k for all n>m; p is the projection
(x(n))_n|->x(0) to the 0-th component, and q is the left shift (x(n))_n|->(x(n+1))_n. The
algebra A is freely generated by the sequence g:=(n)_n of natural numbers; every element of
S can be obtained from g by first shifting left several times and then shifting right with
inserting arbitary natural numbers in the initial component; every n in N can be obtained as
n=pq^n(g).

The free algebra on one element of X is ({0},0,...), where 0 is the empty set. Now the
coproduct A+B is isomorphic to A with injections i:A->A,  j:B->A, where i is the sucessor map
n|->n+1 and j maps 0 to 0 in the first component; this uniquely determines the second
components (because "different versions of the same sequence do not occur".) Now A_n:=n
and i_n:=i for all n in N yields a counterexample to the above statement.

                                                                       Greetings
                                                                       Reinhard