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