------------- Asperti and Longo say it is an open question whether there are non-trivial cartesian closed categories with omega complete retraction categories. (Their book, p.249.) Here I show that any category _C_ with binary coproducts has omega complete retraction category iff _C_ is a preorder. The theorem does not even use all binary coproducts. It is constructive so it applies to categories over any topos. To define the terms: a "chain" in _C_ is an infinite sequence of objects, with an initial member, each with a map to the next. A category is "omega complete" if every chain in it has a colimit. Given any category _C_ the "retraction category" is called _C_^Ret and has the same objects as _C_. An arrow in _C_^Ret from A to A' is a retraction pair making A a retract of A'. If _C_ is a preorder then _C_^Ret is a groupoid and trivially omega complete. To connect this with the converse notice that _C_ is a preorder iff every object of _C_ is subinitial (i.e. it has at most one map to any object). Theorem: Suppose a category _C_ has an object A which is not subinitial but which does have a coproduct with every object of _C_. Then _C_^Ret is not omega complete. Proof: Define a sequence of objects An this way: A0=A and for every n we let A(n+1) be the coproduct (An)+A. "The map of A onto the last summand of An" means the coproduct injection of A into A(n-1)+A if n is not 0, and the identity if n=0. Each An is a retract of A(n+1) in a canonical way where the monic is the coproduct injection and the splitting epic from An+A to A acts as the identity on An and maps A onto the last summand of An. These retractions give a chain in _C_^Ret. Suppose the chain has a colimit B. Then there is a map from A to B so B is a retract of B+A. Define a cone over the chain this way: The vertex is B+A. For each An in the chain the monic to B+A is the colimit monic to B followed by the inclusion into B+A. The splitting epic from B+A to An acts as the colimit epic on B and maps A to the last summand in An. But since A is not subinitial the map of A onto the last summand in A(n+1) does not factor through the inclusion of An. Thus no splitting epic from B+A to B to commutes with the cone projections, and so B is NOT a colimit. Our chain has no colimit. Colin McLarty =====================================================================