It has been known for a long time that all monomorphisms in the category of groups are regular, therefore the regular monos compose. I do not know who proved it first. It can be shown directly by representing any inclusion map, say from H to G, as an equaliser of two arrows from G to the group of all bijections from the set of all left cosets modulo H to itself. Another way is to look at the usual descriptions of amalgamated products (i.e. poushouts of pairs of monos); then every mono m can be seen to be the equaliser ofthe two maps forming the pushout of m with itself. In the category of rings (commutative or not, unital or not), regular monos do not compose. The problem is stdied in detail in John Isbell's paper "Epimorphisms and dominions", La Jolla Proceedings, 1965. Similar arguments work for monoids or semigroups. A nice and easy, but seemingly not very well-known example is given by Gabriel and Ulmer (LNM 221): let N be the additive monoid of non-negative integers, let A be the submonoid generated by 3 and 5 and B the submonoid generated by 3, 5, and 7. Then the inclusions from A to B and from B to N can easily be see to be regular monos, but their composite is not, due to the following nice computation: assume f(3)=g(3) and f(5)=g(5) for some morphism from N to some monoid (multiplicatively written). Then f(7)=f(2)f(5)=f(2)g(5)= f(2)g(3)g(2)=f(2)f(3)g(2)=f(5)g(2)=g(5)g(2)=g(7). Gabriel and Ulmer also prove the (now?) well known fact that in any category with finite limits regular epis compose, if they are stable under pullback, and they even state their result more generally. Only recently, I realized that their result easily implies that the class of regular epimorphisms stable under pullback (i. e. of all morphisms e such that every pullback of e is a regular epi) is aways closed under compostion, even if regular epis do not compose. Greetings Reinhard Boerger +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++