I thought some of the people on the net might be interested in the outcome of the correspondence I've been having with Paul Taylor on regepis. Basically, he wanted the definition extended further so that if there were a multi-coequalizer, then every candidate should also be considered a regepi. It seems that that all works too. One thing was that I also wanted to be able to define stable in the absence of pullbacks and show that the composite of regepis is regepi if they are stable. A couple of loose ends. Does anyone know of a natural example of a category with multicoequalizers that does have pullbacks (or at least kernel pairs) and a candidate for a multicoequalizer that is not itself the coequalizer of its kernel pair? I could have sworn that Joyal once claimed to have shown that if you have an E/M factorization system and if E is stable under pullbacks, then E = regepis (and M = monos). Maybe it was assumed that E incin epis and M incin monos. Why isn't posets with E = surjective on objects and M = full inclusions a counter-example? In fact, why isn't Cat, with the same two classes, a counter-example? Follows, with minor changes, is the letter I wrote to PT: Paul: With help from Dusko Pavlovic at a crucial point, here is the proof that if regular epis in my sense are stable, in my sense, then they are also closed under composition. Dusko also claims that this can even be made to work for candidates in the way you want it to work. I haven't checked, but I guess if he says it works I believe him. First the simple case. For an arrow q: A --> Q, let Kerp(q) (in contrast with kerp(q), the usual kernel pair) denote the class of all pairs of arrows <x0,x1> such that f.x0=f.x1. (Here and later, I will let the domain/codomain requirements be inferred from the composites and equations written down, so that from the above, x0 and x1 have to be parallel and their codomain has to be the domain of q.) Then my defn of regular epi is that q is regular epi iff Kerp(q) incin Kerp(f) ==> E! g such that g.q = f. I will say that regular epis are stable iff given a regepi q and a map f: Q' --> Q, there is a commutative square f.q' = q.g and q' is a regepi. Prop. If regepis are stable, they are closed under composition. Proof. Suppose q: A --> Q and p: Q --> P are regepi. Suppose Kerp(p.q) incin Kerp(f). Then Kerp(q) incin Kerp(p.q) incin Kerp(f) so E! g such that f = g.q. Now suppose that <u0,u1> in Kerp(p). Begin by choosing squares q'0 q'1 A'0 -------> Q' A'1 -------> Q' | | | | | | | | u'0| |u0 u'1| |u1 | | | | | | | | v q v v q v A --------> Q A --------> Q with q'0 and q'1 regepic. Then find a square t0 A' --------> A'0 | | | | t1| |q'0 | | | | v q'1 v A'1 ---------> Q' with t0, and therefore q' = q'0.t0 = q'1.t1 epic. Then we have q.u'0.t0 = u0.q'0.t0 = u0.q and similarly q.u'1.t1 = u1.q. From p.u0 = p.u1, we conclude that p.q.u'0.t0 = p.u0.q = p.u1.q = p.q.u'1.t1 and thus g.u0.q' = g.q.u'0.t0 = f.u'0.t0 = f.u'1.t1 = g.q.u'1.t1 = g.u1.q'. But q' is epi and so g.u0 = g.u1 and so E! h such that h.p = g. Now for the multi-coequalizer case, Dusko suggest the following. Let Coeq(Kerp(q)) be the set of candidates for thhe multicoequalizer, assuming it exists. Define a predicate Z(q;f,g) that is true whenever Kerp(q) incin Kerp(f) & Kerp(g) and f and g are in the same component of the cocone of all arrows with that property. Then by sprinkling Zs at appropriate places in the above, he says that it all works. I haven't tried it, but I see no reason it wouldn't work. Michael =====================================================================