In reply to Frank Piessen's question about the equivalence of [C,Set]/F and [G(F),Set]. I would say that this result is well known, though I am not so familiar with the original literature. A discrete fibration p : E--->C (over C) is a functor for which given any morphism f : A---> A' in C and X' in E with p(X') = A', there is a unique v : X---> X' in E with p(v) = f. Then there is a category DFib/C of discrete fibrations over C, and it is well known that DFib/C <--equiv--- [C^op, SET] : G , this being the analogue of the more commonly cited result for split fibrations. One half of the above equivalence is given by the Grothendieck construction, which induces an equivalence on the slices ( DFib/C ) / ( G(F)---> C) <---equiv--- [C^op,Set] / F. If p = p'' o p' is a composition of discrete fibrations, then p' is a discrete fibration; thus ( DFib/C ) / ( G(F)---> C) equiv DFib / G(F) and this gives the result (modulo op's !) Of course, proving the result directly is easy enough, I guess. Roy Crole +++++++++++++++++++