Dear André Your guess is quite correct. More generally every full and faithful functor is a foliation. Let me call a functor P: X --> S locally full and faithful (lff) iff for each object x of X the obvious functor X/x --> S/P(x) is full and faithful. Such functors are also characterized by : Every map of X is hypercartesian. Thus they are foliatiions. Now every full and faithful functor is lff. Hence is a foliation. I mentioned in my mail that there are many foliations which are not fibrations, this is a typical example. It shows how much more general than (pre) fibrations (pre) foliations can be. To give an easy but important application, let me note that, if X is a groupoid avery functor P: X --> S, where S is arbitrary, is a foliation, because all the maps of X are isos, hence hypercartesian. This gives me the opportunity to explain condition (ii) for cartesian functors in a special case. Suppose P: X --> S, P' : X' --> S and F; X --> X' verify P = P'F, where X, X' and S are groups. Then P and P' are foliations and F preserves cartesian maps. However F need not be cartesian. More precisely F satisfies (ii) iff P and P' have same image in S. In that case the theorem says: F is a mono (resp an epi) iff its restriction Ker(P) --> Ker(P') is a mono (resp an epi). This would be obviously false without (ii) To complete the picture let us see that (i) => (ii) when P is a fibration. In that case P is surjective, i.e. Im(P) = S contains Im(P') . But P = P'F => Im(P) is contained in Im(P'), hence the equality required. Thank you for having given me the occasion to explicit some examples, an in particular to show that (ii) is meaningful. Bien amicalement, Jean Le 28 juil. 2014 à 17:53, Joyal, André a écrit :

Dear Jean,

I apologise for my ignorance of your work.

I guess that an equivalence of categories P:X-->S is always a foliation, but not a fibration, unless it is surjective on objects.

-André

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