Dear Paul, I have received many mails concerning the posting you mentioned and I try to answer all of them with precision. This takes a lot of time. I shall answer yours, in detail, about the general definition of cartesian functors in terms of disdtributors very soon. But the precise answer shall be a bit long, so please be patient. All I can say now is that, if P: X --> S is a prefoliation the general definition reduces to the simple one I have given (the proof is not trivial). Moreover some of the results of the theorem require the assumption that P is a prefoliation. Hardly surprising, if you want to prove stronger results you need stronger assumptions. The ones I make are, in a sense, minimal. You'll have I think noticed that for 7) I need P to be a foliation. Thanks for your interest. Best regards, Jean Le 1 août 2014 à 12:35, Paul Levy a écrit :
On 28 Jul 2014, at 10:54, Jean Bénabou wrote:
2) CARTESIAN FUNCTORS Let P: X --> S, P': X' --> S and F: X --> X' be functors such that P = P'F. For every object s of S ,I denote by F_s : X_s --> X'_s the functor induced by F on the fibers. I have a general definition of F being cartesian, without any assumption on P and P' and without any reference to cartesian maps, but it uses distributors in an essential manner.
Please tell us your general definition using distributors.
Do any of the results in your Theorem 2.3 hold in this more general setting?
Paul
2.3. THEOREM. If P is a pre foliation, P' arbitrary, and F is cartesian, then: (1) F is faithful every F_s is. (2) F is full iff every F_s is. (3) F is essentially surjective iff every F_s is. (4) F is final iff every F_s is. (5) F is flat iff every F_s is. (6) F has a left adjoint iff every F_s has. If moreover P is a foliation, then (7) F is conservative iff every F_s is.
-- Paul Blain Levy School of Computer Science, University of Birmingham +44 121 414 4792 http://www.cs.bham.ac.uk/~pbl
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