Dear Thomas, Thanks for that example (and to someone else who, off-line, gave me the example where B is trivial). Here's the version for categories fibred in groupoids https://stacks.math.columbia.edu/tag/06N7 So I guess this extends your example where the codomain is a discrete fibration, merely having to replace the domain by an equivalent category. This makes the original cartesian functor a Street fibration, I believe. This is all in the context of stacks, and in particular algebraic or other presentable sacks, which is what I'm looking at, though in greater generality than the Stacks Project. David On 21 Sep. 2017 3:19 am, "Thomas Streicher" < streicher@mathematik.tu-darmstadt.de> wrote:
I'm trying to find a reference for the following result, if indeed it is true.
I think the claim is wrong in general. Let P : X->B be a fibration of categories with a terminal object, i.e. P has a right adjoint right inverse One. Then One : Id_B -> P is a cartesian functor though itself not a fibration in general (e.g. B = 1 and X the ordinal 2 then One picks 1 from 2 which has empty fibre over 0).
However, if P is a fibration and Q is a discrete fibration and F is a functor with QF = P then F is a fibration iff F is a cartesian functor from P to Q.
Thomas
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