Cartesian morphism ~~> fibration
Dear all, I'm trying to find a reference for the following result, if indeed it is true. Let X1->B and X2->B be fibrations and F:X1->X2 a cartesian functor over B. Then F factors on the nose as X1 -> X1' -> X2 (as functors over B) such that X1->X1' is an equivalence and X1'->X2 is a fibration. I know it is true if X1 and X2 are fibred in groupoids, this construction is in the Stacks Project. But the general case? Thanks, David [For admin and other information see: http://www.mta.ca/~cat-dist/ ]
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 [For admin and other information see: http://www.mta.ca/~cat-dist/ ]
participants (2)
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David Roberts -
Thomas Streicher