Re: Cartesian morphism ~~> fibration

[David Roberts <droberts.65537@gmail.com>]

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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