Cartesian morphism ~~> fibration

Cartesian morphism ~~> fibration

[David Roberts]

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


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Re: Cartesian morphism ~~> fibration

[Thomas Streicher]

> 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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Re: Cartesian morphism ~~> fibration

[David Roberts]

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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Re: Cartesian morphism ~~> fibration

[David Roberts]

In fact there's a more general construction, recorded in the Stacks Project

https://stacks.math.columbia.edu/tag/08NF

and in the case that the codomain (only!) is fibred in groupoids it shows
that any cartesian functor factors as an equivalence of fibrations followed
by a fibration.

Regards,
David

On 21 Sep. 2017 7:17 pm, "Thomas Streicher" <
streicher@mathematik.tu-darmstadt.de> wrote:

>> 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.
>
> Indeed every functor F between groupoids is a Street fibration and thus
> equivalent to a Grothendieck fibration in the sense that there is a
> fibarion P and an equivalence E such that F = PE.
>
> Interesting that this extends to fibered functors!
>
> Thomas
>

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Re: Cartesian morphism ~~> fibration

[Thomas Streicher]

> 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.

Indeed every functor F between groupoids is a Street fibration and thus
equivalent to a Grothendieck fibration in the sense that there is a
fibarion P and an equivalence E such that F = PE.

Interesting that this extends to fibered functors!

Thomas


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