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Reposted from Math Overflow
Let V=sSet with its cartesian monoidal structure.
Let f:C→Dop be a V-functor, and let π:C→C′ be another V-functor. Let p∈D be an object, and let hp:=D(−,p):Dop→V.
Let π! be the left adjoint of the pullback V-functor π*:VC′→VC induced by π (That is to say, for a V-functor h:C′→V, the functor π!(h) is the left Kan extension of h along π).
Then g:=hp∘f is a functor C→V.
Why is the following true:
In terms of the pushout diagram
D′:=colim(C′op←Cop→D)where ι1:C′op→D′ and ι2:D→D′ are the canonical injections:
We have an isomorphism of V-functors C′→V
π!(g)(−)≅D′(ιop1(−),ι2(p)).I think we can do something like this:
We have a diagram
Cf→Dophp→V↓π↓ιop2C′→ιop1D′opThen we take the left Kan extension of hp by ιop2, but by one of the many variations of Yoneda’s lemma (computing via coends, this follows from Yoneda reduction), this is exactly D′(−,ι2(p)).
Then it’s enough to show that the composite
k:=D′(−,ι2(p))∘ιop1=D′(ιop1(−),ιop2(p))(with the natural map g→kπ) has the required universal property, that is, that it is initial among fillers of the Kan extension diagram:
Cg→V↓πC′which is where I’m not sure how to proceed.
I just noticed that this is closely related to the calculus of exact squares and mates.
I bet that there’s some theorem that proves that pushout squares in V-Cat are exact, since my question is equivalent to:
Is the natural Beck-Chevalley transformation π!f*→ι*1(ι2)! an isomorphism?
Well, arbitrary pushout squares aren’t exact even when V=Set. Consider the pushout of discrete categories 1←2→1, which is 1. I don’t think you can fix that by taking homotopy pushouts either. But cocomma squares are, I believe, always exact for any V, although I could be misremembering.
Ah, thanks! You’re right. My original question was too generalized from the original source (therefore false). I e-mailed Lurie, and the answer hinges on some additional information (i.e. that the point p in my first post is a weak strict sink (or whatever the right term is) in the following sense: D(p,p)=Δ0 and D(p,x)=∅ for all x≠p (description is evil, but the point p is adjoined formally in a strict way as well (up to equality)).
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