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I’ll cut to the chase: Lurie gives two isomorphisms that I don’t understand at all how to derive.
They are in remark 2.2.2.11 and remark 2.2.2.12. The first describes the fiber of the unstraightening functor over a point s as the Q-singular complex of the evaluated functor.
The second one gives an isomorphism between the unstraightening of the product of two functors (different source) (over two other functors) as the product of their unstraightenings (over their respective source functor).
Maybe they’re easy, but the computations I tried to do to verify them were getting very tricky. I’d type it out here, but it would take me a while to do it, and they’re really easy to see immediately in the pdf (sorry!).
Here’s a link for your convenience.
Anyone willing to give it a shot? I’m really stuck, at least on the second one.
I might give it a shot, but probably not before some time tonight. I had taken a quick look earlier, but it was clear that I would need some time to decipher Lurie’s notation (since I am not actively reading his book).
Thanks! If you need any clarification on what’s going on up to that point, I will be around.
Harry, I found I can’t answer this quickly, but would only have time for a quick response, therefore I didn’t reply at all.
But I think the basic idea is that
the unstraightening functor is defined to be the right adjoint of straightening;
the straightening over the point is identified on top of p. 70 (and pages before that) with ;
so by definition is the unstraightening over the point;
so the statement of 2.2.2.11 amounts to saying that the right adjoint of the unstraighening functor on presheaves is computed objectwise as the right adjoint of the functor on presheaves over the point.
So I’d think you see this by using that limits of presheaves are computed objectwise, by which the formula for the right adjoint on presheaves gives objectwise a formula for a right adjoint.
But I don’t feel I have the time right now to substantiate this sketch by a more detailed discussion. Sorry.
Dear Urs, I gave computing it a shot more than once. The problem is that the “product” is of the form [C,sSet] x [C’,sSet] -> [CxC’,sSet] given by the product on the target, so you can’t use any of the tricks while pulling around the adjoints (this is for remark 2.2.2.12). If I had to guess how to do it, it would be looking at Hom(K,Un_{\phi\boxtimes\phi’}(F\boxtimes F’)) for some K->SxS’. Taking adjoints, we look at the straightening of K->SxS’, which we can think of actually as a category with a fixed map , then by some magic, show that that guy decomposes as some kind of product of two things, which then somehow falls apart.
Basically, I’m stuck.
Anyway, thanks for the help. If you can explain it in more detail when you have more time, I’d really appreciate it.
Ah, the product.
Hm, let’s see, is a right adjoint, by definition, so it preserves products of presheaves. Isn’t that all there is to it?
No, because we’re not taking the product of presheaves. In particular, the product of two presheaves means the product of two presheaves with the same source.
If I had to guess, I’d guess that there’s an end hidden somewhere in here.
Alright, I think the trick is to turn into where pt is the terminal diagram (for C and C’ respectively). If we do this, then we can actually take the diagram apart, and each one of the constant factors will kill its half of (in the unstraightening).
This is of course an extension of the technique from analysis: multiplying by 1 creatively.
Does that work?
Yup.
Alright, and the first question comes from the following: The pullback by the vertex {s} has a left adjoint (given by composition with the inclusion), so then we get Hom(K,Un_phi(F) x_S {s}) is in natural bijection with Hom(St_phi o inc(K),F), but then by the stuff stated in 2.2.1, that somehow composes with phi, then we can split the composition apart in a different way to get the right answer.
Basically, it’s an application of 2.2.1.1
Can we also force 2.2.2.12 out of 2.2.1.1? The way I did it works (I checked now), but was Lurie thinking of something different? The way I did it is kinda inelegant.
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