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1. • CommentRowNumber1.
   • CommentAuthorHarry Gindi
   • CommentTimeJun 27th 2010
   • (edited Jun 28th 2010)
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   Author: Harry Gindi
   Format: MarkdownItexI'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](http://arxiv.org/pdf/math/0608040v4) for your convenience.

   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.

2. • CommentRowNumber2.
   • CommentAuthorHarry Gindi
   • CommentTimeJun 28th 2010
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   Author: Harry Gindi
   Format: MarkdownItexAnyone willing to give it a shot? I'm really stuck, at least on the second one.

   Anyone willing to give it a shot? I’m really stuck, at least on the second one.

3. • CommentRowNumber3.
   • CommentAuthorTodd_Trimble
   • CommentTimeJun 28th 2010
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   Author: Todd_Trimble
   Format: MarkdownItexI 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).

   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).

4. • CommentRowNumber4.
   • CommentAuthorHarry Gindi
   • CommentTimeJun 28th 2010
   • (edited Jun 28th 2010)
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   Author: Harry Gindi
   Format: MarkdownItexThanks! If you need any clarification on what's going on up to that point, I will be around.

   Thanks! If you need any clarification on what’s going on up to that point, I will be around.

5. • CommentRowNumber5.
   • CommentAuthorUrs
   • CommentTimeJun 29th 2010
   • (edited Jun 29th 2010)
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   Author: Urs
   Format: MarkdownItexHarry, 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 1. the unstraightening functor is defined to be the right adjoint of straightening; 1. the straightening over the point is identified on top of p. 70 (and pages before that) with $|-|_{Q^\bullet}$; 1. so by definition $Sing_{Q^\bullet}$ is the unstraightening over the point; 1. 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.

   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

   1. the unstraightening functor is defined to be the right adjoint of straightening;

   2. the straightening over the point is identified on top of p. 70 (and pages before that) with $|-|_{Q^\bullet}$;

   3. so by definition $Sing_{Q^\bullet}$ is the unstraightening over the point;

   4. 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.

6. • CommentRowNumber6.
   • CommentAuthorHarry Gindi
   • CommentTimeJun 29th 2010
   • (edited Jun 29th 2010)
   • PermaLink
   Author: Harry Gindi
   Format: MarkdownItexDear 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 $(C\times C')^{op}\to\mathfrak{C}[RightCone(k)]\coprod_{\mathfrak{C}[k]}(C\times C')^{op}$, 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.

   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 $(C\times C')^{op}\to\mathfrak{C}[RightCone(k)]\coprod_{\mathfrak{C}[k]}(C\times C')^{op}$, 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.

7. • CommentRowNumber7.
   • CommentAuthorUrs
   • CommentTimeJun 29th 2010
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   Author: Urs
   Format: MarkdownItexAh, the product. Hm, let's see, $Un$ is a right adjoint, by definition, so it preserves products of presheaves. Isn't that all there is to it?

   Ah, the product.

   Hm, let’s see, $Un$ is a right adjoint, by definition, so it preserves products of presheaves. Isn’t that all there is to it?

8. • CommentRowNumber8.
   • CommentAuthorHarry Gindi
   • CommentTimeJun 29th 2010
   • (edited Jun 29th 2010)
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   Author: Harry Gindi
   Format: MarkdownItexNo, 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.

   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.

9. • CommentRowNumber9.
   • CommentAuthorHarry Gindi
   • CommentTimeJun 29th 2010
   • (edited Jun 29th 2010)
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   Author: Harry Gindi
   Format: MarkdownItexAlright, I think the trick is to turn $F\boxtimes F'$ into $(F\boxtimes pt_{C'})\times (pt_C\boxtimes F')$ 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 $\phi\boxtimes \phi'$ (in the unstraightening). This is of course an extension of the technique from analysis: multiplying by 1 _creatively_. Does that work?

   Alright, I think the trick is to turn $F\boxtimes F'$ into $(F\boxtimes pt_{C'})\times (pt_C\boxtimes F')$ 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 $\phi\boxtimes \phi'$ (in the unstraightening).

   This is of course an extension of the technique from analysis: multiplying by 1 creatively.

   Does that work?

10. • CommentRowNumber10.
    • CommentAuthorUrs
    • CommentTimeJun 29th 2010
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    Author: Urs
    Format: MarkdownItexYup.

    Yup.

11. • CommentRowNumber11.
    • CommentAuthorHarry Gindi
    • CommentTimeJun 29th 2010
    • (edited Jun 29th 2010)
    • PermaLink
    Author: Harry Gindi
    Format: MarkdownItexAlright, 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

    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

12. • CommentRowNumber12.
    • CommentAuthorHarry Gindi
    • CommentTimeJun 29th 2010
    • (edited Jun 29th 2010)
    • PermaLink
    Author: Harry Gindi
    Format: MarkdownItexCan 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.

    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.