nForum - Discussion Feed (Straightening over an interval -> Straightening over a point) 2023-12-03T06:44:34+00:00 https://nforum.ncatlab.org/ Lussumo Vanilla & Feed Publisher Harry Gindi comments on "Straightening over an interval -> Straightening over a point" (17786) https://nforum.ncatlab.org/discussion/2055/?Focus=17786#Comment_17786 2010-11-04T11:28:18+00:00 2023-12-03T06:44:33+00:00 Harry Gindi https://nforum.ncatlab.org/account/34/ Thanks again!

Thanks again!

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Urs comments on "Straightening over an interval -> Straightening over a point" (17780) https://nforum.ncatlab.org/discussion/2055/?Focus=17780#Comment_17780 2010-11-04T10:13:12+00:00 2023-12-03T06:44:33+00:00 Urs https://nforum.ncatlab.org/account/4/ Maybe again more in detail: with the above argument we find first that X &triangleright;&times; &Delta; &triangleright;{0} &triangleright;=(X&times; &Delta; ...

Maybe again more in detail: with the above argument we find first that

$X^{\triangleright} \times_{\Delta^{\triangleright}} \{0\}^{\triangleright} = \left( X \times_{\Delta^{\triangleright}} \{0\}^{\triangleright} \right)^{\triangleright}$

And that remaining pullback is easily seen to be

$X \times_{\Delta^{\triangleright}} \{0\}^{\triangleright} = X \times_{\Delta} \{0\} \,.$

(All equality signs denote isomorphisms.)

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Urs comments on "Straightening over an interval -> Straightening over a point" (17779) https://nforum.ncatlab.org/discussion/2055/?Focus=17779#Comment_17779 2010-11-04T10:00:39+00:00 2023-12-03T06:44:33+00:00 Urs https://nforum.ncatlab.org/account/4/ Is it obvious for any formal reason that the pullback X &triangleright;&times; (&Delta; 1) &triangleright;{0} &triangleright;=(X&times; &Delta; 1{0}) ...

Is it obvious for any formal reason that the pullback $X^\triangleright\times_{(\Delta^1)^\triangleright} \{0\}^\triangleright=(X\times_{\Delta^1}\{0\})^\triangleright$? I mean, I think I can just show it by a computation, but can you derive it again from the fact that colimits are universal?

If we invoke the description of the join in terms of Day convolution we have the coend expression

$X^{\triangleright} : [k] \mapsto \int^{[i],[j] \in \Delta_a} X_i \times Hom_{\Delta_a}([k],[i] \boxplus [j]) \,.$

A coend is just a certain kind of colimit, so on the right this is some colimit of sets (for each $k$) over a diagram whose vertices are sets of the form $X_i \times Hom_{\Delta_a}([k],[i'] \boxplus [j])$.

I think therefore the argument that colimits are stable under pullback applies to this case, too. Yes.

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Harry Gindi comments on "Straightening over an interval -> Straightening over a point" (17774) https://nforum.ncatlab.org/discussion/2055/?Focus=17774#Comment_17774 2010-11-04T06:05:29+00:00 2023-12-03T06:44:33+00:00 Harry Gindi https://nforum.ncatlab.org/account/34/ Meanwhile: Is it obvious for any formal reason that the pullback X &triangleright;&times; (&Delta; 1) &triangleright;{0} &triangleright;=(X&times; &Delta; 1{0}) ...

Meanwhile: Is it obvious for any formal reason that the pullback $X^\triangleright\times_{(\Delta^1)^\triangleright} \{0\}^\triangleright=(X\times_{\Delta^1}\{0\})^\triangleright$? I mean, I think I can just show it by a computation, but can you derive it again from the fact that colimits are universal?

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Harry Gindi comments on "Straightening over an interval -> Straightening over a point" (17766) https://nforum.ncatlab.org/discussion/2055/?Focus=17766#Comment_17766 2010-11-04T02:18:10+00:00 2023-12-03T06:44:33+00:00 Harry Gindi https://nforum.ncatlab.org/account/34/ Oh, by the way, I posted your answer on MO (with attribution and a link, as well as making it community wiki). If you’d prefer to answer it there yourself, I will delete the copy. Thanks again!

Oh, by the way, I posted your answer on MO (with attribution and a link, as well as making it community wiki). If you’d prefer to answer it there yourself, I will delete the copy. Thanks again!

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Harry Gindi comments on "Straightening over an interval -> Straightening over a point" (17763) https://nforum.ncatlab.org/discussion/2055/?Focus=17763#Comment_17763 2010-11-04T00:55:44+00:00 2023-12-03T06:44:33+00:00 Harry Gindi https://nforum.ncatlab.org/account/34/ Thanks!

Thanks!

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Urs comments on "Straightening over an interval -> Straightening over a point" (17762) https://nforum.ncatlab.org/discussion/2055/?Focus=17762#Comment_17762 2010-11-04T00:52:35+00:00 2023-12-03T06:44:33+00:00 Urs https://nforum.ncatlab.org/account/4/ More in detail: Since &Cfr;\mathfrak{C} is left adjoint we can essentially compute the pushout before applying &Cfr;\mathfrak{C}. Let me call the analog of MM obtained this way PP X ...

More in detail:

Since $\mathfrak{C}$ is left adjoint we can essentially compute the pushout before applying $\mathfrak{C}$. Let me call the analog of $M$ obtained this way $P$

$\array{ X &\to& X^{\triangleright} \\ \downarrow && \downarrow \\ \Delta &\to& P }$

We have a canonical map $P \to \Delta^{\triangleright}$ induced from the commutativity of

$\array{ X &\to& X^{\triangleright} \\ \downarrow && \downarrow \\ \Delta &\to& \Delta^{\triangleright} } \,.$

For evaluating $P(0,p)$ we just need the fiber over $\{0\}^{\triangleright}$, hence the pullback of the diagram

$\array{ && P \\ && \downarrow \\ \{0\}^{\triangleright} &\hookrightarrow& \Delta^{\triangleright} } \,.$

Now, since colimits commute with pullbacks in $sSet$, this pullback is the pushout of the corresponding pullbacks of $X$, and $X^{\triangleright}$. But that pullback of $X$ is $X \times_{\Delta} \Delta$. Because you can compute it as this consecutive pullback:

$\array{ X \times_{\Delta} \{0\} &\to& X \\ \downarrow && \downarrow \\ \{0\} &\to& \Delta \\ \downarrow && \downarrow \\ \{0\}^{\triangleright} &\to & P }$ ]]>
Urs comments on "Straightening over an interval -> Straightening over a point" (17761) https://nforum.ncatlab.org/discussion/2055/?Focus=17761#Comment_17761 2010-11-04T00:14:40+00:00 2023-12-03T06:44:33+00:00 Urs https://nforum.ncatlab.org/account/4/ This should be the statement that in sSetsSet (being a Grothendieck topos) we have pullback stability of colimits.

This should be the statement that in $sSet$ (being a Grothendieck topos) we have pullback stability of colimits.

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Harry Gindi comments on "Straightening over an interval -> Straightening over a point" (17759) https://nforum.ncatlab.org/discussion/2055/?Focus=17759#Comment_17759 2010-11-03T23:16:39+00:00 2023-12-03T06:44:33+00:00 Harry Gindi https://nforum.ncatlab.org/account/34/ =(. If someone could pry himself/herself away from the argument over at universe and explain this even for a moment, I’d really appreciate it.

=(. If someone could pry himself/herself away from the argument over at universe and explain this even for a moment, I’d really appreciate it.

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Harry Gindi comments on "Straightening over an interval -> Straightening over a point" (17702) https://nforum.ncatlab.org/discussion/2055/?Focus=17702#Comment_17702 2010-11-03T07:58:10+00:00 2023-12-03T06:44:33+00:00 Harry Gindi https://nforum.ncatlab.org/account/34/ (Copied from MO) Recall: Let FU &bullet;:Cat&rightarrow;Cat &Delta;FU_\bullet:Cat\to Cat_\Delta be the bar construction assigned to the comonad FUFU determined by free-forgetful ...

(Copied from MO)

Recall:

Let $FU_\bullet:Cat\to Cat_\Delta$ be the bar construction assigned to the comonad $FU$ determined by free-forgetful adjunction $F:Quiv\rightleftarrows Cat:U$. The restriction of $FU_\bullet$ to the full subcategory $\Delta$ (which is isomorphic to the category of finite nonempty ordinals) naturally determines a colimit-preserving functor $\mathfrak{C}:Set_\Delta=Set^{\Delta^{op}}\to Cat_\Delta$. The right adjoint of this functor is called $\mathfrak{N}$, the homotopy-coherent nerve.

Identify $Cat$ (not by $FU_\bullet$) with the full subcategory of $Cat_\Delta$ spanned by those simplicially enriched categories with discrete hom-spaces.

Also, recall the definition of the right cone $X^\triangleright$ on a simplicial set $X$ is the join $X\star \Delta^0$. This determines an obvious natural map $X\to X^\triangleright$.

Let $X\to \Delta^1=\mathfrak{N}()$ be an object of $(Set_\Delta\downarrow \Delta^1)$, and let $\epsilon:\mathfrak{C}(\Delta^1)=\mathfrak{C}(\mathfrak{N}()\to $ be the counit (here $$ is the category determined by the ordinal number $2$ (two objects, one nonidentity arrow). Form the pushout $M$ of the span $\mathfrak{C}(X^\triangleright) \leftarrow \mathfrak{C}(X)\to \mathfrak{C}(\Delta^1)\to $ (the two arrows in the same direction are replaced by their composite, so this is $M=\mathfrak{C}(X^\triangleright)\coprod_{\mathfrak{C}(X)} $).

This determines a functor $St_\epsilon X:\to Set_\Delta$ defined as $i\mapsto M(i,p)$ where $p$ is the image of the cone point of $\mathfrak{C}(X^\triangleright).$

Question:

The book I’m reading asserts that $St_\epsilon X(0)$ can be identified with $St_*(X\times_{\Delta^1} \Delta^0)$ (where $\Delta^0\to \Delta^1$ is the map $\mathfrak{N}(\lambda)$ where $\lambda:\to $ is the map choosing the object $0$ of $$) where $St_*S$ is simply defined to be the analogous construction when $\epsilon$ is replaced with the identity $=\mathfrak{C}(\Delta^0)\to $. (Note that here we can identify functors $\to Set_\Delta$ with simplicial sets themselves, and suggestively, that under this identification, $St_\epsilon X(0)=\lambda^*St_\epsilon X$).

Why is this true?

Edit: (Blah, modulo the inevitable sign error here).

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