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1. • CommentRowNumber1.
   • CommentAuthorHarry Gindi
   • CommentTimeNov 3rd 2010
   • (edited Nov 3rd 2010)
   • PermaLink
   Author: Harry Gindi
   Format: MarkdownItex(Copied from [MO](http://mathoverflow.net/questions/44652/reducing-straightening-over-an-interval-to-straightening-over-a-point)) ------ 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 $X\to \Delta^1=\mathfrak{N}([1])$ be an object of $(Set_\Delta\downarrow \Delta^1)$, and let $\epsilon:\mathfrak{C}(\Delta^1)=\mathfrak{C}(\mathfrak{N}([1])\to [1]$ be the counit (here $[1]$ is the category determined by the ordinal number $2$ (two objects, one nonidentity arrow). Form the pushout $M$ of the span $\mathfrak{C}(X^\triangleright) \leftarrow \mathfrak{C}(X)\to \mathfrak{C}(\Delta^1)\to [1]$ (the two arrows in the same direction are replaced by their composite, so this is $M=\mathfrak{C}(X^\triangleright)\coprod_{\mathfrak{C}(X)} [1]$). This determines a functor $St_\epsilon X:[1]\to Set_\Delta$ 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 $\lambda:[0]\to [1]$ is the map choosing the object $0$ of $[1]$) where $St_*S$ is simply defined to be the analogous construction when $\epsilon$ is replaced with the identity $[0]=\mathfrak{C}(\Delta^0)\to [0]$. (Note that here we can identify functors $[0]\to Set_\Delta$ 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).

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

   This determines a functor $St_\epsilon X:[1]\to Set_\Delta$ 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 $\lambda:[0]\to [1]$ is the map choosing the object $0$ of $[1]$) where $St_*S$ is simply defined to be the analogous construction when $\epsilon$ is replaced with the identity $[0]=\mathfrak{C}(\Delta^0)\to [0]$. (Note that here we can identify functors $[0]\to Set_\Delta$ 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).

2. • CommentRowNumber2.
   • CommentAuthorHarry Gindi
   • CommentTimeNov 3rd 2010
   • PermaLink
   Author: Harry Gindi
   Format: MarkdownItex=(. 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.

3. • CommentRowNumber3.
   • CommentAuthorUrs
   • CommentTimeNov 4th 2010
   • PermaLink
   Author: Urs
   Format: MarkdownItexThis should be the statement that in $sSet$ (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.

4. • CommentRowNumber4.
   • CommentAuthorUrs
   • CommentTimeNov 4th 2010
   • (edited Nov 4th 2010)
   • PermaLink
   Author: Urs
   Format: MarkdownItexMore 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 $P$ $$ \array{ X &\to& X^{\triangleright} \\ \downarrow && \downarrow \\ \Delta[1] &\to& P } $$ We have a canonical map $P \to \Delta[1]^{\triangleright}$ induced from the commutativity of $$ \array{ X &\to& X^{\triangleright} \\ \downarrow && \downarrow \\ \Delta[1] &\to& \Delta[1]^{\triangleright} } \,. $$ For evaluating $P(0,p)$ we just need the fiber over $\{0\}^{\triangleright}$, hence the pullback of the diagram $$ \array{ && P \\ && \downarrow \\ \{0\}^{\triangleright} &\hookrightarrow& \Delta[1]^{\triangleright} } \,. $$ 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 $X \times_{\Delta[1]} \Delta[0]$. Because you can compute it as this consecutive pullback: $$ \array{ X \times_{\Delta[1]} \{0\} &\to& X \\ \downarrow && \downarrow \\ \{0\} &\to& \Delta[1] \\ \downarrow && \downarrow \\ \{0\}^{\triangleright} &\to & P } $$

   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 $P$

   $$\array{ X &\to& X^{\triangleright} \\ \downarrow && \downarrow \\ \Delta[1] &\to& P }$$

   We have a canonical map $P \to \Delta[1]^{\triangleright}$ induced from the commutativity of

   $$\array{ X &\to& X^{\triangleright} \\ \downarrow && \downarrow \\ \Delta[1] &\to& \Delta[1]^{\triangleright} } \,.$$

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

   $$\array{ && P \\ && \downarrow \\ \{0\}^{\triangleright} &\hookrightarrow& \Delta[1]^{\triangleright} } \,.$$

   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 $X \times_{\Delta[1]} \Delta[0]$. Because you can compute it as this consecutive pullback:

   $$\array{ X \times_{\Delta[1]} \{0\} &\to& X \\ \downarrow && \downarrow \\ \{0\} &\to& \Delta[1] \\ \downarrow && \downarrow \\ \{0\}^{\triangleright} &\to & P }$$
5. • CommentRowNumber5.
   • CommentAuthorHarry Gindi
   • CommentTimeNov 4th 2010
   • PermaLink
   Author: Harry Gindi
   Format: MarkdownItexThanks!

   Thanks!

6. • CommentRowNumber6.
   • CommentAuthorHarry Gindi
   • CommentTimeNov 4th 2010
   • (edited Nov 4th 2010)
   • PermaLink
   Author: Harry Gindi
   Format: MarkdownItexOh, 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!

7. • CommentRowNumber7.
   • CommentAuthorHarry Gindi
   • CommentTimeNov 4th 2010
   • (edited Nov 4th 2010)
   • PermaLink
   Author: Harry Gindi
   Format: MarkdownItexMeanwhile: 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?

   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?

8. • CommentRowNumber8.
   • CommentAuthorUrs
   • CommentTimeNov 4th 2010
   • (edited Nov 4th 2010)
   • PermaLink
   Author: Urs
   Format: MarkdownItex> 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 <a href="http://ncatlab.org/nlab/show/join+of+simplicial+sets#DayConvolution">join in terms of Day convolution</a> 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.

   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.

9. • CommentRowNumber9.
   • CommentAuthorUrs
   • CommentTimeNov 4th 2010
   • (edited Nov 4th 2010)
   • PermaLink
   Author: Urs
   Format: MarkdownItexMaybe again more in detail: with the above argument we find first that $$ X^{\triangleright} \times_{\Delta[1]^{\triangleright}} \{0\}^{\triangleright} = \left( X \times_{\Delta[1]^{\triangleright}} \{0\}^{\triangleright} \right)^{\triangleright}$$ And that remaining pullback is easily seen to be $$ X \times_{\Delta[1]^{\triangleright}} \{0\}^{\triangleright} = X \times_{\Delta[1]} \{0\} \,. $$ (All equality signs denote isomorphisms.)

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

   $$X^{\triangleright} \times_{\Delta[1]^{\triangleright}} \{0\}^{\triangleright} = \left( X \times_{\Delta[1]^{\triangleright}} \{0\}^{\triangleright} \right)^{\triangleright}$$

   And that remaining pullback is easily seen to be

   $$X \times_{\Delta[1]^{\triangleright}} \{0\}^{\triangleright} = X \times_{\Delta[1]} \{0\} \,.$$

   (All equality signs denote isomorphisms.)

10. • CommentRowNumber10.
    • CommentAuthorHarry Gindi
    • CommentTimeNov 4th 2010
    • PermaLink
    Author: Harry Gindi
    Format: MarkdownItexThanks again!

    Thanks again!