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 functor classifying 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? It's not so clear to me.



Reducing straightening over an interval to straightening over a point
(Community Wiki Answer)


Urs Schreiber answered this question over at the [nForum](http://www.math.ntnu.no/~stacey/Mathforge/nForum/comments.php?DiscussionID=2055&page=1#Item_4), so I'll reproduce his answer here (since I don't think he's going to add it himself)




---


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:


$$
 \array{
 X \times\_{\Delta[1]} \{0\} &\to& X
 \\
 \downarrow && \downarrow
 \\
 \{0\} &\to& \Delta[1]
 \\
 \downarrow && \downarrow
 \\
 \{0\}^{\triangleright} &\to & P
 }
$$


