Reducing straightening over an interval to straightening over a point

Question

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.

Answer 1

(Community Wiki Answer)

Urs Schreiber answered this question over at the nForum, so I'll reproduce his answer here (since I don't think he's going to add it himself)

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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:

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