Proof of Proposition 2.5.9.10.
Fix an integer $n \geq 0$, and let $\sigma $ be an $(n+1)$-simplex of the homotopy coherent nerve $\operatorname{N}^{\operatorname{hc}}_{\bullet }( \operatorname{\mathcal{C}}^{\Delta } )$, which we will identify with a simplicial functor $\sigma : \operatorname{Path}[n+1]_{\bullet } \rightarrow \operatorname{\mathcal{C}}^{\Delta }_{\bullet }$. Set $X = \sigma (0)$, $Y = \sigma (n+1)$, and let $I$ denote the set $\{ 1, 2, \cdots , n \} $, endowed with the opposite of its usual ordering. By virtue of Remark 2.5.3.9, it will suffice to verify the following three assertions:
- $(a)$
If $n=0$ and $\sigma $ is the degenerate edge of $\operatorname{N}^{\operatorname{hc}}_{\bullet }( \operatorname{\mathcal{C}}^{\Delta } )$ determined by the object $X \in \operatorname{\mathcal{C}}$, then $\sigma ( [ \operatorname{\raise {0.1ex}{\square }}^{I}] ) = \operatorname{id}_ X$.
- $(b)$
If $n > 0$ and $\sigma $ is degenerate, then $\sigma ( [ \operatorname{\raise {0.1ex}{\square }}^{I}] ) = 0$.
- $(c)$
Let $n \geq 0$. For $1 \leq i \leq n$, let $I_{< i}$ denote the set $\{ 1, 2, \cdots , i-1 \} $ and let $I_{> i }$ denote the set $\{ i+1, i+2, \cdots , n \} $, which we endow with the reverse of their usual orderings. Then we have
\begin{eqnarray*} \partial \sigma ( [ \operatorname{\raise {0.1ex}{\square }}^{I}] ) & = & \sum _{i=1}^{n} (-1)^{n+1-i} ( \sigma _{\geq i}( [ \operatorname{\raise {0.1ex}{\square }}^{I_{> i}} ] ) \sigma _{\leq i}( [ \operatorname{\raise {0.1ex}{\square }}^{I_{< i} } ] ) - d^{n+1}_ i(\sigma )( [ \operatorname{\raise {0.1ex}{\square }}^{I \setminus \{ i\} } ] ). \end{eqnarray*}
Assertion $(a)$ is immediate from the definition. To prove $(b)$, we observe that $\sigma $ determines a map of simplicial sets
\[ \operatorname{Hom}_{\operatorname{Path}[n+1]}( 0, n+1)_{\bullet } \rightarrow \operatorname{Hom}_{ \operatorname{\mathcal{C}}^{\Delta } }( X,Y )_{\bullet } \simeq \mathrm{K}( \operatorname{Hom}_{\operatorname{\mathcal{C}}}( X,Y)), \]
which we can identify with a chain map $u: \mathrm{N}_{\ast }( \operatorname{Hom}_{\operatorname{Path}[n+1]}( 0, n+1); \operatorname{\mathbf{Z}}) \rightarrow \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Y)_{\ast }$. If $\sigma $ is degenerate, then (as a simplicial functor) it factors as a composition
\[ \operatorname{Path}[n+1]_{\bullet } \rightarrow \operatorname{Path}[n]_{\bullet } \rightarrow \operatorname{\mathcal{C}}^{\Delta }_{\bullet }, \]
where $\rho $ is a simplicial functor satisfying $\rho (0) = 0$ and $\rho (n+1) = n$. For $n > 0$, it follows that the chain map $u$ factors through the complex $\mathrm{N}_{\ast }( \operatorname{Hom}_{ \operatorname{Path}[n]}( 0, n); \operatorname{\mathbf{Z}}) \simeq \mathrm{N}_{\ast }( \operatorname{\raise {0.1ex}{\square }}^{n-1}; \operatorname{\mathbf{Z}})$. Since $\operatorname{\raise {0.1ex}{\square }}^{n-1}$ is a simplicial set of dimension $\leq n-1$, the chain complex $\mathrm{N}_{\ast }( \operatorname{\raise {0.1ex}{\square }}^{n-1}; \operatorname{\mathbf{Z}})$ vanishes in degrees $\geq n$ (see Example 2.5.5.13). In particular, the map $u$ vanishes in degree $n$, so that $\sigma ( [ \operatorname{\raise {0.1ex}{\square }}^{I} ] ) = 0$.
We now prove $(c)$. Using Lemma 2.5.9.13 (and taking into account the order reversal on the set $I$), we obtain the identity
\[ \partial \sigma ([ \operatorname{\raise {0.1ex}{\square }}^{I} ]) = \sum _{i=1}^{n} (-1)^{n+1-i} (\sigma ([ \{ 0\} \times \operatorname{\raise {0.1ex}{\square }}^{I \setminus \{ i\} } ]) - \sigma ([ \{ 1\} \times \operatorname{\raise {0.1ex}{\square }}^{I \setminus \{ i\} }] ). \]
It will therefore suffice to show that, for each $1 \leq i \leq n$, we have equalities
\[ \sigma ( [ \{ 0\} \times \operatorname{\raise {0.1ex}{\square }}^{I \setminus \{ i\} } ] ) = \sigma _{\geq i}( [ \operatorname{\raise {0.1ex}{\square }}^{I_{> i}} ] ) \circ \sigma _{\leq i}( [ \operatorname{\raise {0.1ex}{\square }}^{I_{< i} } ] ) \]
\[ \sigma ([ \{ 1\} \times \operatorname{\raise {0.1ex}{\square }}^{I \setminus \{ i\} }] ) = d^{n+1}_ i(\sigma )( [ \operatorname{\raise {0.1ex}{\square }}^{I \setminus \{ i\} }] ) \]
in the abelian group $\operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Y)_{n-1}$. The second of these identities follows immediately from the definition of $d_ i(\sigma )$. To prove the first, we note that the inclusion $\{ 0\} \times \operatorname{\raise {0.1ex}{\square }}^{I \setminus \{ i\} } \hookrightarrow \operatorname{\raise {0.1ex}{\square }}^{I} \simeq \operatorname{Hom}_{ \operatorname{Path}[n+1]}(0,n+1)_{\bullet }$ factors as a composition
\begin{eqnarray*} \{ 0\} \times \operatorname{\raise {0.1ex}{\square }}^{I \setminus \{ i\} } & \simeq & \operatorname{\raise {0.1ex}{\square }}^{I_{ > i }} \times \operatorname{\raise {0.1ex}{\square }}^{I_{< i}} \\ & \simeq & \operatorname{Hom}_{\operatorname{Path}[n+1]}( i, n+1)_{\bullet } \times \operatorname{Hom}_{ \operatorname{Path}[n+1]}( 0, i)_{\bullet } \\ & \xrightarrow {\circ } & \operatorname{Hom}_{ \operatorname{Path}[n+1]}( 0, n+1)_{\bullet }. \end{eqnarray*}
Set $Z = \sigma (i)$. Using the fact that $\sigma $ is a simplicial functor (and the definition of the simplicial category $\operatorname{\mathcal{C}}^{\Delta }_{\bullet }$), we see that $\sigma ([ \{ 0\} \times \operatorname{\raise {0.1ex}{\square }}^{I \setminus \{ i\} } ] )$ is the image of the fundamental chain $[ \operatorname{\raise {0.1ex}{\square }}^{I \setminus \{ i\} } ]$ under the composite map
\begin{eqnarray*} \mathrm{N}_{\ast }( [ \operatorname{\raise {0.1ex}{\square }}^{I \setminus \{ i\} } ]; \operatorname{\mathbf{Z}}) & \xrightarrow { \mathrm{AW} } & \mathrm{N}_{\ast }( [ \operatorname{\raise {0.1ex}{\square }}^{I_{> i}} ]; \operatorname{\mathbf{Z}}) \boxtimes \mathrm{N}_{\ast }( [ \operatorname{\raise {0.1ex}{\square }}^{I_{< i}} ]; \operatorname{\mathbf{Z}}) \\ & \xrightarrow { \sigma _{\geq i} \boxtimes \sigma _{\leq i} } & \operatorname{Hom}_{\operatorname{\mathcal{C}}}(Z,Y)_{\ast } \boxtimes \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Z)_{\ast } \\ & \xrightarrow {\circ } & \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Z)_{\ast }. \end{eqnarray*}
The desired result now follows from the identity $\mathrm{AW}( [ \operatorname{\raise {0.1ex}{\square }}^{I \setminus \{ i\} } ] ) = [ \operatorname{\raise {0.1ex}{\square }}^{I_{> i}} ] \boxtimes [ \operatorname{\raise {0.1ex}{\square }}^{I_{< i}} ]$ supplied by Lemma 2.5.9.16.
$\square$