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Theorem 2.5.3.10. Let $\operatorname{\mathcal{C}}$ be a differential graded category. Then the simplicial set $\operatorname{N}_{\bullet }^{\operatorname{dg}}(\operatorname{\mathcal{C}})$ is an $\infty $-category.

Proof. Suppose we are given $0 < j < n$ and a map of simplicial sets $\sigma _0: \Lambda ^{n}_{j} \rightarrow \operatorname{N}_{\bullet }^{\operatorname{dg}}(\operatorname{\mathcal{C}})$. Using Remark 2.5.3.9, we see that $\sigma _0$ can be identified with the data of a pair $( \{ X_ i \} _{0 \leq i \leq n}, \{ f_ I \} )$, where $\{ X_ i \} _{0 \leq i \leq n}$ is a collection of objects of $\operatorname{\mathcal{C}}$ and $f_{I} \in \operatorname{Hom}_{\operatorname{\mathcal{C}}}( X_{i_0}, X_{i_ k} )_{k-1}$ is defined for every subset $I = \{ i_0 > i_{1} > \cdots > i_{k} \} \subseteq [n]$ for which $k > 0$ and $[n] \neq I \neq [n] \setminus \{ j\} $, satisfying the identity

2.28
\begin{eqnarray} \label{equation:checking-dg-nerve-is-infty-category} \partial f_{I} = \sum _{a=1}^{k-1} (-1)^{a} ( f_{ \{ i_0 > i_1 > \cdots > i_ a \} } \circ f_{ \{ i_ a > \cdots > i_ k \} } - f_{I \setminus \{ i_ a \} } ) \end{eqnarray}

We wish to show that $\sigma _0$ can be extended to an $n$-simplex of $\operatorname{N}_{\bullet }^{\operatorname{dg}}(\operatorname{\mathcal{C}})$. To give such an extension, we must supply chains $f_{[n]} \in \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X_0, X_ n)_{n-1}$ and $f_{[n] \setminus \{ j\} } \in \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X_0, X_ n)_{n-2}$ which satisfy (2.28) in the cases $I = [n]$ and $I = [n] \setminus \{ j\} $. We claim that there is a unique such extension which also satisfies $f_{[n]} = 0$. Applying (2.28) in the case $I = [n]$, we deduce that $f_{[n] \setminus \{ j\} }$ is necessarily given by

\[ f_{ [n] \setminus \{ j\} } = \sum _{ 0 < b < n} (-1)^{b-j} (f_{ \{ n > \cdots > b \} } \circ f_{ \{ b > \cdots > 0\} }) - \sum _{ 0 < b < n, b \neq j} (-1)^{b-j} f_{[n] \setminus \{ b\} }. \]

To complete the proof, it will suffice to verify that this prescription also satisfies (2.28) in the case $I = [n] \setminus \{ j\} $. In what follows, for $0 \leq a < b \leq n$, let us write $[ba]$ for the set $\{ b > b-1 > \cdots > a \} $. We now compute

\begin{eqnarray*} (-1)^{j} \partial f_{[n] \setminus \{ j\} } & = & \sum _{ 0 < b < n} (-1)^{b} \partial (f_{ [nb]} f_{[b0]}) - \sum _{ 0 < b < n, b \neq j} (-1)^{b} \partial f_{[n] \setminus \{ b\} } \\ & = & \sum _{0 < b < n} (-1)^{b} (\partial f_{[nb]}) f_{[b]} - \sum _{0 < b < n} (-1)^{n} f_{[nb]} (\partial f_{[b]}) \\ & & - \sum _{ 0 < b < n, b \neq j} (-1)^{b} \partial f_{ [n] \setminus \{ b\} } \\ & = & \sum _{ 0 < b < c < n} (-1)^{n-c+b} f_{ [nc] } f_{[cb]} f_{[b0] } - \sum _{ 0 < b < c < n} (-1)^{n-c+b} (f_{[nb] \setminus \{ c\} } f_{[b]}) - \\ & & \sum _{ 0 < a < b < n } (-1)^{n+b-a} f_{[nb]} f_{[ba]} f_{[a0]} + \sum _{ 0 < a < b < n } (-1)^{n+b-a} f_{[nb]} f_{[b0] \setminus \{ a\} } - \\ & & \sum _{ 0 < b < c < n, b \neq j } (-1)^{b+n-c} f_{[nc]} f_{[c0] \setminus \{ b\} } + \sum _{ 0 < b < c < n, b \neq j } (-1)^{b+n-c} f_{[n0] \setminus \{ b,c\} } + \\ & & \sum _{ 0 < a < b < n, b \neq j } (-1)^{b+n-a} f_{[na] \setminus \{ b\} } f_{[a0]} - \sum _{ 0 < a < b < n, b \neq j } (-1)^{b+n-a} f_{[n0] \setminus \{ a,b\} }. \end{eqnarray*}

Here the first and third terms cancel, the seventh term cancels with the second except for those summands with $c=j$, the fifth term cancels with the fourth except for those summands with $a = j$, and the sixth term cancels the eighth except for those terms with $c = j$ and $a = j$, respectively. Multiplying by $(-1)^{j}$, we can rewrite this identity as

\begin{eqnarray*} \partial f_{[n] \setminus \{ j\} } & = & \sum _{ 0 < b < j } (-1)^{n-1-b} (f_{ [nb] \setminus \{ j \} } \circ f_{ [b0] }) + \sum _{ j < b < n} (-1)^{n-b} (f_{[nb]} \circ f_{[b0] \setminus \{ j\} } ) \\ & & - \sum _{0 < b < j} (-1)^{n-1-b} f_{[n] \setminus \{ b,j\} } - \sum _{ j < b < n } (-1)^{n-b} f_{ [n] \setminus \{ b,j \} }, \end{eqnarray*}

which recovers equation (2.28) in the case $I = [n] \setminus \{ j\} $. $\square$