$\Newextarrow{\xhookrightarrow}{10,10}{0x21AA}$
Proposition 2.5.3.5. Let $\operatorname{\mathcal{C}}$ be a differential graded category. Let $m, n \geq 0$ be nonnegative integers and let $\alpha : [n] \rightarrow [m]$ be a nondecreasing function. Then the construction
\[ ( \{ X_ i \} _{0 \leq i \leq m}, \{ f_{I} \} ) \mapsto ( \{ X_{ \alpha (j) } \} _{ 0 \leq j \leq n}, \{ g_{J} \} ), \]
\[ g_{J} = \begin{cases} f_{ \alpha (J) } & \textnormal{ if } \alpha |_{J} \textnormal{ is injective } \\ \operatorname{id}_{ X_ i } & \textnormal{ if } J = \{ j_{0} > j_{1} \} \textnormal{ with } \alpha (j_{0}) = i = \alpha (j_{1}) \\ 0 & \textnormal{ otherwise. } \end{cases} \]
determines a map of sets $\alpha ^{\ast }: \operatorname{N}_{m}^{\operatorname{dg}}(\operatorname{\mathcal{C}}) \rightarrow \operatorname{N}_{n}^{\operatorname{dg}}(\operatorname{\mathcal{C}})$.
Proof.
Let $( \{ X_ i \} _{0 \leq i \leq m}, \{ f_ I \} )$ be an element of $\operatorname{N}_{m}^{\operatorname{dg}}(\operatorname{\mathcal{C}})$. For each subset $J \subseteq [n]$ with at least two elements, define $g_{J}$ as in the statement of Proposition 2.5.3.5. We wish to show that $( \{ X_{ \alpha (j) } \} _{ 0 \leq j \leq n}, \{ g_{J} \} )$ is an element of $\operatorname{N}_{n}^{\operatorname{dg}}(\operatorname{\mathcal{C}})$. For this, we must show that for each subset
\[ J = \{ j_{0} > j_{1} > \cdots > j_{k-1} > j_{k} \} \subseteq [n] \]
having at least two elements, we have an equality
2.26
\begin{eqnarray} \label{equation:functoriality-of-dg-nerve} \partial g_{J} = \sum _{0 < a < k} (-1)^{a} ( g_{ \{ j_0 > j_1 > \cdots > j_ a \} } \circ g_{ \{ j_ a > \cdots > j_ k \} } - g_{J \setminus \{ j_ a \} } ). \end{eqnarray}
We distinguish three cases:
Suppose that the restriction $\alpha |_{J}$ is injective. In this case, we can rewrite (2.26) as an equality
\begin{eqnarray*} \partial f_{\alpha (J)} = \sum _{0 < a < k} (-1)^{a} ( f_{ \{ \alpha (j_0) > \cdots > \alpha (j_ a) \} } \circ f_{ \{ \alpha (j_ a) > \cdots > \alpha (j_ k) \} } - f_{\alpha (J) \setminus \{ \alpha (j_ a) \} } ), \end{eqnarray*}
which follows from our assumption that $( \{ X_ i \} _{0 \leq i \leq m}, \{ f_ I \} )$ is an element of $\operatorname{N}_{m}^{\operatorname{dg}}(\operatorname{\mathcal{C}})$.
Suppose that $J = \{ j_{0} > j_{1} \} $ is a two-element set satisfying $\alpha (j_{0}) = i = \alpha (j_{1} )$ for some $0 \leq i \leq m$. In this case, we can rewrite (2.26) as an equality $\partial (\operatorname{id}_{ X_ i}) = 0$, which follows from Remark 2.5.2.2.
Suppose that $J = \{ j_{0} > j_{1} > \cdots > j_{k-1} > j_{k} \} $ has at least three elements and that $\alpha |_{J}$ is not injective, so that $g_{J} = 0$. We now distinguish three (possibly overlapping) cases:
The map $\alpha $ is not injective because $\alpha (j_{0}) = i = \alpha ( j_{1} )$ for some $0 \leq i \leq m$. In this case, the expressions $g_{ J \setminus \{ j_ a \} }$ and $g_{ \{ j_{0} > \cdots > j_{a} \} }$ vanish for $1 < a < k$. We can therefore rewrite (2.26) as an an equality
\[ g_{J \setminus \{ j_1 \} } = g_{ \{ j_0 > j_1 \} } \circ g_{ \{ j_{1} > \cdots > j_{k} \} }, \]
which follows from the identities $g_{ J \setminus \{ j_1 \} } = g_{ \{ j_1 > \cdots > j_ k \} }$ and $g_{ \{ j_0 > j_1 \} } = \operatorname{id}_{ X_ i }$.
The map $\alpha $ is not injective because $\alpha (j_{k-1} ) = i = \alpha ( j_{k} )$ for some $0 \leq i \leq m$. In this case, the expressions $g_{J \setminus \{ j_ a \} }$ and $g_{ \{ j_{a} > \cdots > j_ k \} }$ vanish for $0 < a < k-1$. We can therefore rewrite (2.26) as an an equality
\[ g_{J \setminus \{ j_{k-1} \} } = g_{ \{ j_0 > \cdots > j_{k-1} \} } \circ g_{ \{ j_{k-1} > j_ k \} }, \]
which follows from the identities $g_{J \setminus \{ j_{k-1} \} } = g_{ \{ j_{0} > \cdots > j_{k-1} \} }$ and $g_{ \{ j_{k-1} > j_{k} \} } = \operatorname{id}_{ X_ i }$.
The map $\alpha $ is not injective because we have $\alpha ( j_{b} ) = \alpha ( j_{b+1} )$ for some $0 < b < k-1$. In this case, the chains $g_{ J \setminus \{ j_{a} \} }$ vanish for $a \notin \{ b, b+1\} $, and the compositions $g_{ \{ j_0 > \cdots > j_ a \} } \circ g_{ \{ j_ a > \cdots > j_ k \} }$ vanish for all $0 < a < k$. We can therefore rewrite (2.26) as an an equality $g_{J \setminus \{ j_ b \} } = g_{ J \setminus \{ j_{b+1} \} }$, which is clear.
$\square$