# Kerodon

$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$ $\newcommand\empty{}$

Lemma 1.2.3.2. Let $\operatorname{\mathcal{C}}$ be a category. Then the simplicial set $\operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$ satisfies condition $(\ast ')$ of Proposition 1.2.3.1.

Proof. Choose integers $0 < i < n$ together with a map of simplicial sets $\sigma _0: \Lambda ^{n}_{i} \rightarrow \operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$; we wish to show that $\sigma _0$ can be extended uniquely to a $n$-simplex of $\operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$. For $0 \leq j \leq n$, let $C_{j} \in \operatorname{\mathcal{C}}$ denote the image under $\sigma _0$ of the $j$th vertex of $\Delta ^ n$ (which belongs to the horn $\Lambda ^{n}_{i}$). We first consider the case where $n \geq 3$. In this case, $\Lambda ^{n}_{i}$ contains every edge of $\Delta ^ n$. For $0 \leq j \leq k \leq n$, let $f_{k,j}: C_ j \rightarrow C_ k$ denote the $1$-simplex of $\operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$ obtained by evaluating $\sigma _0$ on the edge of $\Delta ^ n$ corresponding to the pair $(j,k)$. We claim that the construction

$j \mapsto C_ j \quad \quad (j \leq k) \mapsto f_{k,j}$

determines a functor $[n] \rightarrow \operatorname{\mathcal{C}}$, which we can then identify with an $n$-simplex of $\operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$ having the desired properties. It is easy to see that $f_{j,j} = \operatorname{id}_{ C_ j}$ for each $0 \leq j \leq n$, so it will suffice to show that $f_{\ell ,k} \circ f_{k,j} = f_{\ell ,j}$ for every triple $0 \leq j \leq k \leq \ell \leq n$. The triple $(j,k, \ell )$ determines a $2$-simplex $\tau$ of $\Delta ^ n$. If $\tau$ is contained in $\Lambda ^{n}_{i}$, then $\tau ' = \sigma _0( \tau )$ is a $2$-simplex of $\operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$ satisfying $d_0( \tau ') = f_{\ell ,k}$, $d_1( \tau ') = f_{\ell ,j}$, and $d_2( \tau ') = f_{k,j}$, so that $\tau '$ “witnesses” the identity $f_{\ell ,k} \circ f_{k,j} = f_{\ell ,j}$. It will therefore suffice to treat the case where the simplex $\tau$ does not belong to the $\Lambda ^{n}_{i}$. In this case, our assumption that $n \geq 3$ guarantees that we must have $\{ j, k, \ell \} = [n] \setminus \{ i \}$. It follows that $n = 3$, so that either $i = 1$ or $i = 2$. We will treat the case $i = 1$ (the case $i =2$ follows by a similar argument). Note that $\Lambda ^3_1$ contains all of the nondegenerate $2$-simplices of $\Delta ^3$ other than $\tau$; applying the map $\sigma _0$, we obtain $2$-simplices of $\operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$ which witness the identities

$f_{3,0} = f_{3,1} \circ f_{1,0} \quad \quad f_{3,1} = f_{3,2} \circ f_{2,1} \quad \quad f_{2,0} = f_{2,1} \circ f_{1,0}.$

We now compute

$f_{3,0} = f_{3,1} \circ f_{1,0} = ( f_{3,2} \circ f_{2,1} ) \circ f_{1,0} = f_{3,2} \circ ( f_{2,1} \circ f_{1,0} ) = f_{3,2} \circ f_{2,0}$

so that $f_{\ell ,j} = f_{\ell ,k} \circ f_{k,j}$, as desired.

It remains to treat the case $n =2$, so that we must also have $i = 1$. In this situation, the map $\sigma _0: \Lambda ^ n_ i \rightarrow \operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$ determines a pair of composable morphisms $f_{1,0}: C_0 \rightarrow C_1$ and $f_{2,1}: C_1 \rightarrow C_{2}$. This data extends uniquely to a $2$-simplex $\sigma$ of $\operatorname{\mathcal{C}}$ satisfying $d_1(\sigma ) = f_{2,1} \circ f_{1,0}$ (see Remark 1.2.1.3). $\square$