Kerodon

Proof. For $0 \leq i \leq n$, let $\sigma _ i: \Delta ^{n+1} \rightarrow \Delta ^1 \times \Delta ^ n$ denote the map of simplicial sets given on vertices by the formula $\sigma _ i(j) = \begin{cases} (0,j) & \text{ if } j \leq i \\ (1,j-1) & \text{ if } j > i. \end{cases}$ We define simplicial subsets $X(i) \subseteq \Delta ^1 \times \Delta ^ n$ inductively by the formulae

where $\operatorname{im}( \sigma _{i} )$ denotes the image of the morphism $\sigma _{i}$. Note that $\Delta ^1 \times \Delta ^ n$ is the union of the simplicial subsets $\{ \operatorname{im}(\sigma _ i) \} _{0 \leq i \leq n}$, and is therefore equal to $X(n+1)$. This definition satisfies condition $(a)$ by construction. To verify $(b)$, it will suffice to show that for $0 \leq i \leq n$, the inverse image $A_{} = \sigma _{i}^{-1} X(i)$ is equal to $\Lambda ^{n+1}_{i+1}$ (Lemma 3.1.2.11). Regarding $\sigma _{i}$ as an $(n+1)$-simplex of $\Delta ^1 \times \Delta ^ n$, we are reduced to showing that the faces $d^{n+1}_{j}( \sigma _{i} )$ belong to $X(i)$ if and only if $j \neq i+1$. One direction is clear: the face $d^{n+1}_{j}( \sigma _{i} )$ is contained in $\Delta ^1 \times \operatorname{\partial \Delta }^ n$ for $j \notin \{ i, i+1 \} $, the face $d^{n+1}_{i}(\sigma _ i) = d^{n+1}_{i}( \sigma _{i-1} )$ is contained in $\operatorname{im}( \sigma _{i-1} ) \subseteq X(i)$ for $i > 0$, and $d^{n+1}_0( \sigma _0)$ is contained in $\{ 1\} \times \Delta ^ n$. To complete the proof, it suffices to show that the face $d^{n+1}_{i+1}( \sigma _ i )$ is not contained in $X(i)$, which follows by inspection. $\square$