# Kerodon

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Proof. Fix integers $0 < i < n$ and a map of simplicial sets $\sigma _0: \Lambda ^{n}_{i} \rightarrow X \star Y$; we wish to show that $\sigma _0$ can be extended to an $n$-simplex of $X \star Y$. Let $J \subseteq [n]$ denote the collection of vertices $j$ of $\Lambda ^ n_ i \subset \Delta ^ n$ for which $\sigma _0(j)$ belongs to $X$. We first claim that $J$ is an initial segment of $[n]$. If not, then there exists an integer $0 < j \leq n$ such that $j \in J$ and $j-1 \notin J$. Let $e$ be the edge of $\Lambda ^ n_ i$ having source $j-1$ and target $j$. Then $\sigma _0(e)$ is an edge of $X \star Y$ having source $\sigma _0(j-1) \in Y$ and target $\sigma _0(j) \in X$, which is impossible.

We now consider three cases:

• Suppose that $J = [n]$: that is, the morphism $\sigma _0$ carries each vertex of $\Lambda ^ n_{i}$ to a vertex of the simplicial subset $X \subseteq X \star Y$. Then $\sigma _0$ factors through $X$, so we can extend $\sigma _0$ to a morphism of simplicial sets $\sigma : \Delta ^ n \rightarrow X$ by virtue of our assumption that $X$ is an $\infty$-category.

• Suppose that $J = \emptyset$: that is, the morphism $\sigma _0$ carries each vertex of $\Lambda ^ n_{i}$ to a vertex of the simplicial subset $Y \subseteq X \star Y$. Then $\sigma _0$ factors through $Y$, so we can extend $\sigma _0$ to a morphism of simplicial sets $\sigma : \Delta ^ n \rightarrow Y$ by virtue of our assumption that $Y$ is an $\infty$-category.

• Suppose that $\emptyset \subsetneq J \subsetneq [n]$. Let us identify the nerve $\operatorname{N}_{\bullet }(J)$ with its image in $\operatorname{N}_{\bullet }([n]) \simeq \Delta ^ n$, so that $\operatorname{N}_{\bullet }(J)$ is contained in the horn $\Lambda ^ n_{i}$. Then the composite map

$\operatorname{N}_{\bullet }(J) \hookrightarrow \Lambda ^ n_{i} \xrightarrow { \sigma _0} X \star Y$

factors through $X$, and therefore determines a map of simplicial sets $\sigma _{-}: \operatorname{N}_{\bullet }(J) \rightarrow X$. Similarly, the composite map

$\operatorname{N}_{\bullet }( [n] \setminus J) \hookrightarrow \Lambda ^ n_{i} \xrightarrow {\sigma _0} X \star Y$

determines a map of simplicial sets $\sigma _{+}: \operatorname{N}_{\bullet }( [n] \setminus J) \rightarrow Y$. The map $\sigma _0$ then admits a unique extension $\sigma : \Delta ^ n \rightarrow X \star Y$, given concretely by the composition

$\Delta ^ n = \operatorname{N}_{\bullet }( [n] ) \simeq \operatorname{N}_{\bullet }( J ) \star \operatorname{N}_{\bullet }( [n] \setminus J) \xrightarrow { \sigma _{-} \star \sigma _{+} } X \star Y.$
$\square$