# Kerodon

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

Proposition 5.2.6.18. Let $\pi : \operatorname{\mathcal{C}}\rightarrow \Delta ^ n$ be a cocartesian fibration of $\infty$-categories having fibers $\{ \operatorname{\mathcal{C}}(i) = \{ i\} \times _{\Delta ^ n} \operatorname{\mathcal{C}}\} _{0 \leq i \leq n}$. Then there exists a scaffold

$U: M( \operatorname{\mathcal{C}}(0) \rightarrow \operatorname{\mathcal{C}}(1) \rightarrow \cdots \rightarrow \operatorname{\mathcal{C}}(n) ) \rightarrow \operatorname{\mathcal{C}}.$

Proof. For $1 \leq i \leq n$, Proposition 5.2.2.4 guarantees that there exists a functor $F(i): \operatorname{\mathcal{C}}(i-1) \rightarrow \operatorname{\mathcal{C}}(i)$ and a natural transformation $h(i): \Delta ^1 \times \operatorname{\mathcal{C}}(i-1) \rightarrow \operatorname{\mathcal{C}}$ which witnesses the functor $F(i)$ as given by covariant transport along the edge $\operatorname{N}_{\bullet }( \{ i-1 < i \} ) \subseteq \Delta ^ n$. Let $M$ denote the mapping simplex for the diagram $\operatorname{\mathcal{C}}(0) \xrightarrow {F(1)} \operatorname{\mathcal{C}}(1) \xrightarrow {F(2)} \operatorname{\mathcal{C}}(2) \xrightarrow {F(3)} \cdots \xrightarrow { F(n)} \operatorname{\mathcal{C}}(n)$. Unwinding the definitions, we see that the inclusion maps $\operatorname{\mathcal{C}}(i) \hookrightarrow \operatorname{\mathcal{C}}$ and the natural transformations $\{ h(i) \} _{1 \leq i \leq n}$ determine a morphism of simplicial sets $U_0: \operatorname{Spine}[n] \times _{\Delta ^ n} M \rightarrow \operatorname{\mathcal{C}}$, where $\operatorname{Spine}[n] \subseteq \Delta ^ n$ denotes the spine of the $n$-simplex (see Example 1.4.7.7). Since the inclusion $\operatorname{Spine}[n] \hookrightarrow \Delta ^ n$ is inner anodyne (Example 1.4.7.7), Lemma 5.2.6.17 guarantees that the inclusion $\operatorname{Spine}[n] \times _{\Delta ^ n} M \hookrightarrow M$ is also inner anodyne. Because $\operatorname{\mathcal{C}}$ is an $\infty$-category, we can extend $U_0$ to a morphism of simplicial sets $U: M \rightarrow \operatorname{\mathcal{C}}$ (Proposition 1.4.6.7). Using the criterion of Remark 5.2.6.15, we conclude that $U$ is a scaffold of $\pi$. $\square$