# Kerodon

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

Proposition 4.8.1.7. Let $\operatorname{\mathcal{C}}$ be an $\infty$-category and let $n$ be a positive integer. Then $\operatorname{\mathcal{C}}$ is an $(n,1)$-category (in the sense of Definition 4.8.1.1) if and only if it is weakly $n$-coskeletal and minimal in dimension $n$.

Proof. We proceed as in the proof of Proposition 3.5.5.12. Assume first that $\operatorname{\mathcal{C}}$ is an $(n,1)$-category. Then, for any integer $m > n$, the composition of the restriction maps

$\operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \Delta ^ m, \operatorname{\mathcal{C}}) \xrightarrow {\theta _ m} \operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \operatorname{\partial \Delta }^{m}, \operatorname{\mathcal{C}}) \rightarrow \operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \Lambda ^{m}_{1}, \operatorname{\mathcal{C}})$

is a bijection. In particular, $\theta _ m$ is an injection. To show that $\operatorname{\mathcal{C}}$ is weakly $n$-coskeletal, it will suffice to show that $\theta _{m}$ is surjective for $m \geq n+2$. Fix a morphism $\sigma _0: \operatorname{\partial \Delta }^{m} \rightarrow \operatorname{\mathcal{C}}$; we wish to show that $\sigma _0$ can be extended to an $m$-simplex of $\operatorname{\mathcal{C}}$. Since $\operatorname{\mathcal{C}}$ is an $(n,1)$-category, there is a unique $m$-simplex $\sigma$ of $\operatorname{\mathcal{C}}$ satisfying $\sigma |_{ \Lambda ^{m}_{1} } = \sigma _0|_{ \Lambda ^{m}_{1} }$. We complete the argument by observing that $\sigma |_{ \operatorname{\partial \Delta }^{m} } = \sigma _0$, by virtue of the injectivity of the map $\theta _{m-1}$.

We next show that if $\operatorname{\mathcal{C}}$ is an $(n,1)$-category, then it is minimal in dimension $n$. Let $\sigma _0, \sigma _1: \Delta ^ n \rightarrow \operatorname{\mathcal{C}}$ be $n$-simplices of $\operatorname{\mathcal{C}}$ and let $h: \Delta ^1 \times \Delta ^ n \rightarrow \operatorname{\mathcal{C}}$ be a natural isomorphism from $\sigma _0 = h|_{ \{ 0\} \times \Delta ^ n}$ to $\sigma _1 = h|_{ \{ 1\} \times \Delta ^ n }$ whose restriction to $\Delta ^1 \times \operatorname{\partial \Delta }^ n$ factors through $\operatorname{\partial \Delta }^ n$; we wish to show that $\sigma _0 = \sigma _1$. For $0 \leq i \leq n$, let $\alpha _{i}: [n+1] \rightarrow [1] \times [n]$ denote the nondecreasing function given by the formula

$\alpha _{i}(j) = \begin{cases} (0, j) & \text{ if } j \leq i \\ (1, j-1) & \text{ if } j > i, \end{cases}$

and let $\tau _{i}$ denote the $(n+1)$-simplex of $\operatorname{\mathcal{C}}$ given by the composition

$\Delta ^{n+1} \xrightarrow { \alpha _{i} } \Delta ^1 \times \Delta ^ n \xrightarrow {h} \operatorname{\mathcal{C}}.$

Let $\rho _{i}, \rho '_{i}: \Delta ^{n} \rightarrow \operatorname{\mathcal{C}}$ be the $n$-simplices of $X$ given by $\rho _{i} = d^{n+1}_{i}( \tau _ i )$ and $\rho '_{i} = d^{n+1}_{i+1}( \tau _ i )$; by construction, we have

$\sigma _0 = \rho '_{n} \quad \quad \rho _{n} = \rho '_{n-1} \quad \quad \cdots \quad \quad \rho _{1} = \rho '_{0} \quad \quad \rho _0 = \sigma _1.$

We will complete the proof by showing that $\rho _{i} = \rho '_{i}$ for $0 \leq i \leq n$. We will treat the case $i > 0$ (the case $i < n$ follows by a similar argument). Using our assumption that $h$ is constant along the boundary $\operatorname{\partial \Delta }^{n}$, we see that the degenerate $(n+1)$-simplex $s^{n}_{i}( \rho _{i} )$ coincides with $\tau _ i$ on the horn $\Lambda ^{n+1}_{i} \subset \Delta ^{n+1}$. Since $\operatorname{\mathcal{C}}$ is an $(n,1)$-category, it follows that $\tau _ i = s^{n}_{i}( \rho _{i} )$. Applying the face operator $d^{n+1}_{i+1}$, we obtain $\rho _{i} = \rho '_{i}$.

We now prove the converse. Assume that $\operatorname{\mathcal{C}}$ is a weakly $n$-coskeletal $\infty$-category which is minimal in dimension $n$; we will show that it is an $(n,1)$-category. Fix a pair of integers $0 < i < m$ with $m > n$ and a pair of $m$-simplices $\tau _0, \tau _1: \Delta ^{m} \rightarrow X$ which coincide on the horn $\Lambda ^{m}_{i} \subset \Delta ^{m}$; we wish to show that $\tau _0 = \tau _1$. Since $\operatorname{\mathcal{C}}$ is weakly $n$-coskeletal, it will suffice to prove that $\tau _0$ and $\tau _1$ coincide on the boundary $\operatorname{\partial \Delta }^{m}$: that is, to show that the $(m-1)$-simplices $\sigma _0 = d^{m}_{i}(\tau _0)$ and $\sigma _1 = d^{m}_{i}(\tau _1)$ coincide. Note that $\sigma _0$ and $\sigma _1$ have the same restriction to the boundary $\operatorname{\partial \Delta }^{m-1}$. Consequently, if $m \geq n+2$, the desired result follows from our assumption that $\operatorname{\mathcal{C}}$ is weakly $n$-coskeletal. We may therefore assume that $m = n+1$. Since $\operatorname{\mathcal{C}}$ is minimal in dimension $n$, it will suffice to show that there is an isomorphism from $\sigma _0$ to $\sigma _1$ (in the $\infty$-category $\operatorname{Fun}(\Delta ^ n, \operatorname{\mathcal{C}})$) whose image in $\operatorname{Fun}( \operatorname{\partial \Delta }^{n}, \operatorname{\mathcal{C}})$ is an identity morphism. In fact, we will prove a stronger claim: there is an isomorphism from $\tau _0$ to $\tau _1$ in the $\infty$-category $\operatorname{Fun}( \Delta ^{m}, \operatorname{\mathcal{C}})$ whose image in $\operatorname{Fun}( \Lambda ^{m}_{i}, \operatorname{\mathcal{C}})$ is an identity morphism. This follows from the observation that the restriction map $\operatorname{Fun}( \Delta ^{m}, \operatorname{\mathcal{C}}) \rightarrow \operatorname{Fun}( \Lambda ^{m}_{i}, \operatorname{\mathcal{C}})$ is a trivial Kan fibration; see Proposition 1.5.7.6. $\square$