Proposition 4.8.3.6. Let $\operatorname{\mathcal{C}}$ be an $\infty $-category, let $n$ be an integer, and let $\operatorname{cosk}_{n}^{\circ }(\operatorname{\mathcal{C}})$ denote the weak $n$-coskeleton of $\operatorname{\mathcal{C}}$. Then:
- $(1)$
The simplicial set $\operatorname{cosk}_{n}^{\circ }(\operatorname{\mathcal{C}})$ is an $\infty $-category.
- $(2)$
The tautological map $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{cosk}_{n}^{\circ }(\operatorname{\mathcal{C}})$ is an inner fibration of $\infty $-categories.
- $(3)$
If $n \geq -1$, the functor $F$ exhibits exhibits $\operatorname{cosk}_{n}^{\circ }(\operatorname{\mathcal{C}})$ as a local $(n-1)$-truncation of $\operatorname{\mathcal{C}}$.
- $(4)$
If $n \neq 0$, then $F$ is an isofibration of $\infty $-categories.
Proof.
For $n < 0$, the weak coskeleton $\operatorname{cosk}_{n}^{\circ }(\operatorname{\mathcal{C}})$ is either empty (if $n = -1$ and $\operatorname{\mathcal{C}}$ is empty) or isomorphic to $\Delta ^0$; in either case, assertions $(1)$ through $(4)$ are clear. We may therefore assume that $n \geq 0$. The map $F$ factors as a composition
\[ \operatorname{\mathcal{C}}\xrightarrow {F'} \operatorname{cosk}_{n+1}(\operatorname{\mathcal{C}}) \xrightarrow {F''} \operatorname{cosk}_{n}^{\circ }(\operatorname{\mathcal{C}}), \]
where $F''$ is a trivial Kan fibration (Proposition 3.5.4.22). Since $\operatorname{cosk}_{n+1}(\operatorname{\mathcal{C}})$ is an $\infty $-category (Proposition 4.8.2.13), assertion $(1)$ follows from Proposition 1.5.5.11.
To prove $(2)$, we proceed as in Proposition 3.5.4.26. Suppose we are given a pair of integers $0 < i < m$; we wish to show that every lifting problem
4.80
\begin{equation} \begin{gathered}\label{equation:fibration-to-strong-coskeleton5} \xymatrix@R =50pt@C=50pt{ \Lambda ^{m}_{i} \ar [r]^-{ \sigma _0 } \ar [d] & \operatorname{\mathcal{C}}\ar [d]^{F} \\ \Delta ^{m} \ar@ {-->}[ur]^{\sigma } \ar [r]^-{ \overline{\sigma } } & \operatorname{cosk}^{\circ }_{n}(\operatorname{\mathcal{C}}) } \end{gathered} \end{equation}
admits a solution. We consider two cases:
If $m \leq n+1$, then we can choose an $m$-simplex $\sigma $ of $\operatorname{\mathcal{C}}$ satisfying $F(\sigma ) = \overline{\sigma }$. Since $F$ is bijective on simplices of dimension $\leq n$, the commutativity of the diagram (4.80) guarantees that $\sigma |_{ \Lambda ^{m}_{i} } = \sigma _0$.
If $m \geq n+2$, then our assumption that $\operatorname{\mathcal{C}}$ is an $\infty $-category guarantees that $\sigma _0$ can be extended to an $m$-simplex $\sigma $ of $X$. The commutativity of the diagram (4.80) then guarantees that $F(\sigma )$ and $\overline{\sigma }$ have the same restriction to the horn $\Lambda ^{m}_{i} \subset \Delta ^{m}$. In particular, they have the same restriction to the $n$-skeleton of $\Delta ^{m}$, so $F(\sigma ) = \overline{\sigma }$.
Since $F''$ is an equivalence of $\infty $-categories, assertion $(3)$ follows by combining Proposition 4.8.2.13 with Remark 4.8.2.12. It remains to prove $(4)$. Let $Y$ be an object of $\operatorname{\mathcal{C}}$ and suppose we are given an isomorphism morphism $\overline{u}: \overline{X} \rightarrow Y$ in the $\infty $-category $\operatorname{cosk}_{n}^{\circ }(\operatorname{\mathcal{C}})$. If $n \geq 1$, then $F$ is bijective on vertices and edges; it follows that we can write $\overline{u} = F(u)$ for a unique morphism $u: X \rightarrow Y$ in $\operatorname{\mathcal{C}}$. To complete the proof, it will suffice to show that $u$ is an isomorphism. Equivalently, we wish to show that the homotopy class $[u]$ is an isomorphism in the homotopy category $\mathrm{h} \mathit{\operatorname{\mathcal{C}}}$. This is clear: the nerve $\operatorname{N}_{\bullet }( \mathrm{h} \mathit{\operatorname{\mathcal{C}}} )$ is weakly $1$-coskeletal (Example 3.5.4.6), so the tautological map $\operatorname{\mathcal{C}}\rightarrow \operatorname{N}_{\bullet }( \mathrm{h} \mathit{\operatorname{\mathcal{C}}} )$ factors (uniquely) through $\operatorname{cosk}^{\circ }_{n}(\operatorname{\mathcal{C}})$ (Proposition 3.5.4.18).
$\square$