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

Corollary Let $\operatorname{\mathcal{C}}$ be an $\infty $-category. For every integer $n$, there exists a functor $F: \operatorname{\mathcal{C}}\rightarrow \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}})}$ which exhibits $\mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}})}$ as a homotopy $n$-category of $\operatorname{\mathcal{C}}$. Moreover:


The functor $F$ is bijective on $m$-simplices for $m < n$.


The functor $F$ factors (uniquely) as a composition $\operatorname{\mathcal{C}}\xrightarrow { F' } \operatorname{cosk}_{n}^{\circ }(\operatorname{\mathcal{C}}) \xrightarrow {F''} \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}})}$, where $F'$ is the inner fibration of Proposition


The functor $F''$ is a trivial Kan fibration.


If $n \geq -1$, the functor $F$ exhibits $\mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}})}$ as a local $(n-1)$-truncation of $\operatorname{\mathcal{C}}$. In particular, $\operatorname{\mathcal{C}}$ is locally $(n-1)$-truncated if and only if $F$ is an equivalence of $\infty $-categories.


The functor $F$ is an isofibration.

Proof. The existence of $F$ follows from Example (in the case $n < 0$), Exercise (in the case $n = 0$), and Proposition (in the case $n > 0$). Assertion $(1)$ is vacuous for $n \leq 0$, and follows from Construction for $n > 0$. Since $\mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}})}$ is an $(n,1)$-category, it is weakly $n$-coskeletal, so that assertion $(2)$ follows from Proposition

We next prove $(3)$. For $n < 0$, the morphism $F''$ is an isomorphism (see Example and there is nothing to prove. For $n > 0$, the desired result follows from Corollary We may therefore assume that $n = 0$. We wish to show that every lifting problem

\[ \xymatrix@C =50pt@R=50pt{ \operatorname{\partial \Delta }^{m} \ar [r] \ar [d] & \operatorname{cosk}_{n}^{\circ }(\operatorname{\mathcal{C}}) \ar [d]^{F''} \\ \Delta ^{m} \ar [r] \ar@ {-->}[ur] & \operatorname{\mathcal{C}}' } \]

admits a solution. For $m \geq 2$, this is automatic (since $\operatorname{cosk}_{n}^{\circ }(\operatorname{\mathcal{C}})$ and $\operatorname{\mathcal{C}}'$ are both $1$-coskeletal). The cases $m = 0$ and $m=1$ follow immediately from the construction of $\mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}})}$ given in Exercise

Assertion $(4)$ follows by combining $(3)$ with Proposition We now prove $(5)$. For $n \neq 0$, the morphism $F'$ is an isofibration (Proposition, so the desired result follows from $(3)$. In the case $n = 0$, $\mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}})}$ is isomorphic to the nerve of a partially ordered set, so the result is automatic (Example $\square$