Kerodon

Proof. Assume that $(1)$ is satisfied; we will prove $(2)$ (the reverse implication follows immediately from the definitions). Let $\operatorname{\mathcal{D}}$ be an $(n,1)$-category; we wish to show that precomposition with $F$ induces an isomorphism of simplicial sets from $\operatorname{Fun}( \operatorname{\mathcal{C}}', \operatorname{\mathcal{D}})$ to $\operatorname{Fun}(\operatorname{\mathcal{C}}, \operatorname{\mathcal{D}})$. Equivalently, we wish to show that for every simplicial set $K$, the induced map

This follows by applying condition $(1)$ to the simplicial set $\operatorname{Fun}(K, \operatorname{\mathcal{D}})$, which is an $(n,1)$-category by virtue of Proposition 4.8.1.22. $\square$

Proof. Since $\operatorname{\mathcal{C}}'$ is an $(n,1)$-category, it is locally $(n-1)$-truncated (Example 4.8.2.2). It will therefore suffice to show that, for every locally $(n-1)$-truncated $\infty $-category $\operatorname{\mathcal{D}}$, precomposition with $F$ induces an equivalence of $\infty $-categories $\theta : \operatorname{Fun}( \operatorname{\mathcal{C}}', \operatorname{\mathcal{D}}) \rightarrow \operatorname{Fun}(\operatorname{\mathcal{C}}, \operatorname{\mathcal{D}})$ (Proposition 4.8.2.18). By virtue of Corollary 4.8.3.3, we may assume that $\operatorname{\mathcal{D}}$ is an $(n,1)$-category. In this case, Proposition 4.8.4.7 guarantees that $\theta $ is an isomorphism of simplicial sets. $\square$

Proof. We proceed as in the proof of Proposition 3.5.6.12. Fix a morphism of simplicial sets $g: A \rightarrow \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}})}$. Using Remark 4.8.4.10 (and Proposition 1.1.4.12), we see that $g|_{ \operatorname{sk}_{n}(A) }$ can be lifted to a morphism of simplicial sets $u: \operatorname{sk}_{n}(A) \rightarrow \operatorname{\mathcal{C}}$. We will show that $u$ can be extended to the $(n+1)$-skeleton of $A$ (and is therefore classified by a morphism $f: A \rightarrow \operatorname{cosk}_{n}^{\circ }(\operatorname{\mathcal{C}})$ satisfying $\theta (f) = g$; see Remark 3.5.4.21). By virtue of Proposition 1.1.4.12, this is equivalent to the assertion that for every $(n+1)$-simplex $\sigma $ of $A$ having restriction $\sigma _0 = \sigma |_{ \operatorname{\partial \Delta }^{n+1} }$, the composition $(g \circ \sigma _0): \operatorname{\partial \Delta }^{n+1} \rightarrow \operatorname{\mathcal{C}}$ can be extended to an $(n+1)$-simplex of $\operatorname{\mathcal{C}}$. Choose a lift of $g(\sigma )$ to an $(n+1)$-simplex of $\operatorname{cosk}_{n}^{\circ }(\operatorname{\mathcal{C}})$, which we can identify with a diagram $\tau _0: \operatorname{\partial \Delta }^{n+1} \rightarrow \operatorname{\mathcal{C}}$ which admits an extension $\tau : \Delta ^{n+1} \rightarrow \operatorname{\mathcal{C}}$. By construction, $g \circ \sigma _0$ and $\tau _0$ coincide after composing with the comparison map $\operatorname{\mathcal{C}}\rightarrow \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}})}$. Using Proposition 1.1.4.12 again, we see that $g \circ \sigma _0$ and $\tau _0$ are isomorphic relative to $\operatorname{sk}_{n-1}( \Delta ^{n+1} )$. The desired result now follows from Corollary 4.4.5.3. This completes the proof that $\theta $ is surjective.

Now suppose that we are given a pair of morphisms $f_0, f_1: A \rightarrow \operatorname{cosk}_{n}^{\circ }(\operatorname{\mathcal{C}})$ satisfying $\theta (f_0) = \theta (f_1)$. We wish to show that the associated maps $u_0, u_1: \operatorname{sk}_{n}(A) \rightarrow \operatorname{\mathcal{C}}$ are isomorphic relative to $\operatorname{sk}_{n-1}(A)$ (the converse is immediate from the definitions). Using Remark 4.8.4.10, we deduce that $u_0$ and $u_1$ coincide on $\operatorname{sk}_{n-1}(A)$. By virtue of Proposition 1.1.4.12, we are reduced to showing that for every nondegenerate $n$-simplex $\sigma $ of $A$, the compositions $u_0 \circ \sigma $ and $u_1 \circ \sigma $ are isomorphic relative to $\operatorname{\partial \Delta }^{n}$. This follows from our assumption that the maps $\theta (f_0), \theta (f_1): A \rightarrow \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}})}$ coincide on the simplex $\sigma $. $\square$

Proof. Fix an integer $m \geq 0$; we wish to show that every lifting problem

admits a solution.

Let $\sigma $ be any $m$-simplex of $\operatorname{cosk}_{n}^{\circ }(\operatorname{\mathcal{C}})$ satisfying $q( \sigma ) = \overline{\sigma }$. By virtue of Remark 4.8.4.10, the commutativity of the diagram (4.81) guarantees that $\sigma _0$ and $\sigma $ coincide on the $(n-1)$-skeleton of $\operatorname{\partial \Delta }^{m}$. Consequently, if $m \leq n$, then $\sigma $ is a solution to the the lifting problem (4.81). We will therefore assume that $m > n$. In this case, the boundary $\operatorname{\partial \Delta }^{m}$ contains the $n$-skeleton of $\Delta ^{m}$. It will therefore suffice to show that $\sigma _0$ can be extended to an $m$-simplex $\sigma '$ of $\operatorname{cosk}_{n}^{\circ }(\operatorname{\mathcal{C}})$: the commutativity of the diagram (4.81) guarantees that any such extension satisfies the identity $q( \sigma ' ) = \overline{\sigma }$ (Proposition 4.8.4.11). If $m \geq n+2$, then the existence of $\sigma '$ is automatic (since $\operatorname{cosk}^{\circ }_{n}(\operatorname{\mathcal{C}})$ is $(n+1)$-coskeletal). It will therefore suffice to treat the case $m = n+1$. In this case, we can identify $\sigma _0$ with a diagram $\tau _0: \operatorname{\partial \Delta }^{n+1} \rightarrow \operatorname{\mathcal{C}}$, and we wish to show that $\tau _0$ can be extended to an $(n+1)$-simplex of $\operatorname{\mathcal{C}}$. Note that $\sigma |_{ \operatorname{\partial \Delta }^{n+1} }$ determines another diagram $\tau _1: \operatorname{\partial \Delta }^{n+1} \rightarrow \operatorname{\mathcal{C}}$. Moreover, the commutativity of the diagram (4.81) guarantees that $\tau _0$ and $\tau _1$ are isomorphic relative to $\operatorname{sk}_{n-1}( \Delta ^{n+1} )$ (Proposition 4.8.4.11). Using Corollary 4.4.5.3, we are reduced to showing that $\tau _1$ can be extended to an $(n+1)$-simplex of $\operatorname{\mathcal{C}}$, which follows from the existence of $\sigma $. $\square$

Proof. By virtue of Proposition 4.8.3.6, the weak $n$-coskeleton $\operatorname{cosk}^{\circ }_{n}(\operatorname{\mathcal{C}})$ is an $\infty $-category. Combining Corollary 4.8.4.13 with Proposition 1.5.5.11, we conclude that $\mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}})}$ is also an $\infty $-category. To complete the proof, it will suffice to show that if $\sigma $ and $\tau $ are $m$-simplices of $\mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}})}$ for some $m > n$ which satisfy $\sigma |_{ \Lambda ^{m}_{i} } = \tau |_{ \Lambda ^{m}_{i} }$ for some $0 < i < m$, then $\sigma = \tau $. Choose maps $\widetilde{\sigma }, \widetilde{\tau }: \operatorname{sk}_{n}( \Delta ^{m} ) \rightarrow \operatorname{\mathcal{C}}$ representing $\sigma $ and $\tau $. Using Proposition 4.8.4.11, we can choose an isomorphism $\alpha $ from $\widetilde{\sigma }|_{ \operatorname{sk}_{n}( \Lambda ^{m}_{i}) }$ to $\widetilde{\tau }|_{ \operatorname{sk}_{n}(\Lambda ^{m}_{i} ) }$ whose image in $\operatorname{Fun}( \operatorname{sk}_{n-1}( \Lambda ^{m}_{i} ), \operatorname{\mathcal{C}})$ is an identity morphism. If $m \geq n+2$, then $\alpha $ is also an isomorphism from $\widetilde{\sigma }|_{ \operatorname{sk}_{n}( \Delta ^ m) }$ to $\widetilde{\tau }|_{ \operatorname{sk}_{n}( \Delta ^ m ) }$, so that $\sigma = \tau $ as desired. In the case $m = n+1$, the morphisms $\widetilde{\sigma }$ and $\widetilde{\tau }$ can be extended to diagrams $\overline{\sigma }, \overline{\tau }: \Delta ^{n+1} \rightarrow \operatorname{\mathcal{C}}$. Using Proposition 1.5.7.6, we can extend $\alpha $ to an isomorphism of $\overline{\sigma }$ with $\overline{\tau }$. Restricting to the $n$-skeleton of $\Delta ^{m}$, we again conclude that $\sigma = \tau $. $\square$

Proof. By virtue of Corollary 4.8.4.14, it will suffice to show that for every $(n,1)$-category $\operatorname{\mathcal{D}}$, the composite map

is a bijection. Since $\operatorname{\mathcal{D}}$ is weakly $n$-coskeletal, the map $\theta '$ is a bijection. By construction, $\mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}})}$ is a quotient of the weak coskeleton $\operatorname{cosk}_{n}^{\circ }(\operatorname{\mathcal{C}})$, so $\theta $ is an injection. We will complete the proof by showing that $\theta $ is also a surjection: that is, every diagram $F: \operatorname{cosk}_{n}^{\circ }(\operatorname{\mathcal{C}}) \rightarrow \operatorname{\mathcal{D}}$ factors through $\mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}})}$. Let $\sigma $ and $\sigma '$ be $m$-simplices of $\operatorname{cosk}_{n}^{\circ }(\operatorname{\mathcal{C}})$ satisfying $\sigma \sim _{m} \sigma '$ (see Construction 4.8.4.9); we wish to show that $F(\sigma ) = F(\sigma ')$. This follows from the observation that $\operatorname{\mathcal{D}}$ is minimal in dimension $n$ (Proposition 4.8.1.7). $\square$

Proof. The existence of $F$ follows from Example 4.8.4.5 (in the case $n < 0$), Exercise 4.8.4.4 (in the case $n = 0$), and Proposition 4.8.4.15 (in the case $n > 0$). Assertion $(1)$ is vacuous for $n \leq 0$, and follows from Construction 4.8.4.9 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 3.5.4.18.

We next prove $(3)$. For $n < 0$, the morphism $F''$ is an isomorphism (see Example 4.8.4.5) and there is nothing to prove. For $n > 0$, the desired result follows from Corollary 4.8.4.13. We may therefore assume that $n = 0$. We wish to show that every lifting problem

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 4.8.4.4.

Assertion $(4)$ follows by combining $(3)$ with Proposition 4.8.3.6. We now prove $(5)$. For $n \neq 0$, the morphism $F'$ is an isofibration (Proposition 4.8.3.6), 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 4.4.1.6). $\square$

Proof. If $n \leq 0$, then we can identify $\operatorname{\mathcal{C}}_{01}$ with the full subcategory of $\operatorname{\mathcal{C}}_0 \times \operatorname{\mathcal{C}}_1$ spanned by those objects $(C_0, C_1)$ satisfying $F_0(C_0) = F_1(C_1)$. In this case, the desired result follows by combining Remarks 4.8.4.18 and 4.8.4.19. We may therefore assume without loss of generality that $n > 0$. Fix a simplicial set $A$; we wish to show that the tautological map

is a monomorphism. We first show that $\theta $ is injective. Suppose we are given a pair of maps $u,u': A \rightarrow \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}_{01})}$ satisfying $\theta (u) = \theta (u')$; we wish to show that $u = u'$. Using Remark 4.8.4.12, we can choose representatives of $u$ and $u'$ by morphisms $\widetilde{u}, \widetilde{u}': \operatorname{sk}_{n}(A) \rightarrow \operatorname{\mathcal{C}}_{01}$. Our assumption that $\theta (u) = \theta (u')$ guarantees that there are natural isomorphisms

which are the identity when restricted to $\operatorname{sk}_{n-1}(A)$. It follows from the proof of Proposition 4.8.1.7 shows that $\alpha _0$ and $\alpha _1$ have the same image in $\operatorname{Fun}( \operatorname{sk}_{n}(A), \operatorname{\mathcal{C}})$. We can therefore identify the pair $(\alpha _0, \alpha _1)$ with an isomorphism from $\widetilde{u}$ to $\widetilde{u}'$ (which is the identity when restricted to $\operatorname{sk}_{n-1}(A)$), which proves that $u = u'$.

We now prove that $\theta $ is surjective. Choose an element $(u_0, u_1)$ of the fiber product $\operatorname{Hom}_{\operatorname{Set_{\Delta }}}( A, \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}_0)} ) \times _{\operatorname{Hom}_{\operatorname{Set_{\Delta }}}(A, \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}})} )} \operatorname{Hom}_{\operatorname{Set_{\Delta }}}( A, \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}}_1)} )$. Using Remark 4.8.4.12, we can choose representatives of $u_0$ and $u_1$ by morphisms $\widetilde{u}_0: \operatorname{sk}_{n+1}(A) \rightarrow \operatorname{\mathcal{C}}_0$ and $\widetilde{u}_1: \operatorname{sk}_{n+1}(A) \rightarrow \operatorname{\mathcal{C}}_1$. Since $\operatorname{\mathcal{C}}$ is an $(n,1)$-category, the tautological map $\operatorname{\mathcal{C}}\rightarrow \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}})}$ is an isomorphism. It follows that $F_0 \circ \widetilde{u}_0$ coincides with $F_1 \circ \widetilde{u}_1$, so that the pair $( \widetilde{u}_0, \widetilde{u}_1 )$ determines a morphism $\widetilde{u}: \operatorname{sk}_{n+1}(A) \rightarrow \operatorname{\mathcal{C}}$. This represents a morphism $u: A \rightarrow \mathrm{h}_{\mathit{\leq n}}\mathit{(\operatorname{\mathcal{C}})}$ satisfying $\theta (u) = (u_0, u_1)$. $\square$