Kerodon

Proof. For every integer $m > n$, the $n$-skeleton $\operatorname{sk}_{n}( \Delta ^ m )$ is contained in the boundary $\operatorname{\partial \Delta }^{m}$, and therefore coincides with the $n$-skeleton of $\operatorname{\partial \Delta }^ m$. We therefore have a commutative diagram of restriction maps

If condition $(1)$ is satisfied, then the vertical maps are injective, so the upper horizontal map is also injective. Moreover, if $m \geq n+2$, then $\operatorname{\partial \Delta }^{m}$ contains the $(n+1)$-skeleton of $\Delta ^ m$. In this case, condition $(2)$ guarantees that the vertical maps have the same image, so that the horizontal map is bijective. It follows that $X$ is weakly $n$-coskeletal.

We now prove the converse. Assume that $X$ is weakly $n$-coskeletal, and let $S$ be a simplicial set. Then we can identify $\operatorname{Hom}_{\operatorname{Set_{\Delta }}}(S, X)$ with the inverse limit of the tower of restriction maps

Consequently, to prove $(1)$, it will suffice to show that the restriction map

is injective for $m=n$ and bijective for $m > n$. Using Proposition 1.1.4.12, we can reduce to the case $S = \Delta ^{m+1}$ is a standard simplex, in which case the desired result is immediate from the definition. $\square$

Proof. It follows from Corollary 3.5.3.12 that $\operatorname{Fun}(K,X)$ is $(n+1)$-coskeletal. It will therefore suffice to show that, if $n \geq -1$, then the restriction map $\theta : \operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \Delta ^{n+1}, \operatorname{Fun}(K,X) ) \rightarrow \operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \operatorname{\partial \Delta }^{n+1}, \operatorname{Fun}(K, X) )$ is injective. Note that $\theta $ can be identified with the restriction map

Since $\operatorname{\partial \Delta }^{n+1} \times K$ contains the $n$-skeleton of $\Delta ^{n+1} \times K$, the injectivity of this map follows from Proposition 3.5.4.12. $\square$

Proof. The tautological map $X \rightarrow \operatorname{cosk}_{n+1}(X)$ is bijective on $m$-simplices for $m \leq n+1$, so $f$ is surjective on $m$-simplices for $m \leq n+1$. Moreover, for $m \leq n$, the composite map $X \xrightarrow {f} \operatorname{cosk}_{n}^{\circ }(X) \hookrightarrow \operatorname{cosk}_{n}(X)$ is bijective on $m$-simplices for $m \leq n$, so that $f$ is injective on $m$-simplices. To complete the proof, it will suffice to show that $\operatorname{cosk}_{n}^{\circ }(X)$ is weakly $n$-coskeletal. Fix an integer $m > n$ and a morphism $\sigma _0: \operatorname{\partial \Delta }^{m} \rightarrow \operatorname{cosk}_{n}^{\circ }(X)$. Since $\operatorname{cosk}_{n}(X)$ is $n$-coskeletal, the morphism $\sigma _0$ extends uniquely to an $m$-simplex $\sigma $ of $\operatorname{cosk}_{n}(X)$. To complete the proof, it will suffice to show that if $m \geq n+2$, then $\sigma $ is contained in $\operatorname{cosk}_{n}^{\circ }(X)$: that is, that it can be lifted to an $m$-simplex of $\operatorname{cosk}_{n+1}(X)$. Using Remark 3.5.3.21, we can identify $\sigma $ with a morphism of simplicial sets $u: \operatorname{sk}_{n}( \Delta ^{m} ) \rightarrow X$; we wish to show that $u$ can be extended to the $(n+1)$-skeleton of $\Delta ^{m}$. By virtue of Proposition 1.1.4.12, this is equivalent to the requirement that, for every nondegenerate $(n+1)$-simplex $\tau $ of $\Delta ^{m}$, the composite map

can be extended to an $(n+1)$-simplex of $X$. This follows from our assumption that $\sigma _0 \circ \tau $ factors through $\operatorname{cosk}_{n}^{\circ }(X)$. $\square$

Proof. Condition $(3)$ is equivalent to the requirement that, for every weakly $n$-coskeletal simplicial set $Z$ and every simplicial set $K$, composition with $f$ induces a bijection

By virtue of Corollary 3.5.4.13, we can replace $Z$ by $\operatorname{Fun}(K, Z)$ and thereby reduce to the case $K = \Delta ^0$. This proves the equivalence $(2) \Leftrightarrow (3)$.

We next show that $(1)$ implies $(3)$. Assume that condition $(1)$ is satisfied, and let $Z$ be a weakly $n$-coskeletal simplicial set. We then have a commutative diagram

Since $f$ is bijective on $m$-simplices for $m \leq n$, the lower horizontal map is a bijection. Using Proposition 3.5.4.12, we see that the vertical maps are injective. Consequently, to prove $(3)$, it will suffice to show that their images agree (under the bijection provided by the the lower horizontal map). This follows from Proposition 3.5.4.12, together with our assumption that $f$ is surjective on $(n+1)$-simplices.

We now show that $(3)$ implies $(1)$. Using Proposition 3.5.4.17, we can choose a morphism $u: X \rightarrow \operatorname{cosk}^{\circ }_{n}(X)$ which exhibits $\operatorname{cosk}^{\circ }_{n}(X)$ as a weak $n$-coskeleton of $X$. Then $u$ satisfies condition $(3)$, so $f$ factors (uniquely) as a composition $X \xrightarrow {u} \operatorname{cosk}^{\circ }_{n}(X) \xrightarrow {g} Y$. We can therefore replace $X$ by $\operatorname{cosk}^{\circ }_{n}(X)$ and thereby reduce to the case where $X$ is weakly $n$-coskeletal. Condition $(3)$ then guarantees that $f$ is an isomorphism of (weakly $n$-coskeletal) simplicial sets, so condition $(1)$ is automatic. $\square$

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

admits a solution. We consider two cases:

• If $m \leq n+1$, then $\overline{\sigma }$ can be lifted to an $m$-simplex of $X$. In particular, there exists an $m$-simplex $\sigma $ of $\operatorname{cosk}_{n+1}(X)$ satisfying $q(\sigma ) = \overline{\sigma }$. Since $q$ is bijective on $k$-simplices for $k \leq n$, the commutativity of the diagram (3.73) guarantees that $\sigma |_{ \operatorname{\partial \Delta }^{m} } = \sigma _0$.

• If $m \geq n+2$, then $\sigma _0$ extends uniquely to an $m$-simplex $\sigma $ of $\operatorname{cosk}_{n+1}(X)$. The commutativity of diagram (3.73) guarantees that $q(\sigma )$ and $\overline{\sigma }$ have the same restriction to $\operatorname{\partial \Delta }^{m}$, and therefore coincide (since $\operatorname{\partial \Delta }^ m$ contains the $(n+1)$-skeleton of $\Delta ^ m$).

In either case, the $m$-simplex $\sigma $ is a solution to the lifting problem (3.73). $\square$

Proof. Proposition 3.5.4.22 supplies a trivial Kan fibration

Since $\operatorname{cosk}_{n+1}(X)$ is a Kan complex (Proposition 3.5.3.23), it follows that $\operatorname{cosk}_{n}^{\circ }(X)$ is also a Kan complex (Proposition 1.5.5.11). $\square$

Proof. The equivalence of $(2)$ and $(3)$ follows from Remark 3.5.4.21. The implication $(3) \Rightarrow (1)$ follows from the observation that $f$ factors as a composition

where the outer maps are trivial Kan fibrations (Proposition 3.5.4.22). We will complete the proof by showing that $(2)$ implies $(3)$. Suppose we are given a morphism $\sigma _0: \operatorname{\partial \Delta }^{n+2} \rightarrow \operatorname{\mathcal{C}}$. Since $\operatorname{cosk}_{n}^{\circ }(\operatorname{\mathcal{C}})$ is $(n+1)$-coskeletal we can extend $F \circ \sigma _0$ to an $(n+2)$-simplex $\overline{\sigma }$ of $\operatorname{cosk}_{n}^{\circ }(\operatorname{\mathcal{C}})$. If condition $(2)$ is satisfied, then the lifting problem

admits a solution; in particular, $\sigma _0$ can be extended to an $(n+2)$-simplex of $\operatorname{\mathcal{C}}$. $\square$

Proof of Proposition 3.5.4.26. Fix a pair of integers $0 \leq i \leq m$ with $m > 0$; we wish to show that every lifting problem

admits a solution. We consider two cases:

• If $m \leq n+1$, then we can choose an $m$-simplex $\sigma $ of $X$ satisfying $u(\sigma ) = \overline{\sigma }$. Since $u$ is bijective on simplices of dimension $\leq n$, the commutativity of the diagram (3.74) guarantees that $\sigma |_{ \Lambda ^{m}_{i} } = \sigma _0$.

• If $m \geq n+2$, then our assumption that $X$ is a Kan complex guarantees that $\sigma _0$ can be extended to an $m$-simplex $\sigma $ of $X$. The commutativity of the diagram (3.74) then guarantees that $u(\sigma )$ 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 $u(\sigma ) = \overline{\sigma }$.

In either case, it follows that $\sigma $ is a solution to the lifting problem (3.74). $\square$