Kerodon

Proof. Fix an integer $m \geq 0$, and let $\sigma : \Delta ^ m \rightarrow \Delta ^ m$ be the identity map, which we identify with its image in the normalized chain complex $\mathrm{N}_{\ast }( \Delta ^ m; \operatorname{\mathbf{Z}})$. Then $\partial (\sigma ) = \sum _{i=0}^{m} (-1)^{i} d^{m}_{i}(\sigma )$ is a cycle in the subcomplex $\mathrm{N}_{\ast }( \operatorname{\partial \Delta }^{m}; \operatorname{\mathbf{Z}})$. Suppose we are given a morphism of simplicial sets $\tau _0: \operatorname{\partial \Delta }^{m} \rightarrow X$, which we identify with a chain map $f_0: \mathrm{N}_{\ast }( \operatorname{\partial \Delta }^{m}; \operatorname{\mathbf{Z}}) \rightarrow A_{\ast }$. Then $y = f_0( \partial \sigma )$ is an $(m-1)$-cycle of $A_{\ast }$. Note that, if $m > 0$, then every $(m-1)$-cycle $y$ of $A_{\ast }$ can be obtained in this way (for example, we can take $f_0: \mathrm{N}_{\ast }( \operatorname{\partial \Delta }^{m}; \operatorname{\mathbf{Z}}) \rightarrow A_{\ast }$ to be the map of chain complexes which carries $d^{m}_{0}( \sigma )$ to $y$ and every other nondegenerate simplex to zero).

If $\tau : \Delta ^{m} \rightarrow X$ is an extension of $\tau _0$ corresponding to a map of chain complexes $f: \mathrm{N}_{\ast }( \Delta ^{m}; \operatorname{\mathbf{Z}}) \rightarrow A_{\ast }$, then $x = f( \sigma )$ is an $m$-chain of $A_{\ast }$ satisfying $\partial (x) = y$. This construction induces a bijection

It follows that the simplicial set $X$ is $n$-coskeletal if and only if it satisfies the following condition for each $m > n$:

$(b_ m)$

The boundary map $\partial : A_{m} \rightarrow A_{m-1}$ is a monomorphism whose image is the group of $(m-1)$-cycles $Z_{m-1} = \{ y \in A_{m-1}: \partial y = 0 \} $.

Note that $(b_{n+1})$ is a restatement of $(b)$. Moreover, if condition $(b_ m)$ is satisfied for some integer $m$, then condition $(b_{m+1} )$ is equivalent to the requirement that the abelian group $A_{m+1}$ is trivial. In particular, $(b_ m)$ is satisfied for all $m > n$ if and only if $A_{\ast }$ satisfies conditions $(a)$ and $(b)$. $\square$

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

If condition $(2)$ is satisfied, then the vertical maps are bijections, so the horizontal map is a bijection as well. Allowing $m$ to vary, we deduce that $X$ is $n$-coskeletal.

We now prove the converse. For every simplicial set $S$, we can identify $\operatorname{Hom}_{\operatorname{Set_{\Delta }}}(S, X)$ with the inverse limit of the tower of restriction maps

Consequently, to prove $(2)$, it will suffice to show that the restriction map

is a bijection for $m > n$. Using Proposition 1.1.4.12, we can reduce to the case $S = \Delta ^ m$, in which case the statement reduces to the assertion that $X$ is $n$-coskeletal. $\square$

Proof. By virtue of Proposition 3.5.3.10, it will suffice to show that for every simplicial set $S$, the restriction map $\theta : \operatorname{Hom}_{\operatorname{Set_{\Delta }}}( S, \operatorname{Fun}(K,X) ) \rightarrow \operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \operatorname{sk}_{n}(S), \operatorname{Fun}(K,X) )$ is a bijection. Using Proposition 1.5.3.2, we can identify $\theta $ with the horizontal map in the commutative diagram

It will therefore suffice to show that the vertical maps in this diagram are bijections, which follows from our assumption that $X$ is $n$-coskeletal (Proposition 3.5.3.10). $\square$

Proof. For each $n \geq 0$, we can identify $\theta _{m}$ with the restriction map

where $S_{\bullet }$ denotes the colimit $\varinjlim _{ ( [k], \sigma ) \in \operatorname{\mathcal{C}}} \Delta ^ k$. Using Corollary 1.1.4.8, we can identify $S_{\bullet }$ with the $n$-skeleton $\operatorname{sk}_{n}( \Delta ^ m )$, so the desired result follows from Proposition 3.5.3.10 and Remark 3.5.3.11. $\square$

Proof. Let $\operatorname{cosk}_{n}(X)$ denote the simplicial set given by the construction

and let $f: X \rightarrow \operatorname{cosk}_{n}(X)$ be the morphism of simplicial sets given on $m$-simplices by the restriction map $\operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \Delta ^{m}, X ) \rightarrow \operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \operatorname{sk}_{n}( \Delta ^ m), X) )$. If $m \leq n$, then $\operatorname{sk}_{n}( \Delta ^ m ) = \Delta ^ m$; it follows that $f$ is bijective on $m$-simplices for $m \leq n$. We will complete the proof by showing that $\operatorname{cosk}_{n}(X)$ is $n$-coskeletal. Fix an integer $m > n$; we wish to show that the restriction map

is a bijection. Writing $\operatorname{\partial \Delta }^{m}$ as a colimit of simplices (Remark 1.1.3.13) and applying Corollary 1.1.4.9, we can identify $\theta $ with the restriction map

The desired result now follows from the observation that the $n$-skeleton of $\Delta ^ m$ is contained in $\operatorname{\partial \Delta }^ m$. $\square$

Proof. Assertion $(2)$ is equivalent to the requirement that, for every $n$-coskeletal simplicial set $Z$ and every simplicial set $K$, composition with $f$ induces a bijection

By virtue of Corollary 3.5.3.12, we can replace $Z$ by $\operatorname{Fun}(K, Z)$ and thereby reduce to the case $K = \Delta ^0$. This proves the equivalence $(2) \Leftrightarrow (3)$.

The implication $(1) \Rightarrow (3)$ follows immediately from Proposition 3.5.3.10. We will complete the proof by showing that $(3)$ implies $(1)$. Using Proposition 3.5.3.16, we can choose a morphism $u: X \rightarrow \operatorname{cosk}_{n}(X)$ which exhibits $\operatorname{cosk}_{n}(X)$ as an $n$-coskeleton of $X$. Then $u$ satisfies condition $(3)$, so $f$ factors (uniquely) as a composition $X \xrightarrow {u} \operatorname{cosk}_{n}(X) \xrightarrow {g} Y$. We can therefore replace $X$ by $\operatorname{cosk}_{n}(X)$ and thereby reduce to the case where $X$ is $n$-coskeletal. In this case, condition $(3)$ implies that $f$ is an isomorphism of ($n$-coskeletal) simplicial sets, and therefore bijective on $m$-simplices for $m \leq n$. $\square$

Proof. Let $m$ be a positive integer. Fix an integer $0 \leq i \leq m$ and a morphism of simplicial sets $\sigma _0: \Lambda ^{m}_{i} \rightarrow \operatorname{cosk}_{n}(X)$; we wish to show that $\sigma _0$ can be extended to an $m$-simplex of $\operatorname{cosk}_ n(X)$. Using Remark 3.5.3.21, we can identify $\sigma _0$ with a morphism of simplicial sets $f_0: \operatorname{sk}_{n}( \Lambda ^{m}_{i} ) \rightarrow X$; we wish to show that $f_0$ can be extended to the $n$-skeleton of $\Delta ^{m}$. If $n < m-1$, then $\operatorname{sk}_{n}( \Lambda ^{m}_{i} ) = \operatorname{sk}_{n}( \Delta ^{m} )$ and there is nothing to prove. We may therefore assume that $n \geq m-1$, so that $\operatorname{sk}_{n}( \Lambda ^{m}_{i} ) = \Lambda ^{m}_{i}$. In this case, our assumption that $X$ is a Kan complex guarantees that $f_0$ can be extended to an $n$-simplex of $X$. $\square$