Kerodon

Proof of Proposition 6.3.7.2. By virtue of Proposition 6.3.6.10, we may assume without loss of generality that the simplicial set $S$ is finite. If $S$ is empty, there is nothing to prove. We may therefore assume that $S$ has dimension $n$ for some integer $n \geq 0$. We proceed by induction on $n$ and on the number of nondegenerate $n$-simplices of $S$. Fix a nondegenerate $n$-simplex $\sigma : \Delta ^ n \rightarrow S$. Using Proposition 1.1.3.13, we see that there is a pushout square of simplicial sets

where $S'$ is a simplicial set of dimension $\leq n$ having fewer nondegenerate $n$-simplices than $S$. Applying Proposition 6.3.6.11 to the commutative diagram

we are reduced to showing that the morphisms $\lambda _{S'}$, $\lambda _{\operatorname{\partial \Delta }^{n}}$, and $\lambda _{\Delta ^{n}}$ are universally localizing. In the first two cases, this follows from our inductive hypothesis. We are therefore reduced to proving Proposition 6.3.7.2 in the special case where $S = \Delta ^ n$ is a standard simplex.

Using Example 3.3.3.5, we can identify the subdivision $\operatorname{Sd}(S) = \operatorname{Sd}( \Delta ^ n )$ with the nerve of the partially ordered set $\operatorname{Chain}[n]$ of nonempty subsets $P \subseteq [n]$. We wish to show that, for every morphism of simplicial sets $\alpha : T \rightarrow S$, the projection map $\pi : T \times _{S} \operatorname{Sd}(S) \rightarrow T$ exhibits $T$ as a localization of $T \times _{S} \operatorname{Sd}(S)$ with respect to some collection of edges. By virtue of Proposition 6.3.6.2, we can assume without loss of generality that $T = \Delta ^{m}$ is a standard simplex, so that $\alpha $ can be identified with a nondecreasing map of linearly ordered sets $[m] \rightarrow [n]$. Unwinding the definitions, we can identify $T \times _{S} \operatorname{Sd}(S)$ with the nerve of the partially ordered set $A \subseteq [m] \times \operatorname{Chain}[n]$ consisting of those pairs $(i, P)$ satisfying $\max (P) = \alpha (i)$. Under this identification, the projection map $\pi $ is induced by the morphism of partially ordered sets

It follows that $\pi $ is a reflective localization: it has a fully faithful right adjoint, given by the construction $i \mapsto (i, \{ 0 < 1 < \cdots < \alpha (i) \} )$. The desired result is now a consequence of Proposition 6.3.3.9. $\square$

Proof. Combine Proposition 6.3.7.2 with Remark 6.3.7.10. $\square$

Proof. By virtue of Remark 6.3.7.7, it will suffice to show that the product $S = \prod _{\alpha } S_{\alpha }$ is nonsingular. Let $\sigma : \Delta ^ n \rightarrow S$ be a nondegenerate simplex of $S$; we wish to show that $\sigma $ is a monomorphism of simplicial sets. For each index $\alpha $, Proposition 1.1.3.4 guarantees that there exists a commutative diagram

where $\sigma _{\alpha }$ is a nondegenerate simplex $S_{\alpha }$. Our assumption that $S_{\alpha }$ is nondegenerate guarantees that $\sigma _{\alpha }$ is a monomorphism of simplicial sets, so that the product map

is also a monomorphism. It will therefore suffice to show that $\tau = \{ \tau _{\alpha } \} $ determines a monomorphism of simplicial sets $\Delta ^{n} \rightarrow \prod _{\alpha } \Delta ^{n_{\alpha }}$. Since $\prod _{\alpha } \Delta ^{n_{\alpha }}$ can be identified with the nerve of the partially ordered set $\prod _{\alpha } [n_{\alpha } ]$, it is a nonsingular simplicial set (Example 6.3.7.6). It will therefore suffice to show that $\tau $ is nondegenerate, which follows immediately from our assumption that $\sigma $ is nondegenerate. $\square$

Proof of Proposition 6.3.7.12. Let $S$ be a simplicial set. For each integer $k \geq 0$, let $\operatorname{sk}_{k}(S)$ denote the $k$-skeleton of $S$ (Construction 1.1.3.5). We will construct a commutative diagram

where each of the horizontal maps is a monomorphism, each of the vertical maps is universally localizing, and each of the simplicial sets $\widetilde{\operatorname{sk}}_{k}(S)$ is nonsingular. It then follows from Remark 6.3.7.8 that the colimit $\widetilde{S} = \varinjlim _{k} \widetilde{\operatorname{sk}}_{k}(S)$ is nonsingular. Applying Proposition 6.3.6.10, we conclude that the morphisms $\varphi _{k}$ determine a universally localizing morphism $\varphi : \widetilde{S} \rightarrow S$.

The construction of the morphisms $\varphi _{k}: \widetilde{\operatorname{sk}}_{k}(S) \rightarrow \operatorname{sk}_{k}(S)$ proceeds by induction. If $k=0$, we can take $\widetilde{\operatorname{sk}}_{k}(S) = \operatorname{sk}_ k(S)$ and $\varphi _{k}$ to be the identity morphism. Let us therefore assume that $k > 0$, and that the morphism $\varphi _{k-1}: \widetilde{\operatorname{sk}}_{k-1}(S) \rightarrow \operatorname{sk}_{k-1}(S)$ has already been constructed. Let $S_{k}^{\mathrm{nd}}$ denote the set of nondegenerate $k$-simplices of $S$, let $T$ denote the coproduct $\coprod _{\sigma \in S_ k^{\mathrm{nd}}} \Delta ^{k}$, and let $T_0 \subseteq T$ denote the coproduct $\coprod _{\sigma \in S_{k}^{\mathrm{nd}}} \operatorname{\partial \Delta }^{k}$, so that Proposition 1.1.3.13 supplies a pushout diagram

Note that $T$ is nonsingular (Example 6.3.7.6), so the simplicial subset $T_0 \subseteq T$ is also nonsingular (Remark 6.3.7.7). Let $\widetilde{T}_0$ denote the fiber product $T_0 \times _{ \operatorname{sk}_{k-1}(S)} \widetilde{\operatorname{sk}}_{k-1}(S)$, and we define $\widetilde{\operatorname{sk}}_{k}(S)$ to be the pushout of the diagram

Note that the cone point of $\widetilde{T}_{0}^{\triangleright }$ determines an embedding $\widetilde{\operatorname{sk}}_{k-1}(S) \rightarrow \widetilde{\operatorname{sk}}_{k}(S)$. Moreover, we have a commutative diagram

which determines an extension of $\varphi _{k-1}$ to a map

Since the cone $\widetilde{T}_0^{\triangleright }$ is weakly contractible, it follows from Corollary 6.3.6.9 that the vertical maps in the diagram (6.16) are universally localizing. Applying Proposition 6.3.6.11, we deduce that $\varphi _{k}$ is also universally localizing.

To complete the proof, it will suffice to show that the simplicial set $\widetilde{\operatorname{sk}}_{k}(S)$ is nonsingular. By virtue of Remark 6.3.7.8, it will suffice to show that the simplicial subsets $\widetilde{\operatorname{sk}}_{k-1}(S) \times \widetilde{T}_0^{\triangleright }$ and $T \times \widetilde{T}_0^{\triangleright }$ are nonsingular. Since $\widetilde{\operatorname{sk}}_{k-1}(S)$ is nonsingular (by our inductive hypothesis) and $T$ is nonsingular (Example 6.3.7.6), we are reduced to proving that the cone $\widetilde{T}_{0}^{\triangleright }$ is nonsingular (Lemma 6.3.7.13). By virtue of Remark 6.3.7.9, we can reduce further to showing that $\widetilde{T}_{0}$ is nonsingular. This follows from Remark 6.3.7.7 and Lemma 6.3.7.13, since $\widetilde{T}_0$ can be identified with a simplicial subset of the product $T \times \widetilde{\operatorname{sk}}_{k-1}(S)$. $\square$

Proof of Theorem 6.3.7.1. Let $S$ be a simplicial set. Applying Proposition 6.3.7.12, we can choose a universally localizing morphism $\varphi : \widetilde{S} \rightarrow S$, where $\widetilde{S}$ is a nonsingular simplicial set. Let $A = \operatorname{Sub}_{\operatorname{{\bf \Delta }}}( \widetilde{S} )$ denote the partially ordered set of simplicial subsets of $\widetilde{S}$ which are isomorphic to a standard simplex, so that Corollary 6.3.7.11 supplies a universally localizing morphism $\lambda _{ \widetilde{S} }: \operatorname{N}_{\bullet }(A) \rightarrow \widetilde{S}$. Applying Proposition 6.3.6.8, we deduce that the composite morphism

is also universally localizing. $\square$