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Proposition Let $S$ be a simplicial set. Then there exists a universally localizing morphism $\varphi : \widetilde{S} \rightarrow S$, where $\widetilde{S}$ is nonsingular.

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

\[ \xymatrix@R =50pt@C=50pt{ \widetilde{\operatorname{sk}}_{0}(S) \ar@ {^{(}->}[r] \ar [d]^{\varphi _0} & \widetilde{\operatorname{sk}}_{1}(S) \ar [d]^{ \varphi _1} \ar@ {^{(}->}[r] & \widetilde{\operatorname{sk}}_{2}(S) \ar [d]^{ \varphi _2} \ar@ {^{(}->}[r] & \cdots \\ \operatorname{sk}_0(S) \ar@ {^{(}->}[r] & \operatorname{sk}_1(S) \ar@ {^{(}->}[r] & \operatorname{sk}_2(S) \ar@ {^{(}->}[r] & \cdots } \]

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 that the colimit $\widetilde{S} = \varinjlim _{k} \widetilde{\operatorname{sk}}_{k}(S)$ is nonsingular. Applying Proposition, 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 supplies a pushout diagram

\[ \xymatrix@R =50pt@C=50pt{ T_0 \ar [r] \ar [d] & T \ar [d] \\ \operatorname{sk}_{k-1}(S) \ar [r] & \operatorname{sk}_{k}(S). } \]

Note that $T$ is nonsingular (Example, so the simplicial subset $T_0 \subseteq T$ is also nonsingular (Remark 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

\[ ( \widetilde{\operatorname{sk}}_{k-1}(S) \times \widetilde{T}_{0}^{\triangleright } ) \hookleftarrow \widetilde{T}_0 \hookrightarrow ( T \times \widetilde{T}_0^{\triangleright } ). \]

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

\begin{equation} \begin{gathered}\label{equation:resolution-of-sset} \xymatrix@R =50pt@C=50pt{ \widetilde{\operatorname{sk}}_{k-1}(S) \times \widetilde{T}_0^{\triangleright } \ar [d] & \widetilde{T}_0 \ar [l] \ar [r] \ar [d] & T \times \widetilde{T}_0^{\triangleright } \ar [d] \\ \operatorname{sk}_{k-1}(S) & T_0 \ar [l] \ar [r] & T. } \end{gathered} \end{equation}

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

\[ \varphi _{k}: \widetilde{\operatorname{sk}}_{k}(S) \rightarrow \operatorname{sk}_{k-1}(S) \coprod _{T_0} T \simeq \operatorname{sk}_{k}(S). \]

Since the cone $\widetilde{T}_0^{\triangleright }$ is weakly contractible, it follows from Corollary that the vertical maps in the diagram (6.16) are universally localizing. Applying Proposition, 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, 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, we are reduced to proving that the cone $\widetilde{T}_{0}^{\triangleright }$ is nonsingular (Lemma By virtue of Remark, we can reduce further to showing that $\widetilde{T}_{0}$ is nonsingular. This follows from Remark and Lemma, since $\widetilde{T}_0$ can be identified with a simplicial subset of the product $T \times \widetilde{\operatorname{sk}}_{k-1}(S)$. $\square$