Lemma 6.3.7.15. Let $\{ S_{\alpha } \} $ be a diagram of nonsingular simplicial sets. Then the limit $\varprojlim _{\alpha } S_{\alpha }$ is also nonsingular.
$\Newextarrow{\xhookrightarrow}{10,10}{0x21AA}$
Proof. By virtue of Remark 6.3.7.9, 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 1.1.3.8 guarantees that there exists a commutative diagram
\[ \xymatrix@R =50pt@C=50pt{ \Delta ^{n} \ar [r]^-{\sigma } \ar [d]^{\tau _{\alpha }} & S \ar [d] \\ \Delta ^{n_{\alpha }} \ar [r]^-{ \sigma _{\alpha } } & S_{\alpha }, } \]
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
\[ \prod _{\alpha } \Delta ^{n_{\alpha }} \xrightarrow { \prod _{\alpha } \sigma _{\alpha } } \prod _{\alpha } S_{\alpha } = S \]
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 6.3.7.8). It will therefore suffice to show that $\tau $ is nondegenerate, which follows immediately from our assumption that $\sigma $ is nondegenerate. $\square$