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Proposition Let $S$ be a simplicial set and let $k \geq 0$. Then the construction outlined above determines a pushout square

\[ \xymatrix@R =50pt@C=50pt{ \underset { \sigma \in S_{k}^{\mathrm{nd}} }{\coprod } \operatorname{\partial \Delta }^{k} \ar [r] \ar [d] & \underset { \sigma \in S_{k}^{\mathrm{nd}} }{\coprod } \Delta ^{k} \ar [d] \\ \operatorname{sk}_{k-1}( S ) \ar [r] & \operatorname{sk}_{k}( S ) } \]

in the category $\operatorname{Set_{\Delta }}$ of simplicial sets.

Proof. Unwinding the definitions, we must prove the following:

$(\ast )$

Let $\tau $ be an $n$-simplex of $\operatorname{sk}_{k}( S )$ which is not contained in $\operatorname{sk}_{k-1}(S)$. Then $\tau $ factors uniquely as a composition

\[ \Delta ^{n} \xrightarrow {\alpha } \Delta ^{k} \xrightarrow {\sigma } S, \]

where $\sigma $ is a nondegenerate simplex of $S$ and $\alpha $ does not factor through the boundary $\operatorname{\partial \Delta }^{k}$ (in other words, $\alpha $ is surjective on vertices).

Proposition implies that any $n$-simplex of $S$ admits a unique factorization $\Delta ^{n} \xrightarrow {\alpha } \Delta ^{m} \xrightarrow {\sigma } S$, where $\alpha $ is surjective on vertices and $\sigma $ is nondegenerate. Our assumption that $\tau $ belongs to the $\operatorname{sk}_{k}(S)$ guarantees that $m \leq k$, and our assumption that $\tau $ does not belong to $\operatorname{sk}_{k-1}( S )$ guarantees that $m \geq k$. $\square$