Oriented simplicial subdivision commutes with boundary

For a vertex $v$, $\mathrm{\partial }S[v]=\mathrm{\partial }[\{v\}]=1=S\mathrm{\partial }[v]$. The degree $-1$ equation is zero on each side. On an edge $[a,b]$, the formula is $S[a,b]=[ab,b]-[ab,a]=[a,ab]-[b,ab]$, abbreviating singleton face labels by their vertices and $\{a,b\}$ by $ab$. Its boundary is $[b]-[a]=S\mathrm{\partial }[a,b]$.

Assume boundary compatibility through degree $n-1$. In the cone $\mathrm{sd}\sigma$ the contraction identity gives $\mathrm{\partial }S(s)=\mathrm{\partial }{c}_{\sigma }S(\mathrm{\partial }s)=S(\mathrm{\partial }s)-c_{\sigma }\mathrm{\partial }S(\mathrm{\partial }s)=S(\mathrm{\partial }s)-c_{\sigma }S({\mathrm{\partial }}^{2}s)=S(\mathrm{\partial }s)$. For the augmented boundary square at a one-simplex, $\epsilon ([b]-[a])=0$; in higher degrees use the boundary-square theorem. Induction proves the identity in every degree. Geometrically the cone terms on boundary-of-boundary faces cancel, exactly accounting for internal faces. Setting the degree-zero ordinary boundary to zero preserves the identity on ordinary chains.