Barycentric subdivision is a chain map

In degree $0$, both sides vanish. Assume $\mathrm{\partial }S=S\mathrm{\partial }$ on dimensions below $n$, where $n\ge 1$.

Let ${\iota }_n$ be the identity simplex of ${\mathrm{\Delta }}^n$ and put $z=S(\mathrm{\partial }{\iota }_n)$, the affine lift used in the definition of $S\sigma$. If $n=1$, then $z=[v_1]-[v_0]$, so its augmentation is zero. The degree-zero cone formula gives $\mathrm{\partial }S\sigma ={\sigma }_{\mathrm{\#}}z=S(\mathrm{\partial }\sigma )$.

If $n\ge 2$, the induction hypothesis applied in ${\mathrm{\Delta }}^n$ gives $\mathrm{\partial }z=S({\mathrm{\partial }}^2{\iota }_n)=0$ by The singular boundary squares to zero. The positive-degree cone formula therefore gives $\mathrm{\partial }S\sigma ={\sigma }_{\mathrm{\#}}z-b_{\sigma }(\mathrm{\partial }z)={\sigma }_{\mathrm{\#}}z=S(\mathrm{\partial }\sigma )$. The last equality follows directly from the recursive definition: subdivision commutes with postcomposition by $\sigma$ on each affine face.

Linearity extends these equalities to finite chains, completing the induction.