Simplicial subdivision is a chain map and homology isomorphism

Both $S$ and ${\lambda }_{\mathrm{\#}}$ commute with boundary and augmentation. For a nonempty original face $\sigma$, carry ${\lambda }_{\mathrm{\#}}S$ and $1$ by the full simplex on $\sigma$, a cone with apex its greatest vertex. Indeed $S$ stays over $\sigma$ and $\lambda$ lands in $\sigma$. These carriers are nested under faces. The specified cone contractions and the carried-homotopy lemma give ${\lambda }_{\mathrm{\#}}S\simeq 1$.

For a face chain $\eta =({\sigma }_0<\cdots <{\sigma }_q)$ carry $S{\lambda }_{\mathrm{\#}}$ and $1$ by $\mathrm{sd}{\sigma }_q$, a cone with apex ${\sigma }_q$. The identity lies there; ${\lambda }_{\mathrm{\#}}\eta$ either vanishes or is a face of ${\sigma }_q$, whose subdivision lies there as well. Removing any chain vertex leaves the same or a smaller maximum, so the carriers are nested. The carried-homotopy lemma gives $S{\lambda }_{\mathrm{\#}}\simeq 1$. Its homotopies have $h_{-1}=0$, so restriction also gives ordinary chain homotopies.

Chain-homotopic maps induce equal homology maps, hence $H({\lambda }_{\mathrm{\#}})H(S)=1$ and $H(S)H({\lambda }_{\mathrm{\#}})=1$. These equations hold in every augmented degree and every ordinary degree. If $K$ has no vertices, both augmented degree $-1$ groups are $\mathbb{Z}$ and both maps are the identity. Any other vertex order produces another two-sided inverse $J$ to the same $H(S)$; then $J=JH(S)H({\lambda }_{\mathrm{\#}})=H({\lambda }_{\mathrm{\#}})$, proving independence.