$H_p(K)=H_p(K^{(p+1)},K^{(p-2)})$?

Question

If I have a simplicial complex $K$ and I denote the $p$-skeleton by $K^{(p)}$, then how can I see that $H_{p}(K)=H_{p}(K^{(p+1)},K^{(p-2)})$?

Answer 1

Two steps: show that the simplicial chain groups for $K$ and $K^{(p+1)}$ are the same in dimensions up to and including $p+1$, so their homology groups are isomorphic in dimension $p$ (and below). Then use the long exact sequence in homology to show that $H_p(K^{(p+1)}) \cong H_p(K^{(p+1)}, K^{(p-2)})$.