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Because $L$ is a simplicial subcomplex of $K$, every face of a simplex of $L$ also belongs to $L$. The simplicial boundary operator therefore satisfiesso $C_{\bullet }(L)$ is a chain subcomplex of the simplicial chain complex $C_{\bullet }(K)$.

Define a map on the quotient groups byIf $c^{\prime }=c+\ell$ with $\ell \in C_k(L)$, then $\partial c^{\prime }-\partial c=\partial \ell \in C_{k-1}(L)$, so this definition is independent of the representative. Moreover,Thus $C_{\bullet }(K,L)$ is the relative simplicial chain complex.

For the pair from part (c), every simplex of dimension below $n$ lies in $L_{n-1}$. Consequently the relative simplicial chain complex haswhere the degree-$n$ generator is the class of $[v_0\cdots v_n]$. Its relative boundary is zero because every codimension-one face lies in $L_{n-1}$. All differentials therefore vanish, andThis is the relative homology of a simplex and its boundary. Equivalently, the long exact sequence in relative homology, together with the fact that $|K_n|$ is a contractible space and the homology of a sphere $|L_{n-1}|\cong S^{n-1}$, gives the reduced homology identification $H_k(K_n,L_{n-1})\cong {\tilde{H}}_{k-1}(S^{n-1})$.

For $n\ge 1$,In the relative simplicial chain complex, every proper face vanishes, leaving one generator in degree $n$ and zero chain groups in all other degrees.