 Solution

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.