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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 a subspace $A\subseteq X$, the relative chain complex is the quotient group in each degree,with boundary operator $[c]\mapsto [\partial c]$. This is well defined because $C_{\ast }(A)$ is a chain subcomplex of $C_{\ast }(X)$.