= 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 satisfies
$$
\partial C_k(L)\subseteq C_{k-1}(L),
$$
so $C_\bullet(L)$ is a <chain subcomplex> of the <simplicial chain complex> $C_\bullet(K)$.
Define a map on the <quotient groups> by
$$
\bar\partial_k:C_k(K,L)\longrightarrow C_{k-1}(K,L),
\qquad
\bar\partial_k(c+C_k(L))
=\partial c+C_{k-1}(L).
$$
If $c'=c+\ell$ with $\ell\in C_k(L)$, then $\partial c'-\partial c=\partial\ell\in C_{k-1}(L)$, so this definition is independent of the representative. Moreover,
$$
\bar\partial_{k-1}\bar\partial_k[c]
=[\partial^2c]=0.
$$
\b[Thus $C_\bullet(K,L)$ is the <relative simplicial chain complex>.]
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