 Solution

Inclusion and passage to the quotient give a short exact sequence of chain complexesDefine the connecting homomorphism as follows. If $[c]\in H_k(K,L)$, then $\bar{\partial }[c]=0$, so $\partial c\in C_{k-1}(L)$. Since ${\partial }^{2}c=0$, it is a chain cycle in $L$, and we setChanging $c$ by a relative boundary or a chain in $L$ changes $\partial c$ by a boundary in $L$, so $\delta$ is well defined.

The resulting sequence isWe verify exactness directly. If a cycle of $K$ maps to zero in the relative homology, it differs from a boundary by a cycle in $L$, so it lies in $\mathrm{im}{i}_{\ast }$. If a relative cycle $[c]$ satisfies $\delta [c]=0$, then $\partial c=\partial \ell$ for some $\ell \in C_k(L)$; hence $c-\ell$ is a cycle in $K$ mapping to $[c]$. Finally, if a cycle $z\in C_{k-1}(L)$ becomes a boundary in $K$, say $z=\partial c$, then $[c]$ is a relative cycle with $\delta [c]=[z]$. The reverse containments follow immediately from consecutive composites being zero. This proves the long exact sequence in relative homology.