Let $X$ be a (possibly infinite) CW-complex, with a CW-subcomplex $X'$. The cellular chain complex of $X$, with coefficients in an abelian group $A$, is given by $\tilde{C}_n(X;A)=H_n(X_n,X_{n-1};A)$. I know that one can prove that the map $H_n(X, X';A)\to H_n(X/X',X'/X';A)$ is an iso, by showing that $(X,X')$ is a "good pair". However, I was wondering if there exists a more sophisticated proof using cellular homology, using the isomorphism $H_n(\tilde{C}(X,A))\to H_n(X;A)$. My idea was to find a chain complex $D_\bullet$ such that $0\to \tilde{C}_\bullet(X')\to \tilde{C}_{\bullet}(X)\to D_\bullet\to 0$ is a short exact sequence of chain complexes, then take homology to obtain a LES $$\cdots\to H_n(X',A)\to H_n(X, A)\to H_n(D_\bullet)\to H_{n-1}(X; ,A)\to \cdots$$ and compare this LES to the LES for the pair of spaces $(X, X'):$ $$ \cdots\to H_n(X',A)\to H_n(X,A)\to H_n(X, X',A)\to H_{n-1}(X',A)\to \cdots, $$to relate $H_n(X, X',A)$ and $H_n(D_\bullet)$. My only problem is how to choose $D_\bullet$. For the singular chain complex, the inclusion $X'\subseteq X$ allows for taking the quotient $C_n(X.A)/C_n(X',A)$, but can we do the same for the cellular chain complex, and why? I don't see a priori why $H_n(X_n',X_{n-1}';A)$ can be seen as a subspace of $H_n(X_n,X_{n-1};A)$. Moreover, if I set $$ D_n=\tilde C_n(X)/\tilde C_n(X') $$ how do I relate $H_n(D_\bullet)$ to the relative homology $H_n(X/X',X'/X';A)$?
EDIT THANKS TO TYRONE: I reduced it to the case that I have to show that $$H^{CW}_n(\tilde{C}_n(X)/\tilde{C}_n(X'))\cong \tilde{H}_n(X/X')$$ But I am an unaware how I have to deal with the cellular homology of the quotient of two cellular complexes. My gut feeling tells me that the quotient complex $\textbf{should}$ be isomorphic to $\tilde{C}_\bullet(X/X')$, which has a natural CW-structure, but I am unable to prove this iso.