at the moment I am learning a little bit algebraic topology.
So let $(H_\ast, d_\ast)$ be a ordinary homology theory satisfying the additivy axiom in the sense of the Eilenberg Steenrod axioms.
Let $X$ be a CW complex. Then we can define the cellular chain complex of X associated to the $(H_\ast, d_\ast)$ by $C_n^{cell} (X) = H_n(X^n, X^{n-1})$. The boundary operator is given by the boundary operator of the long exact sequence of the triple $(X^n, X^{n-1}, X^{n-2})$. The homology of this chain complex is called the cellular homology associated to $(H_\ast, d_\ast)$ and is denoted by $H_n^{CW}(X)$.
I already proved the following theorem:
If A is a CW-subcomplex of X, then $C_\ast^{cell}(A) \subseteq C_\ast^{cell}(X)$ is a subcomplex.
Now we have an induced long exakt sequence of chain complexes and define the relative cellular homology $H_n^{CW}(X,A)$ by the homology of the quotient complex.
Now I want to prove, that there is an isomorphism $H_n(X,A) \to H_n^{CW}(X,A)$.
I already understand the absolute case (see e.g. Hatcher, p.140). But I do not know, how to extend this proof to the relative case.
My idea was to look at the long exact sequences an use the five lemma
$$ \begin{matrix} H_n^{CW}(A)& \to & H_n^{CW}(X) & \to & H_n^{CW}(X,A) & \to & H_{n-1}^{CW}(A) & \to & H_{n-1}^{CW}(X)\\ \downarrow & & \downarrow & & & & \downarrow & & \downarrow\\ H_n(A)& \to & H_n(X) & \to & H_n(X,A) & \to & H_{n-1}(A) & \to & H_{n-1}(X) \end{matrix} $$
But I do not know how to define the map in the middle of the diagram above.
Thanks in advance, Felix
Edit due to Connor Malin's Answer:
I was able to prove that $$C^{cell}_{n}(X,A) \cong H_n(X^n/A^n, X^{n-1}/A^{n-1})$$ for all $n \geq 1$ and $$C^{cell}_{0}(X,A) \cong H_0(X^0/A^0, A^0/A^0)$$ and the boundary operator is just the boundery operator from the long exact sequence of the tripel $(X^n/A^n, X^{n-1}/A^{n-1}, X^{n-2}/A^{n-2})$ resp. $(X^1/A^1, X^0/A^0, A^0/A^0)$
Now I want to adapt the absolute case to proof that $H_n(X/A, A/A) \cong H_n^{CW}(X,A)$.
So, I considered an aanalog diagram as in Hatcher (p.139):
Edit due to Connor Malin's Answer:
I was able to prove that $$C^{cell}_{n}(X,A) \cong H_n(X^n/A^n, X^{n-1}/A^{n-1})$$ for all $n \geq 1$ and $$C^{cell}_{0}(X,A) \cong H_0(X^0/A^0, A^0/A^0)$$ and the boundary operator is just the boundery operator from the long exact sequence of the tripel $(X^n/A^n, X^{n-1}/A^{n-1}, X^{n-2}/A^{n-2})$ resp. $(X^1/A^1, X^0/A^0, A^0/A^0)$
Now I want to adapt the absolute case to proof that $H_n(X/A, A/A) \cong H_n^{CW}(X,A)$.
So I considered the following diagram as in Hatcher (p.139):
$$\begin{array}{rclrl} &&&& 0\\ &&& \nearrow\\ && H_n(X^{n+1}/A^{n+1}, A/A) &\cong& H_n(X/A, A/A) \\ &&\nearrow \\ &H_n(X^n/A^n, A/A)\\ \nearrow&& \searrow\\ H_{n+1}(X^{n+1}/A^{n+1}, X^n/A^n) & \rightarrow & H_n(X^n/A^n, X^{n-1}/A^{n-1}) & \rightarrow &H_{n-1}(X^{n-1}/A^{n-1}, X^{n-2}/A^{n-2}) \end{array}$$
But I do not know how to choose ?. I already tried $H_n(X^n/A^n, X^{n-1}/A^{n-1})$ and $H_n(X^n/A^n, A^{n-1}/A^{n-1})$. But both seems not to work, since I was not able to show that $H_n(X^{n+1}/A^{n+1}, X^{n-1}/A^{n-1})$ or $H_n(X^{n+1}/A^{n+1}, A^{n-1}/A^{n-1})$ are isomorphic to $H_n(X/A, A/A)$.
And I also see, why this commutes with cellular maps of pairs of CW-complexes. But why does the connecting morphism commute with $d$, i.e. why does
$$\begin{matrix} H_n(X,A) & \overset{d_n}{\rightarrow} & H_{n-1}(A)\\ \downarrow & & \downarrow\\ H_n^{CW}(X,A) & \overset{d_n^{CW}}{\rightarrow} & H_{n-1}^{CW}(A) \end{matrix}$$
commute?
From the identification of $X^n / X^{n-1}$ with a wedge of spheres, you can see that your relative chain complex $C^{cell}(X,A)$ is exactly $C^{cell}(X/A)$ with the exception of ${C_0}^{cell}$ which is generated by all the cells except the point $A$ was quotiented to. This means that ${H_*}^{CW}(X,A) \cong {\bar{H}_*}(X/A)$