I have studied the proof of the homotopy invariance of singular homology. If I understood correctly, the key step is to transform a homotopy $F : X \times I \to Y$ between the maps $f, g : X \to Y$, into a chain homotopy between the induced chain maps $f_n, g_n : X_n \to Y_n$. For this purpose, we define the prism natural transformation $P_n : \mathcal C_n \to \mathcal D_n$ between the functors $\mathcal C_n(X) = X_n$ and $\mathcal D_n(X) = (X \times I)_{n+1}$. This allows us to form the following chain homotopy:
Now I am trying to obtain a similar result for relative homology. Let $A \subset X$ and $B \subset Y$ be subspaces, and let $F : (X \times I, A \times I) \to (Y,B)$ be a homotopy between $f, g : (X,A) \to (Y,B)$. I want to construct the following commutative parallelepiped:
The homotopy invariance of absolute homology gives me the left and middle faces. If two chains in $X_n$ differ by a chain in $A_n$, their prisms differ by a chain in $(A \times I)_{n+1}$, so the right face is well-defined as well. This gives me a copy of the first diagram, with $X_n$, $Y_n$, $(X \times I)_n$ replaced with $X_n / A_n$, $Y_n / B_n$, $(X \times I)_n / (A \times I)_n$ respectively. The diagonals
$$\frac {X_n} {A_n} \longrightarrow \frac {(X \times I)_{n+1}} {(A \times I)_{n+1}} \longrightarrow \frac {Y_{n+1}} {B_{n+1}}$$
constitute a chain homotopy between $f$ and $g$, so the relative homology groups are equal. Does this work? Is there anything missing?
The missing step is the actual calculation. By the homotopy invariance of absolute homology, we have
$$\partial_{n+1} F_{n+1} P_n \gamma + F_n P_{n-1} \partial_n \gamma = g_n \gamma - f_n \gamma$$
Let $\gamma \in X_n$ be a relative cycle. In other words, suppose that $\partial_n \gamma \in A_{n-1}$. Then $F_n P_{n-1} \partial_n \gamma \in B_n$, hence $g_n \gamma - f_n \gamma$ is a relative boundary, hence $f$ and $g$ have the same relative homology at the level $n$.