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?
Best Answer
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 $n$.