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I'd like to prove that given $f,g : (X,A) \longrightarrow (Y,B)$ homotopic as map of pair, i.e $H(A\ \times I) \subset B$ then they induce the same homology.

I already know the theorem which states that homotopic map induce the same homology, what I'm interested in understanding how $f_*,g_* : H_n(X,A) \longrightarrow H_n(Y,B)$ are the same thanks to this in order to finish the proof. I should be able to conclude from the fact that called $H_{\sharp} : C_n(X) \longrightarrow C_{n+1}(Y)$ the natural homotopy between $f_{\sharp},g_{\sharp}$, this induces a natural homotopy $C_n(A) \longrightarrow C_{n+1}(B)$ which factors to an homotpy $C_n(X,A) \longrightarrow C_n(Y,B)$.

I don't understand how to prove the Lemma since I don't understand how the term natural is used here, even to obtain an homotopy from $C_n(A) \longrightarrow C_{n+1}(B)$.

My definition of natural trasformation is exactly the given on Wikipedia, any help would be appreciated.

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"Natural" here means that for the two chain homotopies $T:C_n(X) \rightarrow C_{n+1}(Y)$ and $T':C_n(A) \rightarrow C_{n+1}(B)$ we have the identity $Ti_{A,X} = i_{B,Y} T'$ where $i_{A,X}:C_*(A) \rightarrow C_*(X)$ and $i_{B,Y}:C_*(B) \rightarrow C_*(Y)$.

This would imply that we could find maps $S:C_n(X)/C_n(A) \rightarrow C_{n+1}(Y)/C_{n+1}(B)$ such that $S \pi_{X,A} = \pi_{Y,B} T$ for purely group theoretic reasons.

$\pi_{X,A}:C_*(X) \rightarrow C_*(X,A)$ and $\pi_{Y,B}:C_*(Y) \rightarrow C_*(Y,B)$ being the projection maps.

$S$ would then be the desired chain homotopy.

That is, $S$ is a chain homotopy between $f_{\#}$ and $g_{\#}:C_*(X,A) \rightarrow C_*(Y,B)$ since $T$ is a chain homotopy between $f_{\#},g_{\#}:C_*(X) \rightarrow C_*(Y)$

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