# pair homotopic maps induce the same homology

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.

"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 $$f_{\#},g_{\#}:C_*(X) \rightarrow C_*(Y)$$