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If $f,g:(X,A)\to (Y,B)$ are homotopic functions under a homotopy $H:X\times [0,1]\to Y$ such that $H(a,t)\in B$ for all $a\in A$ and $t\in [0,1]$, then $f_*=g_*:H_n(X,A)\to H_n(Y,B)$.

This question has already been posted here Homotopic maps induce the same homomorphism for reduced homology groups and here Homotopic maps $(X, A) \to (Y, B)$ induce homotopic maps $X/A \to X/B$ but I still have many doubts, I would like to know where we are using that $f,g:(X,A)\to (Y,B)$ are homotopic in Homotopic maps $(X, A) \to (Y, B)$ induce homotopic maps $X/A \to X/B$.

I think it is enough to show that $f,g:X/A\to Y/B$ are homotopic, because in this case we already know that $f_*=g_*:H_n(X/A)\to H_n(Y/B)$ where $H_n(X/A)=H_n(X,A)$ and $H_n(Y/B)=H_n(Y,B)$. Now if $f,g:(X,A)\to (Y,B)$ are homotopic, how do I prove that $f,g:X/A\to Y/B$ are homotopic? Thank you.

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    $\begingroup$ To see why the answer to this question is wrong, read the comment by Jason DeVito following the answer. $\endgroup$
    – Lee Mosher
    Jul 3, 2019 at 18:46
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    $\begingroup$ @Nash in the linked question $\hat{F}$ is defined from the initial homotopy $F:I\times X\to Y$. $\endgroup$
    – freakish
    Jul 3, 2019 at 20:31
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    $\begingroup$ @Nash also that idea won't lead you to a general solution. Because $H_n(X/A)=H_n(X, A)$ is not true in general. You need some additional assumptions on $(X,A)$. Like $A$ closed with an open neighbourhood that deformation retracts onto $A$ (a.k.a. a good pair). $\endgroup$
    – freakish
    Jul 3, 2019 at 20:35
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    $\begingroup$ The first linked question is actually a good reference. There's a whole proof in the question itself (ignore the incorrect answer). $\endgroup$
    – freakish
    Jul 3, 2019 at 20:43
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    $\begingroup$ Oh, and there's one more issue with $H_n(X/A)=H_n(X,A)$ approach: there is no equality. There is an isomorphism. And in order to pass from maps $H_n(X,A)\to H_n(Y,B)$ to maps $H_n(X/A)\to H_n(Y/B)$ (and back) you would need to know that this isomorphism is natural. Which I'm not even sure if it's true even for good pairs. So as you can see there are lots of subtleties here. I advice not going down that road. $\endgroup$
    – freakish
    Jul 3, 2019 at 20:49

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Your main goal is to prove that if $f,g:(X,A)\to (Y,B)$ are homotopic functions under a homotopy $H:X\times [0,1]\to Y$ such that $H(a,t)\in B$ for all $a\in A$ and $t\in [0,1]$, then $f_*=g_*:H_n(X,A)\to H_n(Y,B)$.

This is usually done by working on the level of chain complexes. See any book treating singular homology.

Your idea was to reduce the proof for pairs to the "already known absolute case" by using that $H_n(X,A) = H_n(X/A)$. Unfortunately this is not true in general (see freakish's comments).

However, we can answer your final question: If $f,g:(X,A)\to (Y,B)$ are homotopic, how to show that $f,g:X/A\to Y/B$ are homotopic?

Let $H$ be a homotopy as above and let $p : X \to X/A, q : Y \to Y/B$ denote the quotient maps. Then also $p \times id : X \times I \to (X/A) \times I$ is a quotient map. This is a well-known fact from general toplogy. Now consider the map $H' = q \circ H : X \times I \to Y/B$. It has the property that $H'(a,t) = [B]$ for all $a \in A$, where $[B]$ denotes the common equivalence class of all points in $B$. Hence the function $$H'' : (X/A) \times I \to Y/B, H''([x],t) = H'(x,t)$$ is well-defined and satisfies $H'' \circ (p \times id) = H'$. By the universal property of quotient maps this equation implies that $H''$ is continuous. But now you have $H''([x],0) = H'(x,0) = qf(x) = f''([x])$ where $f'' : X/A \to Y/B$ denotes the map induced by $f$. Similarly $H''([x],1) = g''([x])$. This proves $f'' \simeq g''$.

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