I'm currently reading a paper where the authors have a map of pairs $(X,A) \to (Y,B)$ which induces an equivalence of cofibers $X/A \simeq Y/B$. They also have a fibration $Y' \to Y$, obtaining by pullback a map of pairs $(X',A') \to (Y',B')$. They conclude that the map $X'/A' \to Y'/B'$ is also an equivalence.
Question: Why does this implication hold?
I can see how it would follow from a stronger hypothesis. Namely, one good reason we might have $X/A \simeq Y/B$ would be if we had a pushout $Y = X \cup_A B$. Then because pushouts are stable under pullback we would have $Y' = X' \cup_{A'} B'$ and thus $X'/A' \simeq Y'/B'$. But in the paper I'm reading it appears that this is not the case.
EDIT: Not sure how helpful this will be, but the place I'm stuck is in the guts of the proof of Prop 3.10 of Devinatz-Hopkins-Smith. In order to read that proposition in isolation, what you need to know is that $J_j S^{2m}$ denotes the $j$th stage of the James construction and that $E_r \to J_r S^{2m}$ is an arbitrary fibration, with $E_j \to J_j S^{2m}$ defined to be the pullback for $j\leq r$. Note that the cofiber of $J_{j-1} S^{2m} \to J_j S^{2m}$ is $S^{2mj}$. The other potentially relevant information is the precise nature of the map $J_j S^{2m} \to S^{2mj} \vee J_j S^{2m}$ (not the wedge summand inclusion) which I don't fully understand, but which comes from Construction 3.7.
Thanks to Tyrone's suggestion, I revisited the idea that $Y$ might be the pushout $Y = X \cup_A B$ even though in my case it's not the most obvious possible pushout diagram between the spaces in question, and it appears that this does hold after all. In fact, the following criterion applies in this case:
Proposition: Let $$\require{AMScd} \begin{CD} A @>>> X\\ @VVV @VVV\\ B @>>> Y \end{CD}$$
be a square (with cofibrations on the rows, say) such that $X/A \to Y/B$ is an equivalence. Suppose further that $Y$ is simply-connected (or more generally, a simple space) and $\pi_1(Y)$ is the free product $\pi_1(Y) = \pi_1(X) \ast_{\pi_1(A)} \pi_1(B)$ (e.g. this holds if all these spaces are simply-connected). Then the above square is a homotopy pushout.
Proof: Let $P$ be the homotopy pushout. Then we have a map of cofiber sequences:
$$\require{AMScd} \begin{CD} B @>>> P @>>> P/B = X/A \\ @VVV @VVV @VVV \\ B @>>> Y @>>> Y/B \end{CD}$$
where the first and third maps are equivalences. Applying the 5-lemma to the long exact sequence in homology, we conclude that $P \to Y$ is a homology equivalence. Since $Y$ is simply-connected or at least simple) and the map is also a $\pi_1$-isomorphism by the van Kampen theorem, it is an equivalence by the homology Whitehead theorem.