Does a "good" homotopy equivalence between pairs imply homotopy equivalence between quotient spaces?

Question

If $(X,A)$ and $(Y,B)$ are (good) pairs of topological spaces, and $f:X\rightarrow Y$ is a homotopy equivalence such that the restriction $f\restriction_A$ is a homotopy equivalence between $A$ and $B$, is the corresponding quotient map a homotopy equivalence between the quotient spaces $X/A$ and $Y/B$?

I am a bit vague about the assumption:

• The stronger version is requiring that if $f,g$ is the homotopy equivalence, then $f\restriction_A, g\restriction_B$ is homotopy equivalence between the subspaces. I am primarily interested in this version.
• The weaker version is that there just exist another $g':B\rightarrow A$. I suspect this version might cause issues related to biretractibility vs homotopy equivalence.

If the answer is positive, a relevant reference would be appreciated.

Answer 1

Sometimes quotients behave poorly with respect to homotopy. Thatfor it is better to take the homotopy quotient, e.g. the mapping cone of the inclusion. Its homotopy type only depends on the homotopy class of the inclusion and changing the source and the target by composing with a homotopy equivalence also does not change the homotopy type.

Next the question arises, when the homotopy quotient and the quotient are actually homotopy equivalent. This is for example the case if $A\subset X$ has a regular neighborhood, say a compact neighborhood $A\subset K\subset X$ together with a retraction $K\to A$ such that $K$ is the mapping cone of $\delta K \to A$. Examples of inclusions that have regular neighborhoods are smooth submanifolds or sub-CW-complexes. This might be the notion of 'good' that this question asks for.