I know that there's already another post that asked the same question. But I am trying to verify my understanding of the correct approach to this problem.
The problem: let $Y$ and $Y'$ be spaces obtained from attaching an n-cell to $X$ by homotopic attaching maps. Prove that $Y$ and $Y'$ are homotopy equivalent.
The approach I have in mind, described pictorially:
Let $f$ and $g$ be the corresponding attaching maps. The author gives the hint of considering the adjunction space $Z =:X \cup_{H} (\bar B^n \times I)$, where $H$ is the homotopy from $f$ to $g$.
This space is basically the disjoint union of different spaces obtained from attaching an n-cell, with the attaching map being a $H(u,t)$ with a fixed $t$.
We try to find a retraction from $Z$ to $X \cup_{H} (\bar B^n \times {0})$ which can be identified with $Y$ (Assuming $H(u,0) = f$).
And to define such a map, my idea was, every points that are in $\bar B^n \times I$ that are not on the boundary of $\bar{B^n}$, are not identified with any points in $X$, So any points that are either on the boundary, or on $X$, are identified with some points in $X$.
so for every fixed $t$, can we not just map $(x, (u,t)) \in Z$ to the points on $x \in X$ they are identified with in $X \cup_{H} (\bar B^n \times {0})$?
so in some sense, can we not define a retraction from $X \cup_{H} (\bar B^n \times I)$ to $X \cup_{H} (\bar B^n \times {0})$ by defining an "inclusion map" for each $X \cup_{H} (\bar B^n \times {t})$ into $X \cup_{H} (\bar B^n \times {0})$?
Sorry for the vague explanation of my approach, this is because I don't even know how to begin to construct such a map, or even know if this approach makes any sense.