I am self studying hatchers book however I have been stuck on some questions. This is one of them.
Let $(X,A)$ satisfy homotopy extension property. Let $f: A \rightarrow B$ be a homotopy equivalance. Show that the natural map $X \rightarrow B \cup_{f}X$ is a homotopy equivalance.
Hint: Consider $X \cup M_{f}$ (mapping cylinder of f) and apply the theorem stating if $(X,A)$ satisfies homotopy extension then $A \times I \cup X \times \{0\}$ is a deformation retraction of $X \times I$.
My idea was to show that the subset of $B \cup_{f} X$ correspoding to $X$ is deformation retract of $B \cup_{f} X$. I have some scattered ideas which I can not combine to make use of the hint above.
1- Since $A$ is homotopy equivalent to $B$, the mapping cylinder deformation retracts to both $A$ and $B$. This means $X \cup M_{f}$ deformation retracts to $X$.
2- We can attach $X$ to the mapping cylinder $M_{f}$ by the map sending $A$ to $A\times\{0\}$. Thus we have a space $M_{f} \cup_{g} X$. This space also deformation reracts to $X$.
If I show that $B \cup_{f} X$ is homotopy equivalent to $X \cup_{g} M_{f}$ which deformation retracts to $X$ I guess I will be able to finish off the proof from here. The first statement seems intiutively true when you imagine $X \cup_{g} M_{f}$ embedded in some big space and contract the cylinder in between X and $B \subset M_{f}$, but I have not been able to do so rigorously.
Thanks a lot.