# Hatcher Question 0.27

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 x I $\cup$ X x {0} is a deformation retraction of X x 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 Ax0. 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 alot

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We have $B\cup_{f}X$ is the deformation retraction of $X\cup M_{f}$ because we can deform the part $A\times I$. We can consider the set $X\times \{0\}\cup A\times I\cup M_{f}$ with $M_{f}$ attached to $A\times I$ on $A\times \{1\}$ side. Then this space is homotopical equivalent to $X\times{0}\cup A\times 2I\cong X\times I\cong X$ by the homotopic extension property. On other hand it is obvious that $X\times \{0\}\cup A\times I\cup M_{f}$ is homotopically equivalent to $X\cup M_{f}$. Thus we have $B\cup_{f}X\cong X$.
For your question notice that $B\cup_{f}X=X\cup [A\cup _{f}B]$, and $A\cup_{f}B$ is a deformation retraction of $M_{f}=A\times I\cup_{f}B$.
On the first line shouldnt you say something like $X\times \{0\}\cup A\times I\cup M_{f} \cong X\times \{0\}\cup A\times I$ since $M_f \cong A$. Where does 2I come from? – Sina Aug 1 '12 at 20:33
I attach $M_{f}$ to $A\times I$'s $A\times \{1\}$ side. – Bombyx mori Aug 2 '12 at 6:25
Yes and when we deformation retract $M_f$ to A (or to say A$\times$1) then arent we left with $X\times0 \cup A\times I$? – Sina Aug 3 '12 at 7:13
Yes. I use the picture in mind but you can prove it more directly. I used the deformation of $M_{f}$ to $A\times I$ to construct $A\times 2I$, but maybe this is not needed. – Bombyx mori Aug 3 '12 at 11:44
I should comment that now I realize in general $M_{f}$ cannot be retracted to $A$, since the cone over a circle is not homeormophic to the circle. So your argument does not carry over. – Bombyx mori Aug 7 '12 at 7:49