# 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 $$A \times I \cup X \times \{0\}$$ 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.

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

In this question, $$f:X\rightarrow Y$$ is a homotopy equivalence is needed to make sure $$M_f$$ deformation retracts to $$A$$.

Hatcher's Algebraic Topology Corollary 0.21 says:

Corollary 0.21: A map $$f:X\rightarrow Y$$ is a homotopy equivalence iff $$X$$ is a deformation retract of the mapping cylinder $$M_f$$.