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 $M_{f}=A\times I\cup_{f}B$.

| cite | improve this answer | |
  • $\begingroup$ 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? $\endgroup$ – Sina Aug 1 '12 at 20:33
  • $\begingroup$ I attach $M_{f}$ to $A\times I$'s $A\times \{1\}$ side. $\endgroup$ – Bombyx mori Aug 2 '12 at 6:25
  • $\begingroup$ Yes and when we deformation retract $M_f$ to A (or to say A$\times$1) then arent we left with $X\times0 \cup A\times I$? $\endgroup$ – Sina Aug 3 '12 at 7:13
  • $\begingroup$ 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. $\endgroup$ – Bombyx mori Aug 3 '12 at 11:44
  • $\begingroup$ 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. $\endgroup$ – 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 $M_f$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.