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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.

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2 Answers 2

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

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  • $\begingroup$ 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, 2012 at 20:33
  • $\begingroup$ I attach $M_{f}$ to $A\times I$'s $A\times \{1\}$ side. $\endgroup$ Aug 2, 2012 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, 2012 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. $\endgroup$ Aug 3, 2012 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. $\endgroup$ Aug 7, 2012 at 7:49
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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 $M_f$.

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