1
$\begingroup$

In reading the proof of corollary 0.21 of Hatcher's algebraic toplology. I can not understand how the existence of homotopy equivalence $f:X\rightarrow Y$ implies that the inclusion $X\rightarrow M_f$ is a homotopy equivalence, where $M_f$ is a mapping cylinder. Clearly, $X$ is homotopy equivalent to $M_f$ by the map $if$, where $i: Y\rightarrow M_f$ is an inclusion.

$\endgroup$
3
$\begingroup$

It is because of the "2-out-of-3" property of homotopy equivalences: If $f:X\to Y$ and $g:Y\to Z$ are maps, then if any two of the maps $f,g,gf$ are homotopy equivalences, so is the third map. Now for any map $f:X\to Y$ you have a factorization $$X\stackrel i \hookrightarrow M_f\stackrel r\to Y$$ where $i$ is the cofibration which sends $x$ to $(x,1)$, and $r$ is the map $$ \begin{cases} r(x,t)=f(x), &\text{for } (x,t)\in X\times I \\ r(y)=y, &\text{for } y\in Y \end{cases} $$ Then $r$ is a deformation retraction. Since $f=ri$, the map $i$ must be a homotopy equivalence.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I think the point I can not get through is the "2-out-of-3" property of homotopy equivalences. Why homotopy equivalence has this property? I don't know how to construct a homotopy inverse of $i$ from $f$ and $r$, for $r$ does not have a inverse in general. $\endgroup$ – Lewis Zhang Feb 1 '15 at 1:38
  • 2
    $\begingroup$ @LewisZhang: In this case, $r$ has a homotopy inverse, the inclusion $s:Y\hookrightarrow M_f$. Since $f$ has a homotopy inverse $g$, can you use these homotopy inverses to show that $gr$ and $i$ are homotopy inverses to each other? $\endgroup$ – Stefan Hamcke Feb 1 '15 at 10:05
  • 2
    $\begingroup$ I got it. Since $gri\simeq id_X$, $gr$ is the left homotopy inverse of $i$. Since $igr\simeq (sr)igr=s(ri)gr=s(fg)r\simeq$$sr\simeq id_{M_f}$, $gr$ is the right homotopy inverse of $i$. $\endgroup$ – Lewis Zhang Feb 1 '15 at 12:02
  • $\begingroup$ @LewisZhang: Yes, very good :-) $\endgroup$ – Stefan Hamcke Feb 1 '15 at 12:32

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.