Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This question is very similar to this one, but the difference is that I'm asking for a strong deformation retraction.

Notation/Definitions: (all maps are by definition continuous) A homotopy between maps $f,g\!:X\rightarrow Y$ is a map $H(x,t)\equiv H_t(x):X\!\times\!I\rightarrow Y$, $H_0\!=\!f$, $H_1\!=\!g$; denoted $H\!:\!f\!\simeq\!g$. For $A\!\subseteq\!X$, $f|_A\!=\!g|_A$, a homotopy relative to $A$ is a homotopy $H\!:\!f\!\simeq\!g$, $\forall t\!:H_t|_A\!=\!f|_A\!=\!g|_A$; denoted $H\!:\!f\!\simeq\!g \,(\mathrm{rel}\,A)$. Spaces $X$ and $Y$ are homotopy equivalent, denoted $X\!\simeq\!Y$, if there is an $f\!:X\rightarrow Y$ and $g\!:Y\rightarrow X$, such that $f\circ g\simeq id_Y$ and $g\circ f\simeq id_X$; then $f$ is an homotopy equivalence and $g$ is its homotopy inverse. A deformation retraction of $X$ onto $A\!\subseteq\!X$, denoted $H\!:X\searrow A$, is $H\!:id_X\!\simeq\!r$, where $r\!:X\rightarrow X$ is a retraction, i.e. $r(X)\!=\!A$, $r|_A\!=\!id_A$. A strong deformation retraction of $X$ onto $A\!\subseteq\!X$, denoted $H\!:X\searrow\searrow A$, is $H\!:id_X\!\simeq\!r\,(\mathrm{rel}\,A)$, where $r$ is a retraction.

Furthermore, $H^-$ denotes the inverse homotopy, i.e. $H(x,1-t)$ and $H\ast H'$ denotes the product homotopy, i.e. $H(x,2t)$ for $0\leq t\leq1/2$ and $H'(x,2t-1)$ for $1/2\leq t\leq1$ (under the condition that $H_1=H'_0$, i.e. "the endfunctions match"), read from left to right (contrary to composition of maps). Lastly, $\big(\! \begin{smallmatrix} \scriptstyle x&\!\scriptstyle \mapsto \! &\scriptstyle f(x)\\ \scriptstyle y&\!\scriptstyle \mapsto \! &\scriptstyle g(y) \end{smallmatrix}\!\big)$ simply denotes a mapping, defined on two spaces.

Proposition: $X\!\simeq\!Y$ $\;\;\;\Leftrightarrow\;\;\;$ $\exists Z\supseteq X,Y:$ $\;X\swarrow Z\searrow Y$

Proof: $(\Leftarrow):$ If $r\!:Z\rightarrow X$ is a retraction, $i\!:X\hookrightarrow Z$ the inclusion, and $r\!\simeq\!id_Z$, then $r\circ i\!=\!id_X$ and $i\circ r\!\simeq\!id_Z$, so $Z\!\simeq\!X$. Similarly $Z\!\simeq\!Y$, and by transitivity, $X\!\simeq Y$.

$(\Rightarrow):$ Let $f\!:X\!\rightarrow\!Y$ be the homotopy equivalence with homotopy inverse $g$ and define $Z\!:=\!Z_f\!=\!(X\!\times\!I)\coprod Y/_{(x,0)\sim f(x)}$, the mapping cylinder of $f$. enter image description here Clearly $$\left(\! \begin{matrix} (x,s,t)&\!\!\!\! \mapsto \!\!\!\! & (x,s(1-t))\\ (y,t) &\!\!\!\! \mapsto \!\!\!\! & y \end{matrix}\!\right):Z_f\searrow\searrow Y.$$ Let $\widetilde{H}\!\!:f\!\circ\!g\!\simeq\!id_Y$ and $\widehat{H}\!\!:g\!\circ\!f\!\simeq\!id_X$. Then define $r\!:Z_f\!\rightarrow\!Z_f$, $r\!:=\! \big(\! \begin{smallmatrix} \scriptstyle(x,s)&\!\scriptstyle \mapsto \! &\scriptstyle(\widehat{H}(x,s),1)\\ \scriptstyle y &\!\scriptstyle \mapsto \! &\scriptstyle (g(y),1) \end{smallmatrix}\!\big)$. We see that $r$ is well defined (respects $(x,0)=f(x)$), $r(Z_f)\!=\!X\!\times\!\{1\}$, $r|_{X\!\times\!\{1\}}\!=\!id_{X\!\times\!\{1\}}$, which makes $r$ the retraction from $Z_f$ to $Y$. Finally, we construct a homotopy $$H\!:=\! \big(\! \begin{smallmatrix} (x,s,t)&\! \mapsto \! &(x,s(1-t))\\ (y,t) &\! \mapsto \! & y \end{smallmatrix}\!\big) \ast \big(\! \begin{smallmatrix} (x,s,t)&\! \mapsto \! & \widetilde{H}^-(f(x),t)\\ (y,t) &\! \mapsto \! & \widetilde{H}^-(y,t) \end{smallmatrix}\!\big) \ast \big(\! \begin{smallmatrix} (x,s,t)&\! \mapsto \! &(\widehat{H}(x,st),t)\\ (y,t) &\!\mapsto \! &(g(y),t) \end{smallmatrix}\!\big) :id_{Z_f}\!\simeq\!r.$$ Notice that each of the three homotopies is well defined on $Z_f$ (respects $(x,0)=f(x)$) and "their endfunctions match", so $H$ is well defined. Therefore $H\!\!:Z_f\,\searrow\,X\!\times\!\{1\}\!\approx\!X$. $\blacksquare$

Question: How can I prove: $X\!\simeq\!Y$ $\;\;\;\Rightarrow\;\;\;$ $\exists Z\supseteq X,Y:$ $\;X\swarrow\swarrow Z\searrow\searrow Y$?

Comment: I was really hoping that it is possible to change this proof in order to get a homotopy $H$ that doesn't move the points of $X\times\{1\}$, i.e. get a strong deformation retraction.

Request: Please, no advanced answers, since I'm just starting to learn Algebraic Topology. I don't know any obstruction theory/(co)fibrations/...

share|improve this question
    
So you want to prove that two spaces are homotopic iff they are both strong deformation retracts of a space? –  JSchlather May 8 '11 at 1:03
    
Yes. The left implication is obvious, but the right is troubling me. –  Leon May 8 '11 at 1:16
5  
This is corollary 0.21 in Hatcher's book (Hatcher uses "deformation retraction" to mean "strong deformation retraction"). –  Chris Eagle May 8 '11 at 5:57
2  
Leon: Perhaps you can convert @Chris's comment into an answer so that this question appears as answered in the future. –  Srivatsan Nov 7 '11 at 0:29

2 Answers 2

up vote 1 down vote accepted

I saw this post a while ago and did not think I had anything to add. However, purely by chance I found this paper today: A Short Note On Mapping Cylinders. There your question seems to be answered precisely as you wanted. That is, a formula is provided for a strong deformation retraction of of the mapping cylinder $Z_{f}$ onto its top $X\times\{1\}$.

It is quite a big formula if you ask me!

share|improve this answer
    
Very interesting, I didn't expect the formula to be so long :). Thank you! –  Leon Jun 9 '12 at 17:04

This is a copied answer from Hatcher's text. I wrote it only so that this question can be labelled as answered, as desired by @Srivatsan in the comments.

At the moment I am too "out" of algebraic topology, so I will write a proper answer (minimized, that deals only with the given question) later, but how much later I can't say. I must study Morse theory at the moment...

Also, if anyone is willing to convert this in a proper answer, I will delete mine and accept his, as soon as I notice a new answer has appeared (for some reason, I don't get any notifications on this site, or don't notice them or smth.).

enter image description here enter image description here enter image description here enter image description here

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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