# $X$ and $Y$ are homotopy equivalent $\Leftrightarrow$ $\exists Z:$ $X,Y$ are strong deformation retracts of $Z$

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 $f$. 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/...

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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 Lampret May 8 '11 at 1:16
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
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

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!

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