deformation retract is transitive

Proposition.

Let $Z\mathrm{\subset }Y\mathrm{\subset }X$ be nested topological spaces. If there exist a deformation retraction (http://planetmath.org/DeformationRetraction) of $X$ onto $Y$ and a deformation retraction of $Y$ onto $Z$, then there also exists a deformation retraction of $X$ onto $Z$. In other words, “being a deformation retract of” is a transitive relation.

Proof.

Since $Y$ is a deformation retract of $X$, there is a homotopy $F:I\times X\to X$ between ${\mathrm{id}}_X$ and a retract $r:X\to Y$ of $X$ onto $Y$. Similarly, there is a homotopy $G:I\times Y\to Y$ between ${\mathrm{id}}_Y$ and a retract $s:Y\to Z$ of $Y$ onto $Z$.

First notice that since both $r$ and $s$ fix $Z$, the map $sr:X\to Z$ is a retraction.

Now define a map $\tilde{G}:I\times X\to X$ by $\tilde{G}=iG({\mathrm{id}}_I\times r)$, where $i:Y\hookrightarrow X$ is inclusion. Observe that

• $\tilde{G}(0,x)=r(x)$ for any $x\in X$;

• $\tilde{G}(1,x)=sr(x)$ for any $x\in X$; and

• $\tilde{G}(t,a)=a$ for any $a\in Z$.

Hence $\tilde{G}$ is a homotopy between the retractions $r$ and $sr$.

Finally we must glue together the homotopies (http://planetmath.org/GluingTogentherContinuousFunctions) $F$ and $\tilde{G}$ to get a homotopy between ${\mathrm{id}}_X$ and $sr$. To do this, define a function $H:I\times X\to X$ by

Since $F(1,x)=\tilde{G}(0,x)=r(x)$, the gluing yieds a continuous map. By construction,

• $H(0,x)=x$ for all $x\in X$;

• $H(1,x)=sr(x)$ for all $x\in X$; and

• $H(t,a)=a$ for any $a\in Z$.

Hence $H$ is a homotopy between the identity map on $X$ and a retraction of $X$ onto $Z$. We conclude that $H$ is a deformation retraction of $X$ onto $Z$. ∎

Title	deformation retract is transitive
Canonical name	DeformationRetractIsTransitive
Date of creation	2013-03-22 15:43:59
Last modified on	2013-03-22 15:43:59
Owner	mps (409)
Last modified by	mps (409)
Numerical id	4
Author	mps (409)
Entry type	Result
Classification	msc 55Q05