Notes:Types of retractions

Definitions

Here $(X,\mathcal{ J })$ is a topological space

Source	Retraction	Deformation Retraction	Strong Deformation Retraction
Topology and Geometry	Subspace $A$ of $(X,\mathcal{ J })$ and a continuous map, $f:X\rightarrow A$ such that $f(a)=a$ for all $a\in A$ is called a retraction and $A$ is the retract of $X$.	Subspace $A$ of $(X,\mathcal{ J })$ is a deformation retract if there is a homotopy, $H:X\times I\rightarrow X$ - called a deformation such that: $\begin{align*} H(x,0) &= x & \\ H(x,1) &\in A \\ H(a,1) &= a & \forall a\in A \end{align*}$	Subspace $A$ of $(X,\mathcal{ J })$ is a strong deformation retract if there is a homotopy, $H:X\times I\rightarrow X$ - called a deformation such that: $\begin{align*} H(x,0) &= x & \\ H(x,1) &\in A \\ H(a,t) &= a & \forall a\in A\forall t\in I \end{align*}$
An Introduction to Algebraic Topology	A subspace $A$ of $(X,\mathcal{ J })$ is a retract of $X$ is there exists a continuous map, $r:X\rightarrow A$ such that $r\circ i=\text{Id}_A$ (where $i:A\hookrightarrow X$ is the inclusion map).	A subspace $A$ of $(X,\mathcal{ J })$ is a deformation retract of $X$ if $r\circ i=\text{Id}_A$ and $i\circ r\simeq\text{Id}_X$. Again $r:X\rightarrow A$ is a continuous map.	A subspace $A$ of $(X,\mathcal{ J })$ is a strong deformation retract of $X$ if there is a continuous map, $r:X\rightarrow A$ such that $r\circ i=\text{Id}_A$ and $i\circ r\simeq\text{Id}_X\text{ rel } A$. $r$ is a strong deformation retraction.

Distinguishing examples given

Introduction to Algebraic Topology

Consider the closed vertical strip in $\mathbb{R}^2$ given by $[0,1]\times\mathbb{R}$, take a subspace, $X$, of this which is the union of $I:=[0,1]\subseteq\mathbb{R}$ (interpreting as $[0,1]\times\{0\}\subseteq\mathbb{R}^2$) and all the line segments through the origin having slope $\frac{1}{n}$ for $n\in\mathbb{N}_{\ge 1}$.

It can be shown $I\times\{0\}$ is a deformation retract of $X$, but not a strong one.

Apparently....

Deciphering the example

I am not the first to be troubled by this, this person also tried. The reply they got looks rather like a $\epsilon$-$\delta$-continuity argument, treating $X$ as a metric subspace.

However if we just take the claim as "true":

• "The key is that if one want a continuous function r:X-->I ,then r can not be id on I."

Then we have a problem:

• We have neither a deformation retraction nor a retraction! Both of these require that $r\circ i_A=\text{Id}_A$, which means $r\vert_A=\text{Id}_A$, so what's going on!

I have managed to "reverse engineer" what I think the proof does, however it involves constructing a totally different metric. However it involves taking a different metric to the one inherited from $\mathbb{R}^2$, but agrees with it on $\mathbb{I}$

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Reverse engineering notes

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I have found a way to get the result, however it requires abuse, if we get rid of a map from a space to the same space, we can define two metrics: A new metric, which is the Euclidean distance between points which would be (for points in different line segments), the path from a point, to the origin, $(0,0)$ then towards the second point, and ALSO keep the subspace metric (then it isn't a map between the same spaces). The negation of continuity is:

• $\exists\epsilon>0\forall\delta>0[d_1(x,a)<\delta\wedge d_2(r(x),r(a))\ge\epsilon]$ (recall from negation of implies that $¬(A\implies B)\iff A\wedge(¬B)$)

Now when $x$ is VERY close to $a$ (for some point $x=(1,\frac{1}{n})$ - for a large $n$) the $\epsilon$ approaches $2$ from above.

This is a very dubious thing to do