Notes:Types of retractions

Definitions

Here [ilmath](X,\mathcal{ J })[/ilmath] is a topological space

Source	Retraction	Deformation Retraction	Strong Deformation Retraction
Topology and Geometry	Subspace [ilmath]A[/ilmath] of [ilmath](X,\mathcal{ J })[/ilmath] and a continuous map, [ilmath]f:X\rightarrow A[/ilmath] such that [ilmath]f(a)=a[/ilmath] for all [ilmath]a\in A[/ilmath] is called a retraction and [ilmath]A[/ilmath] is the retract of [ilmath]X[/ilmath].	Subspace [ilmath]A[/ilmath] of [ilmath](X,\mathcal{ J })[/ilmath] is a deformation retract if there is a homotopy, [ilmath]H:X\times I\rightarrow X[/ilmath] - called a deformation such that: [ilmath]\begin{align*} H(x,0) &= x & \\ H(x,1) &\in A \\ H(a,1) &= a & \forall a\in A \end{align*}[/ilmath]	Subspace [ilmath]A[/ilmath] of [ilmath](X,\mathcal{ J })[/ilmath] is a strong deformation retract if there is a homotopy, [ilmath]H:X\times I\rightarrow X[/ilmath] - called a deformation such that: [ilmath]\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*}[/ilmath]
An Introduction to Algebraic Topology	A subspace [ilmath]A[/ilmath] of [ilmath](X,\mathcal{ J })[/ilmath] is a retract of [ilmath]X[/ilmath] is there exists a continuous map, [ilmath]r:X\rightarrow A[/ilmath] such that [ilmath]r\circ i=\text{Id}_A[/ilmath] (where [ilmath]i:A\hookrightarrow X[/ilmath] is the inclusion map).	A subspace [ilmath]A[/ilmath] of [ilmath](X,\mathcal{ J })[/ilmath] is a deformation retract of [ilmath]X[/ilmath] if [ilmath]r\circ i=\text{Id}_A[/ilmath] and [ilmath]i\circ r\simeq\text{Id}_X[/ilmath]. Again [ilmath]r:X\rightarrow A[/ilmath] is a continuous map.	A subspace [ilmath]A[/ilmath] of [ilmath](X,\mathcal{ J })[/ilmath] is a strong deformation retract of [ilmath]X[/ilmath] if there is a continuous map, [ilmath]r:X\rightarrow A[/ilmath] such that [ilmath]r\circ i=\text{Id}_A[/ilmath] and [ilmath]i\circ r\simeq\text{Id}_X\text{ rel } A[/ilmath]. [ilmath]r[/ilmath] is a strong deformation retraction.
Hatcher's Algebraic Topology	Retraction is a map [ilmath]r:X\rightarrow X[/ilmath] such that [ilmath]r(X)=A[/ilmath] and [ilmath]r\vert_A=\text{Id}_A[/ilmath]. He then says we "could equally well" regard a retraction as [ilmath]r:X\rightarrow A[/ilmath] with [ilmath]r\vert_A=\text{Id}_A[/ilmath]. N.b: this is why I dislike Hatcher's book. "Could equally well" vs "equivalent to".	Deformation retraction of a space [ilmath](X,\mathcal{ J })[/ilmath] onto a subspace [ilmath]A[/ilmath] is a family of maps: [ilmath]f_t:X\rightarrow X[/ilmath] for [ilmath]t\in I:=[0,1]\subset\mathbb{R}[/ilmath] such that: [ilmath]f_0=\text{Id}_X[/ilmath] [ilmath]f_1(X)=A[/ilmath] and [ilmath]f_t\vert_A=\text{Id}_A[/ilmath] for all [ilmath]t\in I[/ilmath] This should define a homotopy, [ilmath]H:X\times I\rightarrow X[/ilmath] given by: [ilmath]H(x,t):=f_t(x)[/ilmath] Everyone else calls this a strong deformation retract	(No mention)
Munkres - Elements of Algebraic Topology	Continuous map with [ilmath]r(a)=a[/ilmath] for all [ilmath]a\in A[/ilmath]	(See Topology and Geometry's strong deformation retraction) Munkres notes: A deformation retraction has a retraction, the map: [ilmath]r(x)=H(x,1)[/ilmath]	(no mention)

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TODO: Intro to top. manifolds

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Distinguishing examples given

Introduction to Algebraic Topology

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

It can be shown [ilmath]I\times\{0\} [/ilmath] is a deformation retract of [ilmath]X[/ilmath], 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 [ilmath]\epsilon[/ilmath]-[ilmath]\delta[/ilmath]-continuity argument, treating [ilmath]X[/ilmath] 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 [ilmath]r\circ i_A=\text{Id}_A[/ilmath], which means [ilmath]r\vert_A=\text{Id}_A[/ilmath], so what's going on!

By the author's own definition (if this claim is true) we don't have even a retraction

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 [ilmath]\mathbb{R}^2[/ilmath], but agrees with it on [ilmath]\mathbb{I} [/ilmath]

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, [ilmath](0,0)[/ilmath] 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:

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

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

This is a very dubious thing to do