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See picture below enter image description here the following picture. According to Hatcher, homotopy extension property implies that for a pair $(X,A)$ where $A$ is a subspace of $X$,

$X\times I$ should retract to $X\times\{0\}\cup A\times I$ .

My question is whether the retract given in the picture is possible ? If yes, how would we do it. Seems to me it is impossible to do the retraction continuously.

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  • $\begingroup$ What exactly are $X$ and $A$ here? A big disk and a small disk or something else? $\endgroup$
    – Dejan Govc
    May 31, 2013 at 22:20
  • $\begingroup$ Yeah, X is the big disk and A is the smaller disk. $\endgroup$ May 31, 2013 at 22:28

1 Answer 1

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It's definitely doable. Let's consider a simpler example first: let $X=[0,1]$, and let $A=\{0\}$.

You can retract $X\times I$ (a square) to $(X\times\{0\})\cup(A\times I)$ (the union of the "bottom" and "left" sides of the square) by projecting each point along the ray from $(2,2)$:

enter image description here

To move this intuition to your example of $X=$ a disk and $A=$ a smaller disk inside $X$, just "swing this around" (as one would to form a solid of revolution) and leave the interior of $A$ alone.


For fun:

enter image description here

PlotACylinder[RadiusOfA_, Height_, theta_, u_] :=
  {RadiusOfA*Cos[theta], RadiusOfA*Sin[theta], Height*u}

PlotATop[RadiusOfA_, Height_, theta_, u_] :=
  {RadiusOfA*u*Cos[theta], RadiusOfA*u*Sin[theta], Height}

PlotX[RadiusOfX_, theta_, u_] :=
  {RadiusOfX*u*Cos[theta], RadiusOfX*u*Sin[theta], 0}

PlotTopSurface[RadiusOfA_, RadiusOfX_, Height_, t_, theta_, u_] := 
 Module[{x, y},
  x = RadiusOfA + (RadiusOfX - RadiusOfA) u;
  y = 2 Height*(1 - (2 RadiusOfX - 2 RadiusOfA)/(2 RadiusOfX - RadiusOfA - x))
      + Height*(2 RadiusOfX - 2 RadiusOfA)/(2 RadiusOfX - RadiusOfA - x);
  {(x (1 - t) + RadiusOfA*t)*Cos[theta], (x (1 - t) + RadiusOfA*t)*Sin[theta], 
  Height (1 - t) + y*t}]

PlotSideSurface[RadiusOfA_, RadiusOfX_, Height_, t_, theta_, u_] := 
 Module[{x, y},
  y = Height*u;
  x = (2 RadiusOfX - RadiusOfA)*(1 - (2 Height/(2 Height - y))) 
      + RadiusOfX (2 Height/(2 Height - y));
  {(RadiusOfX (1 - t) + x*t)*Cos[theta], (RadiusOfX (1 - t) + x*t)*Sin[theta], 
  y (1 - t)}]

PlotRetract[RadiusOfA_, RadiusOfX_, Height_, t_] := ParametricPlot3D[
  {PlotACylinder[RadiusOfA, Height, theta, u],
   PlotATop[RadiusOfA, Height, theta, u],
   PlotX[RadiusOfX, theta, u],
   PlotTopSurface[RadiusOfA, RadiusOfX, Height, t, theta, u],
   PlotSideSurface[RadiusOfA, RadiusOfX, Height, t, theta, u]},
  {theta, 0, 2 Pi}, {u, 0, 1}, Mesh -> None, Axes -> None, 
  Boxed -> False, PlotPoints -> 30, 
  Lighting -> {{"Directional", White, {{1, 1, 1}, {0, 0, 0}}}}, 
  PlotStyle -> {Gray, Gray, Gray, Directive[Blue, Opacity[0.5]], 
  Directive[Blue, Opacity[0.5]]}]

Export["animation.gif", Table[PlotRetract[1, 3, 4, Max[0, t]],
  {t, -0.1, 0.98, 0.02}], "DisplayDurations" -> {0.125}]
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    $\begingroup$ Glad I could help! Just for fun I cooked up an animation of the retraction. $\endgroup$ Jun 1, 2013 at 2:07
  • $\begingroup$ Awesome, thanks. $\endgroup$ Jun 1, 2013 at 14:51

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