# Retract and Homotopy extension property

See picture below 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.

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

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 $(2,2)$:

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:

PlotACylinder[RadiusOfA_, Height_, theta_, u_] :=

Module[{x, y},
{(x (1 - t) + RadiusOfA*t)*Cos[theta], (x (1 - t) + RadiusOfA*t)*Sin[theta],
Height (1 - t) + y*t}]

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)}]

{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}]

• Glad I could help! Just for fun I cooked up an animation of the retraction. Commented Jun 1, 2013 at 2:07
• Awesome, thanks. Commented Jun 1, 2013 at 14:51