Relative cubical disk model and compression

Write cube coordinates $(u,t)\in I^{n-1}\times I$. On the complement of J, send each $u_i\in (0,1)$ homeomorphically to $v_i=(2u_i-1)\mathrm{/}(1-\mathrm{\mid }2u_i-1\mathrm{\mid })$ and send $t\in [0,1)$ to $v_n=t\mathrm{/}(1-t)$. This identifies that complement with the closed upper half-space in ${\mathbb{R}}^n$. Approaching J is precisely escaping every bounded subset. The inverse stereographic map $v\mapsto (2v\mathrm{/}(1+\mathrm{\parallel }v{\mathrm{\parallel }}^2),(\mathrm{\parallel }v{\mathrm{\parallel }}^2-1)\mathrm{/}(1+\mathrm{\parallel }v{\mathrm{\parallel }}^2))$ therefore extends over J collapsed to the north pole. Its image is the closed hemisphere where coordinate n is nonnegative; projection dropping coordinate n identifies that hemisphere homeomorphically with a disk, with inverse inserting the nonnegative square root. Its boundary comes from F and its marked boundary point from J. For n=1 this is the compactified half-line, an interval.

Quotient descent and pullback identify representatives in the two models. The same holds for homotopies because the quotient times I is quotient. In particular a relative nullhomotopy in the disk model is $H:D^n\times I\to X$ with $H(z,0)=f(z)$, $H(z,1)=x_0$, $H(S^{n-1}\times I)\subseteq A$, and $H(b,t)=x_0$.

For $z\in D^n$ put $r=\mathrm{\parallel }z\mathrm{\parallel }$, $v(z)=z\mathrm{/}\max (1\mathrm{/}2,r)$ and $h(z)=\min (1,2-2r)$. Both are continuous, $v(z)\in D^n$ and $h(z)\in I$. The map $R_s(z)=((1-s)z+sv(z),sh(z))$ lies in $D^n\times I$ and fixes every rim point $(z,0)$ with r=1. At s=0 it is the bottom disk. At s=1, points with $r\le 1\mathrm{/}2$ lie in the top disk and points with $r\ge 1\mathrm{/}2$ lie in the side boundary. Thus $H(R_s(z))$ is a homotopy fixed on the whole boundary from f to a map into A. The formula has no singularity at z=0 and agrees on r=1/2.

Conversely suppose a boundary-fixed homotopy joins f to $g:D^n\to A$. The formula $g((1-s)z+sb)$ contracts g to $g(b)=x_0$ through maps into A fixing b, because the disk is convex. Concatenating this with the given homotopy produces a relative nullhomotopy. The two constructions prove both implications, including n=1, where the rim has two points.