Relative smoothing fixes the endpoints of a path

Regard $[0,1]$ as ${\mathrm{\Delta }}^1$. Its two faces are the separate points $0$ and $1$. Prescribe their smooth zero-simplices with values $p,q$ and the constant homotopies $H_0(0,s)=p$, $H_1(1,s)=q$. The faces have empty intersection, so compatibility is vacuous. Under [F2], [F1] supplies a strict smooth path $g$ and a continuous homotopy $H$ satisfying $H(0,s)=p$ and $H(1,s)=q$ for every $s$, as required.

The neighbourhood hypothesis behind [F1] is concrete in this dimension: disjoint small affine neighbourhoods of the two endpoint faces carry the constant smooth maps $p$ and $q$. The relative-smoothing proof first changes the continuous path, keeping the endpoints fixed, to agree with such a smooth neighbourhood extension near the endpoint union; only then does it apply relative Whitney approximation. Thus no claim is made that mere equality of endpoint values already means smoothness near those endpoints.

In the explicit real example define $H(t,s)=(1-s)\mathrm{\mid }t-1\mathrm{/}2\mathrm{\mid }+s\mathrm{/}2$. This is continuous, equals $f$ at $s=0$, equals the constant smooth path $g$ at $s=1$, and has $H(0,s)=H(1,s)=(1-s)\mathrm{/}2+s\mathrm{/}2=1\mathrm{/}2$ at every time. Its middle value is $H(1\mathrm{/}2,s)=s\mathrm{/}2$, exhibiting the actual change of the path. If the original path is smooth, [F1] also permits the constant homotopy with $g=f$, including constant paths and a one-point target. An empty target admits no path. The general assertion inherits only countable choice; the displayed real formula needs none.