An embedded height-folded arc has no configuration-loop slices

Construct the folded arc and vertical strand. Let $a=h\mathrm{/}4$. In ${\mathbb{R}}^2\times I$, define $P_0=(-h,0,0),P_1=(-h+a,0,3\mathrm{/}4),P_2=(-h+2a,a,1\mathrm{/}4),P_3=(-h,0,1)\mathrm{.}$ Let $A$ be the polygonal path $P_0{P}_1\cup P_1{P}_2\cup P_2{P}_3$, and let $B=\{(h,0,t):t\in I\}$. The endpoint sets of $A\cup B$ at heights $0$ and $1$ are both $\{q_1,q_2\}$. The height of $A$ has a local maximum $3\mathrm{/}4$ at $P_1$ and a local minimum $1\mathrm{/}4$ at $P_2$, so $A$ is not height-monotone.

Verify that these are disjoint embedded arcs in the cylinder. The segment $P_0{P}_1$ has spatial coordinate $y=0$, while $P_1{P}_2$ has $y>0$ except at $P_1$ and $P_2{P}_3$ has $y>0$ except at $P_3$. The only possible intersection of $P_0{P}_1$ and $P_2{P}_3$ at $y=0$ would be $P_3$, whose height is $1$ while $P_0{P}_1$ has height at most $3\mathrm{/}4$, so those segments are disjoint. Along $P_1{P}_2$ the spatial coordinates satisfy $x+h-y=a$; along $P_2{P}_3$ they satisfy $x+h=2y$. A common point must therefore have $y=a$, which occurs on both segments only at $P_2$. Thus the three segments form an embedded arc. Its spatial points lie in the convex ball of radius $\sqrt{5}a$ about $q_1$, since that ball contains all four vertices. As $\sqrt{5}a<3a=3h\mathrm{/}4<2h=\mathrm{\mid }{q}_2-q_1\mathrm{\mid }$, the arc misses $B$. Also $\mathrm{\mid }{q}_1\mathrm{\mid }+\sqrt{5}a<h+3a=7h\mathrm{/}4=7\mathrm{/}48<1$, so $A$ lies in $D^{\circ }\times I$; $B$ lies there because $\mathrm{\mid }{q}_2\mathrm{\mid }=h<1$.

Compute the slice at height $1\mathrm{/}2$. The height coordinate is strictly monotone on each segment of $A$, so each segment meets that plane once. Linear interpolation gives the three spatial points $(-h+2a\mathrm{/}3,0),(-h+3a\mathrm{/}2,a\mathrm{/}2),(-h+4a\mathrm{/}3,2a\mathrm{/}3)\mathrm{.}$ The first has $y=0$ and the other two have different positive $y$ coordinates; their $x$ coordinates are also different, so these three points are distinct. The strand $B$ contributes the fourth point $(h,0)=q_2$, which is not on $A$ by step 2.1. The slice therefore contains four distinct points.

By [L2], a value of $C_2(D^{\circ })$ must have exactly two points, whereas the horizontal slice at $1\mathrm{/}2$ has four. Hence this embedded picture does not define a path $I\to C_2(D^{\circ })$ by slicing, despite having the same endpoint set $\{q_1,q_2\}$ at heights $0$ and $1$. Its folded arc also fails the one-point-per-height condition in [L3], so the level-preserving braid slicing lemma [L4] does not apply.