Hanani-Tutte for Radial Planarity

NEW

Problem: Reconstructed statement: In the proposed extension of radial drawings to a PL embedded sphere $S\subset \mathbb R^{3}$ with levels given by the $z$-coordinate, does prescribing, for each vertex $v$ at level $i$, the connected component of

$$S\cap \{z=i\}$$

containing $v$, change the resulting level-planar drawing/embedding problem?

Here an $S$-level drawing maps each vertex $v$ to $S\cap\{z=\ell(v)\}$, and each edge to an arc on $S$ whose $z$-coordinate is strictly monotone between its endpoint levels.

Result: Yes, prescribing components makes a difference.

Let $S$ be the boundary of a sufficiently small regular neighborhood of a $Y$-shaped tree in the $xz$-plane: one trunk rises to height $1$, then splits into two disjoint upward branches, a left branch and a right branch. This boundary is a PL 2-sphere. For every regular level $z=t>1$ away from the split, the section $S\cap\{z=t\}$ has two connected components, one around the left branch and one around the right branch.

Consider the graph $G$ consisting of one edge $uv$, with

$$\ell(u)=2,\qquad \ell(v)=3.$$

Without prescribing components, $G$ has an $S$-level embedding: place both $u$ and $v$ on the right branch components at levels $2$ and $3$, and draw $uv$ along the right tube with strictly increasing $z$.

Now prescribe instead that $u$ lies on the left component of $S\cap\{z=2\}$, while $v$ lies on the right component of $S\cap\{z=3\}$. Any $z$-monotone edge from $u$ to $v$ must remain in $S\cap\{z\ge 2\}$. But $S\cap\{z\ge 2\}$ has two connected components, the left branch component and the right branch component. Hence no connected arc contained in $S\cap\{z\ge2\}$ can join $u$ to $v$. Therefore no prescribed-component $S$-level drawing exists.

Thus the unprescribed and component-prescribed versions are not equivalent; the component prescription genuinely changes the problem.

Citation: The question is posed in Fulek–Pelsmajer–Schaefer, “Hanani-Tutte for Radial Planarity,” JGAA 21(1), 2017, §5. The counterexample above is self-contained.