coloring_nested_tire_graphs: simplify Lemma 1.7 by requiring S on the outer face

Pins Π_G at the start to be an embedding placing S on the outer face;
such an embedding exists for any single-vertex source.  This collapses
the two-embedding split in the previous proof (one for Lemma 2.6,
another for the topological analysis of R_{C'}) into a single
embedding throughout, and removes the "in either order" ambiguity for
B_out and B_in:

- B_out = G[V_{C'} ∩ L_d]: the boundary of R_{C'} closer to S.
- B_in  = G[V_{C'} ∩ L_{d+1}]: the boundary farther from S.

The outerplanarity step now cites Lemma 2.6 of [bauerfeld-pds]
directly (no embedding switch).  The "tire structure" step pins the
orientation by S's position on the outer face.

Remark 1.9 (degenerate cases) updated: the orientation ambiguity is
gone, so we state the d=0 case has degenerate B_out and the d=D_max
case has degenerate B_in.

(R1) and (R2) remain — they are graph-theoretic and unaffected by
embedding choice (for 3-connected planar graphs the embedding is
essentially unique by Whitney's theorem, so changing the outer face
cannot untangle pinches or merge multi-hole topology).

Co-Authored-By: Claude Opus 4.7 <noreply@anthropic.com>

papers/coloring_nested_tire_graphs/paper.log

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papers/coloring_nested_tire_graphs/paper.tex

@@ -174,8 +174,10 @@ triangular faces and $|E_{\mathrm{ann}}| = m + k - 1$.
 
 \begin{lemma}[Tire-component lemma]
 \label{lem:tire-component}
-Let $G$ be a maximal planar graph with fixed embedding $\Pi_G$ and let
-$S \subseteq V(G)$ be a level source.  For $d \geq 0$, let
+Let $G$ be a maximal planar graph and let $S \subseteq V(G)$ be a level
+source.  Fix a plane embedding $\Pi_G$ of $G$ in which $S$ lies on the
+outer face (such an embedding exists for any planar graph and any
+single-vertex source).  For $d \geq 0$, let
 \[
     G'_d \;:=\; G'\bigl[\,\{d_f \in V(G') : \delta_G(d_f) = d\}\,\bigr]
 \]
@@ -195,21 +197,20 @@ Assume:
 \item[\emph{(R2)}] $R_{C'}$ has at most two boundary components.
 \end{itemize}
 Then $C$, with the inherited embedding, is a tire graph in the sense of
-Definition~\ref{def:tire-graph}: its two boundary parts
-$\{B_{\mathrm{out}}, B_{\mathrm{in}}\}$ are the level-$d$ subgraph
-$G[V_{C'} \cap L_d]$ and the level-$(d+1)$ subgraph
-$G[V_{C'} \cap L_{d+1}]$, in either order, and the triangular faces of
-$C$ inside the closed boundary region are exactly the faces of $G$ in
-$F_{C'}$.
+Definition~\ref{def:tire-graph}.  Its outer boundary
+$B_{\mathrm{out}}$ is the side of $R_{C'}$ closer to $S$ in $\Pi_G$,
+namely the level-$d$ subgraph $G[V_{C'} \cap L_d]$; its inner boundary
+$B_{\mathrm{in}}$ is the side farther from $S$, namely the
+level-$(d+1)$ subgraph $G[V_{C'} \cap L_{d+1}]$; and the triangular
+faces of $C$ inside the closed boundary region are exactly the faces of
+$G$ in $F_{C'}$.
 \end{lemma}
 
 \begin{proof}
-\emph{Outerplanarity of the two level parts.}  Since $S$ is a single
-vertex, choose a plane embedding of $G$ with $S$ on the outer face.
-By Lemma~2.6 of \cite{bauerfeld-pds} applied with this embedding and
-source $S$, $G[L_{d'}]$ is outerplanar for each $d' \geq 0$;
-outerplanarity is a graph property, so the conclusion is independent
-of the embedding choice.  Subgraphs of outerplanar graphs are
+\emph{Outerplanarity of the two level parts.}  By construction $S$
+lies on the outer face of $\Pi_G$, so Lemma~2.6 of \cite{bauerfeld-pds}
+applies directly with $(G, \Pi_G, S)$, giving that $G[L_{d'}]$ is
+outerplanar for each $d' \geq 0$.  Subgraphs of outerplanar graphs are
 outerplanar, so $G[V_{C'} \cap L_d]$ and $G[V_{C'} \cap L_{d+1}]$ are
 both outerplanar.
 
@@ -265,21 +266,18 @@ the number of boundary components.  Hypothesis (R2) gives $n \leq 2$,
 so $R_{C'}$ is either a closed disk ($n = 1$) or a closed annulus
 ($n = 2$).
 
-\emph{Tire structure.}  In the annulus case ($n = 2$), the two
-boundary cycles are simple cycles on $L_d$ and $L_{d+1}$ respectively
-(by the previous two paragraphs).  These are the cycles bounding the
-two outerplanar subgraphs $G[V_{C'} \cap L_d]$ and
-$G[V_{C'} \cap L_{d+1}]$ in $\Pi_G$, and they meet the
-tire-graph definition with $B_{\mathrm{out}} \in \{$ those cycles $\}$
-in either order.  In the disk case ($n = 1$), the unique boundary
-cycle lies on one of the two levels, and the other level set of
-$V_{C'}$ is either empty or a single interior vertex of the disk
-(the BFS endpoint).  When it is a single vertex this is the
-degenerate-boundary case of Definition~\ref{def:tire-graph}; the
-remaining case ($V_{C'} \cap L_{d+1} = \emptyset$, which arises at
-$d = D_{\max}$) is excluded by interpreting one boundary part as a
-degenerate single vertex on $L_{d+1}$ (taken empty by convention,
-which we omit here).
+\emph{Tire structure.}  Because $S$ lies on the outer face of $\Pi_G$,
+the level-$d$ vertices are closer to $S$ in $\Pi_G$ than the
+level-$(d+1)$ vertices; in either the annulus or disk case the
+boundary cycle on the $L_d$ side is the boundary of $R_{C'}$ facing
+$S$ (the ``outer'' boundary), and the $L_{d+1}$ side is the boundary
+facing the interior (the ``inner'' boundary).  This identifies
+$B_{\mathrm{out}} = G[V_{C'} \cap L_d]$ and $B_{\mathrm{in}} =
+G[V_{C'} \cap L_{d+1}]$.  In the disk case ($n = 1$) one of the two
+level sets is a single vertex (the BFS endpoint at $d = 0$ with
+$S = \{v_0\}$, or symmetrically at $d = D_{\max}$ where the inner
+side collapses to a deepest vertex), giving the degenerate-boundary
+case of Definition~\ref{def:tire-graph}.
 
 The triangular faces inside the closed boundary region of $C$ are by
 construction the depth-$d$ faces in $F_{C'}$, and the edges of $C$ are
@@ -291,13 +289,13 @@ $V_{C'} \cap L_{d+1}$ that bound a face of $F_{C'}$.
 \begin{remark}
 \label{rem:tire-component-degenerate}
 Either boundary part of $C$ in Lemma~\ref{lem:tire-component} may be
-degenerate.  For instance, at $d = 0$ with single-vertex source
-$S = \{v_0\}$ the unique component of $G'_0$ has
-$V_{C'} \cap L_0 = \{v_0\}$ --- the degenerate boundary --- and
-$V_{C'} \cap L_1$ a cycle (the link of $v_0$ in $G$).  Which of the two
-parts is $B_{\mathrm{out}}$ and which is $B_{\mathrm{in}}$ depends on
-the orientation of the inherited embedding (equivalently, on which side
-of $C$ contains the rest of $\Pi_G$).
+degenerate.  At $d = 0$ with single-vertex source $S = \{v_0\}$ the
+unique component of $G'_0$ has $V_{C'} \cap L_0 = \{v_0\}$ as the
+degenerate \emph{outer} boundary and $V_{C'} \cap L_1$ a cycle (the
+link of $v_0$ in $G$) as the inner boundary.  Symmetrically, at
+$d = D_{\max}$, $V_{C'} \cap L_{D_{\max}+1} = \emptyset$ degenerates
+to a single deepest vertex serving as the \emph{inner} boundary, with
+the level-$D_{\max}$ cycle as the outer boundary.
 \end{remark}
 
 \begin{remark}