%% filename: amsart-template.tex
%% American Mathematical Society
%% AMS-LaTeX v.2 template for use with amsart
%% ====================================================================

\documentclass{amsart}

\usepackage{amssymb}
\usepackage{graphicx}
\usepackage{tikz}
\usetikzlibrary{backgrounds}

\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{conjecture}[theorem]{Conjecture}

\theoremstyle{definition}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{xca}[theorem]{Exercise}

\theoremstyle{remark}
\newtheorem{remark}[theorem]{Remark}

\numberwithin{equation}{section}

\begin{document}

\title{Coloring Nested Tire Graphs}

%    author one information
\author{Eric Bauerfeld}
\address{}
\curraddr{}
\email{}
\thanks{}

\subjclass[2010]{Primary }

\keywords{plane graph, triangulation, plane depth, level edge, dual graph, tire graph}

\date{}

\dedicatory{}

\begin{abstract}
We establish the foundational structure of nested
level-induced tire decompositions of a plane triangulation $G$.
A \emph{level source} of $G$ induces a BFS layering of $G$ and
endows the inner planar dual $G'$ with a \emph{dual depth}
grading.  The basic object of study is the \emph{tire graph}
$T$ --- a plane graph whose outer and inner boundaries bound a
closed planar region, the \emph{tire tread} $R$, triangulated by
the \emph{annular edges} $E_{\mathrm{ann}}$.  Our main structural
result, the \emph{tire-component lemma}, exhibits each connected
component of $G'_d$ as a tire graph; the \emph{tire-tread
partition theorem} consequence shows the resulting tire treads
partition the bounded faces of $G$.  Coloring questions on
$G$ thereby factor through coloring questions on the
individual treads.
\end{abstract}

\maketitle

\section{Introduction}

A classical theorem of Tait recasts the Four Colour Theorem in dual,
edge-colouring terms: a plane triangulation $G$ is properly $4$-vertex-colourable
if and only if its dual cubic graph $G'$ is properly $3$-edge-colourable. Thus a
minimal counterexample to the Four Colour Theorem -- a smallest triangulation
admitting no proper $4$-colouring -- corresponds to a smallest cubic plane graph
admitting no proper $3$-edge-colouring.

The structural study of such a minimal counterexample is the
overarching motivation for the present line of work.  This first
paper establishes the foundational vocabulary --- level sources,
dual depth, tire graphs, and partial tire duals --- on which
subsequent papers in the series build.  In particular, the
companion paper \cite{bauerfeld-nested-tire-duals} uses these
definitions to develop nested-cycle structure theorems and
chain-pigeonhole conjectures for tire annular subgraphs of $G'$.

Throughout, $G = (V, E)$ is a plane maximal planar graph (a triangulation)
with a fixed planar embedding $\Pi_G$. We write $|V| = n$, so $|E| = 3n - 6$
and $G$ has $2n - 4$ triangular faces.

\begin{definition}[Level source]
A \emph{level source} of $G$ is any vertex $v \in V$; we write
$S = \{v\}$ for the level-0 source.
\end{definition}

\begin{definition}[Levels]
Given a level source $S \subseteq V$, the \emph{level} of $v \in V$ is
$\ell_G(v) = \mathrm{dist}_G(v, S)$, the graph distance from $v$ to the nearest
source vertex.
\end{definition}

\begin{definition}[Dual]
\label{def:dual}
The \emph{dual} of $G$, written $G'$, is the inner (weak) planar dual of $G$ with
respect to the embedding $\Pi_G$: it has one vertex $d_f$ for each bounded face
$f$ of $G$, and an edge joining $d_f$ and $d_{f'}$ for each edge of $G$ shared by
two bounded faces $f$ and $f'$. The unbounded outer face contributes no vertex,
and edges of $G$ on the outer boundary contribute no dual edge. Since $G$ is a
triangulation, each vertex $d_f \in V(G')$ corresponds to a triangular face $f$
of $G$, and we write $V(f) \subseteq V$ for its three incident vertices.
\end{definition}

\begin{definition}[Dual depth]
\label{def:dual-depth}
Given a level source $S \subseteq V$, the \emph{dual depth} of a dual vertex
$d_f \in V(G')$ is
\[
    \delta_G(d_f) = \min_{v \in V(f)} \ell_G(v)
        = \min_{v \in V(f)} \mathrm{dist}_G(v, S),
\]
the smallest level among the three vertices of $G$ bounding the face $f$.
\end{definition}

\begin{figure}[h]
\centering
\includegraphics[width=0.7\textwidth]{fig_dual_depth.png}
\caption{Dual depth in a stacked-ring triangulation $G$ with level source
$S = \{0\}$. Each $G$ vertex is labelled by its level $\ell$. Each bounded face
carries a dual vertex (square, joined by dashed dual edges) coloured by its dual
depth $\delta(d_f) = \min_{v \in V(f)} \ell(v)$: the central fan has depth $0$,
the inner annulus depth $1$, and the outer annulus depth $2$. The outer face
(the level-$3$ triangle) is excluded from the inner dual and carries no dual
vertex.}
\label{fig:dual-depth}
\end{figure}

\begin{definition}[Tire graph]
\label{def:tire-graph}
A \emph{tire graph} consists of a plane graph $T$ together with an
\emph{outer boundary} $B_{\mathrm{out}} \subseteq T$ and an \emph{inner
outerplanar graph} $O \subseteq T$ with $V(B_{\mathrm{out}}) \cap V(O)
= \emptyset$, where
\begin{itemize}
  \item $B_{\mathrm{out}}$ is either a simple cycle of length $\geq 3$
        or a single vertex (a \emph{degenerate outer boundary});
  \item $O$ is an outerplanar graph; its \emph{inner boundary}
        $B_{\mathrm{in}}$ is the closed walk in $O$ that traces the
        boundary of $O$'s outer face in the inherited embedding,
        which is a simple cycle when $O$ is $2$-connected and a
        non-simple closed walk in general (visiting bridges twice and
        cut-vertices multiple times); if $|V(O)| = 1$, we say $T$ has
        a \emph{degenerate inner boundary}.
\end{itemize}
At most one of $B_{\mathrm{out}}, B_{\mathrm{in}}$ may be degenerate.
The vertex and edge sets of $T$ are
\[
    V(T) = V(B_{\mathrm{out}}) \cup V(O),
    \qquad
    E(T) = E(B_{\mathrm{out}}) \cup E(O) \cup E_{\mathrm{ann}},
\]
where $E_{\mathrm{ann}}$ --- the \emph{annular edges} --- has the
property that, in the plane embedding of $T$, the closed planar
region $R$ bounded externally by $B_{\mathrm{out}}$ and internally
by $B_{\mathrm{in}}$ is partitioned into triangular faces of $T$
whose union is $R$.  We call $R$ the \emph{tire tread} of $T$.

When $B_{\mathrm{out}}$ is a simple cycle and $O$ is $2$-connected,
the tread is a closed annulus.  More generally, $R$ is a closed
planar region that may fail to be a $2$-manifold at cut-vertices of
$O$ (where two ``lobes'' of the depth-$d$ region meet at a single
vertex); the inner boundary $B_{\mathrm{in}}$ is then a non-simple
closed walk that visits the cut-vertex multiple times.  The relaxed
definition accommodates outerplanar inner graphs with bridges,
cut-vertices, or multiple connected components.  When either
boundary is degenerate, the tread is a closed disk with that vertex
as apex.
\end{definition}

\begin{figure}[h]
\centering
\includegraphics[width=0.78\textwidth]{fig_tire_example.png}
\caption{A tire graph with non-degenerate boundaries: outer boundary
$B_{\mathrm{out}}$ a $6$-cycle on vertices $0,\dots,5$ (blue), inner
boundary $B_{\mathrm{in}}$ a $4$-cycle on vertices $6,\dots,9$ (red),
inner outerplanar graph $O = B_{\mathrm{in}} \cup \{7\text{--}9\}$
(with one chord, orange), and $E_{\mathrm{ann}}$ (grey) tiling the
annulus between $B_{\mathrm{out}}$ and $B_{\mathrm{in}}$ by ten
triangular faces.}
\label{fig:tire-example}
\end{figure}

\begin{remark}
\label{rem:tire-counts}
Let $m = |V(B_{\mathrm{out}})|$ and $k = |V(B_{\mathrm{in}})|$.  By
Euler's formula on the tire tread $R$, the tire graph has $m + k$
triangular faces inside $R$ and $|E_{\mathrm{ann}}| = m + k$
annular edges when neither boundary is degenerate; when exactly one
boundary is degenerate (so $\min(m, k) = 1$), there are $m + k - 1$
triangular faces and $|E_{\mathrm{ann}}| = m + k - 1$.
\end{remark}

\begin{proposition}[Source-side simple-cycle property]
\label{prop:no-level-d-pinch}
Let $G$ be a maximal planar graph with planar embedding $\Pi_G$ and
single-vertex source $v_0$.  Let $d \geq 1$, $v \in L_d$, and let
$C'$ be a connected component of $G'_d$ such that $v$ is incident to
some face in $F_{C'}$.  Then the depth-$d$ faces in $F_{C'}$ incident
to $v$ form a single contiguous arc in $v$'s rotation in $\Pi_G$.

Equivalently: for any such component, the source-side boundary of
$R_{C'}$ is a simple cycle in $L_d$ (no cut-vertices at level $d$).
\end{proposition}

\begin{proof}
Suppose for contradiction that the depth-$d$ faces in $F_{C'}$ at $v$
lie in two or more disjoint arcs of $v$'s rotation.  Adjacent vertices
in $G$ differ in level by at most $1$, so a face at $v$ has depth
exactly $d$ iff both other vertices have level $\geq d$, and depth
$\leq d-1$ iff at least one has level $d-1$.  Hence the gaps between
the depth-$d$ arcs at $v$ are populated by level-$(d-1)$ neighbours of
$v$, occurring in at least two disjoint arcs of $v$'s rotation.  Pick
$p$ in one such gap and $q$ in another.

The BFS ball $G[L_{<d}]$ is connected, so there exists a simple path
$P$ in $G[L_{<d}]$ from $p$ to $q$.  Define the closed walk
\[
    W \;:=\; v \to p \to P \to q \to v.
\]
Every vertex of $P$ lies in $L_{<d}$, while $\ell(v) = d$, so $v$ is
distinct from every vertex of $P$; $P$ is simple, so its internal
vertices are distinct; and $p \neq q$ since they lie in different gaps.
Hence $W$ is a simple cycle in $G$.

By the Jordan curve theorem, the planar embedding of $W$ divides $\Pi_G$
into two regions.  In $v$'s rotation, the edges $v-p$ and $v-q$ lie at
two specific positions, and they split the rotation into two arcs;
each arc lies in one of the two regions determined by $W$.  By choice
of $p, q$, the two arcs of depth-$d$ faces at $v$ in $F_{C'}$ lie in
different regions of $W$ (i.e., one arc on each side).

Since $C'$ is connected in $G'$ and contains depth-$d$ faces in both
arcs, there is a dual path $f_1, f_2, \ldots, f_k$ in $G'_d$ with
$f_1, f_k \in F_{C'}$ incident to $v$ in different arcs, and with the
intermediate faces $f_2, \ldots, f_{k-1}$ not incident to $v$ (a
shortest such dual path).  Consecutive faces $f_i, f_{i+1}$ share an
edge $e_i$ of $G$; for $i \geq 2$, both endpoints of $e_i$ lie in
$L_{\geq d}$ (since neither $f_i$ nor $f_{i+1}$ is incident to $v$,
all six vertices of these two triangles lie in $L_{\geq d}$).  In
particular, $e_i$ shares no endpoint with $W$ except possibly $v$ ---
and $v$ is excluded from $f_2, \ldots, f_{k-1}$.

A planar edge with neither endpoint on a simple closed planar curve
$W$ has both of its incident faces on the same side of $W$.  Applying
this to each $e_i$ ($i \geq 2$) inductively: starting from $f_2$ on
the same side of $W$ as $f_1$ (their shared edge $e_1 = w-w'$ opposite
to $v$ in $f_1$ has $w, w' \in L_{\geq d}$ and hence is not on $W$),
the path $f_2 \to f_3 \to \cdots \to f_{k-1} \to f_k$ stays on one
side of $W$.

But $f_1$ and $f_k$ lie on different sides of $W$ (by construction),
contradicting the conclusion that the entire path lies on one side.
\end{proof}

\begin{lemma}[Tire-component lemma]
\label{lem:tire-component}
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
\[
    G'_d \;:=\; G'\bigl[\,\{d_f \in V(G') : \delta_G(d_f) = d\}\,\bigr]
\]
be the inner-dual subgraph on dual vertices of dual depth $d$, and let
$C'$ be a connected component of $G'_d$.  Write
$F_{C'} := \{f : d_f \in V(C')\}$,
$V_{C'} := \bigcup_{f \in F_{C'}} V(f)$, and let $C := G[V_{C'}]$ inherit
its embedding from $\Pi_G$.  Set $R_{C'} := \bigcup_{f \in F_{C'}} f
\subseteq |\Pi_G|$.

Then $C$, with the inherited embedding, is a tire graph in the sense of 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 $G[V_{C'} \cap L_d]$ (a simple cycle or
single vertex); its inner outerplanar graph is $O = G[V_{C'} \cap
L_{d+1}]$, and its inner boundary $B_{\mathrm{in}}$ is the outer-face
boundary closed walk of $O$ in the inherited embedding (a simple cycle
when $O$ is $2$-connected, a non-simple closed walk in general).  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.}  By construction $S$
lies on the outer face of $\Pi_G$, so the outerplanarity lemma of
\cite{bauerfeld-depth} 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.

\emph{Layer containment.}  Each $f \in F_{C'}$ has at least one vertex
at level $d$, and adjacent vertices in $G$ differ in level by at most
$1$; combined with $\delta_G(d_f) = d$, this forces
$V(f) \subseteq L_d \cup L_{d+1}$.  Hence $V_{C'} \subseteq L_d \cup
L_{d+1}$, and $C$ has vertex partition
$V_{C'} = (V_{C'} \cap L_d) \sqcup (V_{C'} \cap L_{d+1})$.

\emph{Boundary edges are monochromatic in level.}  Each edge $e$ on
$\partial R_{C'}$ separates a face $f \in F_{C'}$ from a face
$f' \notin F_{C'}$.  Because $f$ and $f'$ share the edge $e$, their
dual vertices are adjacent in $G'$; if both had depth $d$ they would
lie in the same component of $G'_d$, contradicting $d_{f} \in C'$ and
$d_{f'} \notin C'$.  Hence $\delta_G(d_{f'}) \neq d$; combined with
the bounded-step property of $\delta$ across $G'$-adjacent faces,
$\delta_G(d_{f'}) \in \{d-1, d+1\}$.
\begin{itemize}
\item If $\delta_G(d_{f'}) = d - 1$, the third vertex $w$ of
      $f' = \{u, v, w\}$ (where $u, v$ are the endpoints of $e$) has
      $\ell(w) = d - 1$.  Each of $u, v$ has $\ell \in \{d, d+1\}$
      (from $V(f) \subseteq L_d \cup L_{d+1}$) and is adjacent to $w$,
      forcing $\ell(u), \ell(v) \in \{d-2, d-1, d\} \cap \{d, d+1\} =
      \{d\}$.
\item If $\delta_G(d_{f'}) = d + 1$, then all three vertices of $f'$
      lie in $L_{\geq d+1}$, so in particular $\ell(u) = \ell(v) =
      d + 1$.
\end{itemize}
Each connected boundary component thus carries a single type at every
edge: any vertex on a boundary component has two boundary edges
incident to it (by R1, see below), both of the same type, so its
level is fixed.  Therefore each boundary component of $\partial R_{C'}$
is monochromatic in level.

\emph{Boundary structure.}  Each connected component of
$\partial R_{C'}$ traces a closed walk in $G$ that, by the
monochromaticity above, lies entirely in $L_d$ or entirely in
$L_{d+1}$.  By Proposition~\ref{prop:no-level-d-pinch}, the depth-$d$
faces of $F_{C'}$ at any $v \in L_d \cap V_{C'}$ form a single
contiguous arc in $v$'s rotation, so the source-side boundary walk
visits each $L_d$-vertex of $V_{C'}$ exactly once: it is a simple
cycle.  At vertices $v \in L_{d+1} \cap V_{C'}$ the depth-$d$ faces
may split into multiple arcs of $v$'s rotation; this corresponds
exactly to $v$ being a cut-vertex of $O$, and the inner-side
boundary walk visits $v$ correspondingly many times --- which is
already accommodated by Definition~\ref{def:tire-graph} (where
$B_{\mathrm{in}}$ is the outer-face boundary closed walk of $O$, not
necessarily a simple cycle).

\emph{Outer boundary.}  Because $S$ lies on the outer face of $\Pi_G$,
the boundary curve(s) of $R_{C'}$ on the $L_d$ side are closer to $S$
in the embedding.  In the inherited embedding of $C$, the unique
unbounded face is the merged region containing the rest of $\Pi_G$
outside $R_{C'}$ on the $S$ side, so its boundary --- a simple cycle
on $L_d$ (or a single vertex when $V_{C'} \cap L_d = \{v_0\}$, the
$d = 0$ case) --- serves as $B_{\mathrm{out}}$.  We set
$B_{\mathrm{out}} := G[V_{C'} \cap L_d]$ if this is a cycle, and
the single vertex $\{v_0\}$ in the degenerate case.

\emph{Inner outerplanar graph.}  By the outerplanarity lemma of
\cite{bauerfeld-depth}, $G[V_{C'} \cap L_{d+1}]$ is outerplanar.  We set $O :=
G[V_{C'} \cap L_{d+1}]$.  The boundary curve(s) of $R_{C'}$ on the
$L_{d+1}$ side are exactly the boundary of $O$'s outer face in the
inherited embedding; this outer-face boundary is a single closed walk
that traces around $O$ from the outside, traversing any bridge edge
twice and visiting cut-vertices multiple times.  This walk is the
inner boundary $B_{\mathrm{in}}$.  No further restriction on $O$'s
internal structure is needed: when $R_{C'}$ has more than two
boundary components in the surface-classification sense (i.e.\
several disjoint simple cycles on $L_{d+1}$), these correspond
precisely to the multiple connected components or bridge crossings of
$O$, and the outer-face boundary closed walk of $O$ captures them
collectively.

\emph{Tire structure.}  The triangular faces of $C$ inside the closed
boundary region are by construction the depth-$d$ faces in $F_{C'}$,
and the edges of $C$ are $E(B_{\mathrm{out}}) \cup E(O) \cup
E_{\mathrm{ann}}$ where $E_{\mathrm{ann}}$ are the edges of $G$
between $V_{C'} \cap L_d$ and $V_{C'} \cap L_{d+1}$ that bound a face
of $F_{C'}$.
\end{proof}

\begin{theorem}[Tire treads partition the bounded faces]
\label{thm:tread-partition}
Let $G$ be a maximal planar graph with planar embedding $\Pi_G$ and
let $S \subseteq V(G)$ be a level source lying on the outer face.
For each $d \ge 0$ and each connected component $C'$ of $G'_d$, let
$T^{(d, C')}$ denote the tire graph supplied by
Lemma~\ref{lem:tire-component}, with tire tread
$R_{C'} \subseteq |\Pi_G|$.  Then the collection of treads
\[
   \mathcal{R}(G, S) \;:=\;
   \bigl\{\, R_{C'} \,:\, d \ge 0,\;
   C' \text{ a connected component of } G'_d \,\bigr\}
\]
partitions the bounded part of $|\Pi_G|$:
\begin{enumerate}
  \item[(i)] every bounded face $f$ of $G$ is contained in exactly
        one tread $R_{C'} \in \mathcal{R}(G, S)$;
  \item[(ii)] distinct treads in $\mathcal{R}(G, S)$ have disjoint
        interiors and may share only boundary edges or vertices.
\end{enumerate}
\end{theorem}

\begin{proof}
\emph{Existence and uniqueness.}  Each bounded face $f \in F(G)$
has a uniquely-defined dual depth $\delta_G(d_f) \in \mathbb{Z}_{\ge
0}$, so the dual vertex $d_f$ lies in $G'_d$ for $d =
\delta_G(d_f)$ and in no other $G'_{d'}$.  Within $G'_d$, the
vertex $d_f$ belongs to exactly one connected component $C'$.  By
Lemma~\ref{lem:tire-component}, $F_{C'}$ is precisely the set of
faces $f' \in F(G)$ with $d_{f'} \in V(C')$; in particular $f \in
F_{C'}$, hence $f \subseteq R_{C'}$.

For any other tread $R_{C''} \in \mathcal{R}(G, S)$, the
component $C''$ is either at a different depth $d' \ne d$ (in
which case $F_{C''}$ consists of depth-$d'$ faces and $f \notin
F_{C''}$) or at depth $d$ but a different component $C'' \ne C'$
(in which case the two components are vertex-disjoint in $G'_d$,
so again $f \notin F_{C''}$).  In both cases $f \notin R_{C''}$
(more precisely, $f$ is not one of the triangular faces of $G$ in
$F_{C''}$, so $f$'s interior is not contained in $R_{C''}$).

\emph{Disjoint interiors.}  Each tread $R_{C'}$ is the union of
its triangular faces $F_{C'} \subseteq F(G)$; distinct treads
correspond to disjoint $F_{C'}$ (by the argument above), and the
interiors of distinct $G$-faces are disjoint.  Hence interiors of
distinct treads are disjoint.

\emph{Coverage.}  Conversely, every bounded $f \in F(G)$ has $d_f
\in V(G')$ with some dual depth $d$, and thus lies in $R_{C'}$
where $C'$ is its component of $G'_d$.  So $\bigcup_{R \in
\mathcal{R}(G, S)} R$ contains every bounded face of $G$.
\end{proof}

\begin{remark}
\label{rem:tire-component-degenerate}
Either boundary part of $C$ in Lemma~\ref{lem:tire-component} may be
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}
\label{rem:tire-no-extra-hypotheses}
Two structural features of $R_{C'}$ that might at first appear to
obstruct the tire-graph conclusion are both already accommodated by
Definition~\ref{def:tire-graph}:

\emph{Cut-vertices of $O$.}  A vertex $v \in V_{C'} \cap L_{d+1}$ may
have the faces of $F_{C'}$ incident to it split into two or more
arcs in $v$'s rotation in $\Pi_G$, separated by faces of higher
depth.  This corresponds exactly to $v$ being a cut-vertex of
$O = G[V_{C'} \cap L_{d+1}]$, and the inner boundary closed walk
$B_{\mathrm{in}}$ then visits $v$ multiple times --- once for each
arc.  No additional hypothesis is needed.

\emph{Multi-hole topology of $R_{C'}$.}  Even when $R_{C'}$ encloses
several disjoint depth-$>d$ sub-regions, the inner outerplanar graph
$O$ captures the multi-hole structure as a disconnected or
non-$2$-connected outerplanar graph (with bridges or multiple
components), and its outer-face boundary closed walk serves as
$B_{\mathrm{in}}$ traversing bridges twice and visiting cut-vertices
multiple times.

In the special case $d = 0$ with single-vertex source $S = \{v_0\}$,
$R_{C'}$ is the star of $v_0$, a topological closed disk with one
boundary cycle (the link of $v_0$); the corresponding tire graph has
degenerate outer boundary $\{v_0\}$.
\end{remark}

\begin{theorem}[Inner dual of a tire tread is outerplanar]
\label{thm:inner-dual-outerplanar}
Let $T = (B_{\mathrm{out}}, O, E_{\mathrm{ann}})$ be a tire graph,
and let $\Gamma$ be the graph on vertex set
$\{d_f : f \in F_{\mathrm{ann}}\}$ with an edge $d_f d_{f'}$ for
each interior annular edge of $T$ (= each edge of $T$ whose two
incident faces both lie in $F_{\mathrm{ann}}$).  Then $\Gamma$ is
outerplanar.

Moreover, $\Gamma$ admits a planar embedding as a (possibly
non-simple) Hamilton walk through every $d_f$, plus zero or more
non-crossing chords.
\end{theorem}

\begin{proof}
We argue by cases on whether the tire tread $R$ is a disk or an
annulus.

\medskip
\emph{Case 1: $R$ is a closed disk} (one of $B_{\mathrm{out}},
B_{\mathrm{in}}$ degenerate, by Definition~\ref{def:tire-graph}).
Let $v_0$ be the degenerate-boundary vertex (the apex) and let
$k = |B_{\mathrm{non-deg}}|$ be the length of the non-degenerate
boundary cycle.  The triangulation of $R$ is a \emph{fan} of $k$
triangles around $v_0$: each triangle has the form $\{v_0, u_i,
u_{i+1}\}$ where $u_1, \dots, u_k$ are the boundary-cycle vertices
in cyclic order.  Each triangle has two spoke edges (= the two
edges incident to $v_0$, shared with the two neighbouring fan
triangles) and one boundary edge (in $B_{\mathrm{non-deg}}$,
contributing a leaf in $D(T)$ but no edge in $\Gamma$).  Hence
every $d_f$ has $\Gamma$-degree exactly $2$, and $\Gamma$ is a
single cycle of length $k$.  Cycles are outerplanar.

See Figure~\ref{fig:inner-dual-disk-case} for the disk case
($k = 6$).

\begin{figure}[h]
\centering
\begin{tikzpicture}[scale=1.4]
  \def\R{1.8}
  % apex
  \node[circle, fill=black, inner sep=1.6pt, label={right:$v_0$}] (apex) at (0, 0) {};
  % boundary vertices (hexagon)
  \foreach \i in {0,...,5} {
    \pgfmathsetmacro{\ang}{60*\i + 90}
    \node[circle, fill=black, inner sep=1.3pt] (u\i) at (\ang:\R) {};
  }
  % boundary cycle edges (non-degenerate boundary)
  \foreach \i in {0,...,5} {
    \pgfmathtruncatemacro{\j}{mod(\i+1,6)}
    \draw[red, thick] (u\i) -- (u\j);
  }
  % spoke edges (annular interior)
  \foreach \i in {0,...,5} {
    \draw[gray] (apex) -- (u\i);
  }
  % dual vertices at triangle centroids + dual cycle
  \foreach \i in {0,...,5} {
    \pgfmathsetmacro{\angmid}{60*\i + 90 + 30}
    \pgfmathsetmacro{\rmid}{0.62*\R}
    \node[circle, fill=blue!70!black, inner sep=1.6pt] (d\i) at (\angmid:\rmid) {};
  }
  \foreach \i in {0,...,5} {
    \pgfmathtruncatemacro{\j}{mod(\i+1,6)}
    \draw[blue!70!black, very thick] (d\i) -- (d\j);
  }
  % Labels
  \node[red] at (0, -\R - 0.3) {\small non-degenerate boundary $B_{\mathrm{non-deg}}$};
  \node[blue!70!black] at (\R + 1.0, 0.5) {\small dual cycle $\Gamma \cong C_6$};
  \node[blue!70!black] at (\R + 1.0, 0.2) {\small (outerplanar)};
  \node[gray] at (-\R - 0.4, 0.3) {\small spokes};
\end{tikzpicture}
\caption{Case 1 ($R$ = disk, $k = 6$).  The apex $v_0$ sits at the
centre; the non-degenerate boundary $B_{\mathrm{non-deg}}$ (red)
is the hexagonal outer cycle; spokes (grey) triangulate the disk
into a fan of $6$ triangles around $v_0$.  Each triangle has two
spoke edges (interior, contributing $\Gamma$-edges) and one
boundary edge (contributing a leaf in $D(T)$, no $\Gamma$-edge).
The inner dual $\Gamma$ (blue) is the cycle $C_6$ formed by the
six annular face centroids, a manifestly outerplanar graph.}
\label{fig:inner-dual-disk-case}
\end{figure}

\medskip
\emph{Case 2: $R$ is an annulus} (both $B_{\mathrm{out}}$ and
$B_{\mathrm{in}}$ non-degenerate).  We construct an explicit
outerplanar embedding of $\Gamma$ as a Hamilton walk plus
non-crossing chords.

\medskip
\noindent\emph{Step 1: Cyclic ordering of $F_{\mathrm{ann}}$.}
The boundary of the annular tread is the disjoint union
$\partial R = B_{\mathrm{out}} \sqcup \overline{B_{\mathrm{in}}}$
(viewing $B_{\mathrm{in}}$ as a closed walk traced in the
appropriate orientation).  Each boundary edge of $R$ is incident to
exactly one annular face: walking around $B_{\mathrm{out}}$ in
cyclic order produces a sequence
$f^{\mathrm{out}}_1, f^{\mathrm{out}}_2, \dots, f^{\mathrm{out}}_n$
of (not necessarily distinct) annular faces, one per
$B_{\mathrm{out}}$-edge; similarly walking around
$B_{\mathrm{in}}$ produces a sequence
$f^{\mathrm{in}}_1, \dots, f^{\mathrm{in}}_{m_\partial}$ where
$m_\partial$ is the length of the inner-boundary walk.  Pick any
spoke $e^\star = u w \in E_{\mathrm{ann}}$ with $u \in
V(B_{\mathrm{out}})$ and $w \in V(B_{\mathrm{in}})$; cut $R$ along
$e^\star$.  This converts the annulus into a closed disk
$\tilde R$ whose boundary walks once around $B_{\mathrm{out}}$,
once along $e^\star$, once around $B_{\mathrm{in}}$ in reverse,
and once back along $e^\star$.  Concatenating the two boundary
sequences (in the order dictated by this disk traversal) yields a
single cyclic sequence
\[
   \mathcal{S} = (f^{\mathrm{out}}_1, \dots, f^{\mathrm{out}}_n,
       f^{\mathrm{in}}_1, \dots, f^{\mathrm{in}}_{m_\partial})
\]
of annular faces with multiplicities.

\medskip
\noindent\emph{Step 2: The Hamilton walk.}  Consecutive entries of
$\mathcal{S}$ correspond either to the same annular face (when two
adjacent boundary edges meet at a vertex incident to a single
annular face) or to two annular faces sharing an interior edge of
$E_{\mathrm{ann}}$.  In the former case the walk stays at one
$\Gamma$-vertex; in the latter it uses one $\Gamma$-edge.  The
resulting closed walk in $\Gamma$ visits every face that appears
in $\mathcal{S}$ at least once.

If every $f \in F_{\mathrm{ann}}$ appears in $\mathcal{S}$ (i.e.\
every annular face has at least one boundary edge of $R$), the walk
is a Hamilton walk in $\Gamma$, and we are done up to Step 3.  Each
annular face with two boundary edges contributes a vertex visited
twice; each with three contributes a vertex visited three times.

If some $f \in F_{\mathrm{ann}}$ does not appear in $\mathcal{S}$
(i.e.\ has no boundary edge of $R$), then all three edges of $f$
are interior annular edges, so $d_f$ has degree $3$ in $\Gamma$.
Such a face is ``trapped'' in the interior of the dual graph and
appears as the endpoint of a chord.  Extend the walk by:
whenever it crosses an interior annular edge $e$ shared with a
boundary-free face $f$, detour through $f$ and back.  After
finitely many such detours (one per boundary-free face), the walk
becomes a Hamilton walk visiting every $d_f$.

\medskip
\noindent\emph{Step 3: Non-crossing chords.}  The $\Gamma$-edges
not used by the Hamilton walk constructed in Step~2 are the
remaining interior annular edges.  Each such edge $e \in
E_{\mathrm{ann}}$ corresponds to a chord between two non-adjacent
positions of $\mathcal{S}$.  In the inherited planar embedding of
$\Gamma$ in $R$, these chords are drawn as straight segments
between annular triangle centroids; \emph{they do not cross} because
the underlying $E_{\mathrm{ann}}$ edges they cross are themselves
non-crossing in the planar embedding of $T$.

\medskip
\noindent\emph{Step 4: Outerplanar embedding.}  We now lay out
$\Gamma$ as follows: place the $|F_{\mathrm{ann}}|$ vertices on a
circle in the cyclic order given by $\mathcal{S}$ (treating
multiply-visited faces as single circle vertices).  Connect
consecutive vertices on the circle by the Hamilton-walk edges,
which forms the closed walk.  Draw the remaining edges as chords
inside the circle.  Because the chords were non-crossing in $T$'s
planar embedding, they remain non-crossing here.  All vertices lie
on the outer face (the unbounded region outside the circle),
making $\Gamma$ outerplanar.  $\square$
\end{proof}

\begin{figure}[h]
\centering
\begin{tikzpicture}[scale=1.25]
  % Outer hexagon vertices u_i at angles 90, 30, -30, -90, -150, 150
  \node[circle, fill=black, inner sep=1.2pt, label={[label distance=-1pt]90:\scriptsize $u_0$}]  (u0) at (0,     2.5)   {};
  \node[circle, fill=black, inner sep=1.2pt, label={[label distance=-1pt]30:\scriptsize $u_1$}]  (u1) at (2.17,  1.25)  {};
  \node[circle, fill=black, inner sep=1.2pt, label={[label distance=-1pt]-30:\scriptsize $u_2$}] (u2) at (2.17, -1.25)  {};
  \node[circle, fill=black, inner sep=1.2pt, label={[label distance=-1pt]-90:\scriptsize $u_3$}] (u3) at (0,    -2.5)   {};
  \node[circle, fill=black, inner sep=1.2pt, label={[label distance=-1pt]-150:\scriptsize $u_4$}](u4) at (-2.17,-1.25)  {};
  \node[circle, fill=black, inner sep=1.2pt, label={[label distance=-1pt]150:\scriptsize $u_5$}] (u5) at (-2.17, 1.25)  {};
  % Barbell vertices
  \node[circle, fill=black, inner sep=1.2pt, label={[label distance=-1pt]180:\scriptsize $a_1$}] (a1) at (-0.9,  0.7) {};
  \node[circle, fill=black, inner sep=1.2pt, label={[label distance=-1pt]180:\scriptsize $a_2$}] (a2) at (-0.9, -0.7) {};
  \node[circle, fill=black, inner sep=1.2pt, label={[label distance=-1pt]90:\scriptsize $a_3$}]  (a3) at (-0.25, 0)   {};
  \node[circle, fill=black, inner sep=1.2pt, label={[label distance=-1pt]90:\scriptsize $b_1$}]  (b1) at (0.25,  0)   {};
  \node[circle, fill=black, inner sep=1.2pt, label={[label distance=-1pt]0:\scriptsize $b_2$}]   (b2) at (0.9,   0.7) {};
  \node[circle, fill=black, inner sep=1.2pt, label={[label distance=-1pt]0:\scriptsize $b_3$}]   (b3) at (0.9,  -0.7) {};
  % B_out edges (red)
  \draw[red, thick] (u0) -- (u1) -- (u2) -- (u3) -- (u4) -- (u5) -- (u0);
  % O edges: left triangle, right triangle, bridge (light red)
  \draw[red!55!white, thick] (a1) -- (a2) -- (a3) -- (a1);
  \draw[red!55!white, thick] (b1) -- (b2) -- (b3) -- (b1);
  \draw[red!55!white, thick] (a3) -- (b1);
  % Spokes (gray)
  \foreach \p/\q in {u0/a1, u0/b2, u0/a3, u0/b1,
                     u1/b2, u1/b3,
                     u2/b3,
                     u3/a2, u3/b3, u3/a3, u3/b1,
                     u4/a2,
                     u5/a1, u5/a2} {
    \draw[gray, thin] (\p) -- (\q);
  }
  % Dual centroids -- 14 annular triangles
  \coordinate (T1)  at (barycentric cs:u0=1,u1=1,b2=1);  % outer cap u0-u1
  \coordinate (T2)  at (barycentric cs:u1=1,u2=1,b3=1);  % outer cap u1-u2
  \coordinate (T3)  at (barycentric cs:u2=1,u3=1,b3=1);  % outer cap u2-u3
  \coordinate (T4)  at (barycentric cs:u3=1,u4=1,a2=1);  % outer cap u3-u4
  \coordinate (T5)  at (barycentric cs:u4=1,u5=1,a2=1);  % outer cap u4-u5
  \coordinate (T6)  at (barycentric cs:u5=1,u0=1,a1=1);  % outer cap u5-u0
  \coordinate (I1)  at (barycentric cs:u5=1,a1=1,a2=1);  % inner cap a1-a2
  \coordinate (I2)  at (barycentric cs:u3=1,a2=1,a3=1);  % inner cap a2-a3
  \coordinate (I3)  at (barycentric cs:u0=1,a1=1,a3=1);  % inner cap a1-a3
  \coordinate (I4)  at (barycentric cs:u0=1,b1=1,b2=1);  % inner cap b1-b2
  \coordinate (I5)  at (barycentric cs:u1=1,b2=1,b3=1);  % inner cap b2-b3
  \coordinate (I6)  at (barycentric cs:u3=1,b1=1,b3=1);  % inner cap b1-b3
  \coordinate (Bup) at (barycentric cs:u0=1,a3=1,b1=1);  % bridge cap upper
  \coordinate (Bdn) at (barycentric cs:u3=1,a3=1,b1=1);  % bridge cap lower
  \foreach \p in {T1,T2,T3,T4,T5,T6,I1,I2,I3,I4,I5,I6,Bup,Bdn} {
    \node[circle, fill=blue!70!black, inner sep=1.4pt] at (\p) {};
  }
  % Hamilton cycle of length 14 (going clockwise from Bup)
  \draw[blue!70!black, very thick] (Bup) -- (I4);   % share u0-b1
  \draw[blue!70!black, very thick] (I4)  -- (T1);   % share u0-b2
  \draw[blue!70!black, very thick] (T1)  -- (I5);   % share u1-b2
  \draw[blue!70!black, very thick] (I5)  -- (T2);   % share u1-b3
  \draw[blue!70!black, very thick] (T2)  -- (T3);   % share u2-b3
  \draw[blue!70!black, very thick] (T3)  -- (I6);   % share u3-b3
  \draw[blue!70!black, very thick] (I6)  -- (Bdn);  % share u3-b1
  \draw[blue!70!black, very thick] (Bdn) -- (I2);   % share u3-a3
  \draw[blue!70!black, very thick] (I2)  -- (T4);   % share u3-a2
  \draw[blue!70!black, very thick] (T4)  -- (T5);   % share u4-a2
  \draw[blue!70!black, very thick] (T5)  -- (I1);   % share u5-a2
  \draw[blue!70!black, very thick] (I1)  -- (T6);   % share u5-a1
  \draw[blue!70!black, very thick] (T6)  -- (I3);   % share u0-a1
  \draw[blue!70!black, very thick] (I3)  -- (Bup);  % share u0-a3
  % Chord: bridge dual edge
  \draw[blue!70!black, very thick, dashed] (Bup) -- (Bdn);
  % Labels
  \node[red] at (0, 3.05) {\small $B_{\mathrm{out}}$ (hexagon)};
  \node[red!55!white] at (2.85, 0.0) {\small barbell $O$};
  \node[blue!70!black] at (-3.15, 0.5) {\small Hamilton cycle};
  \node[blue!70!black] at (-3.15, 0.18) {\small (length 14)};
  \node[blue!70!black] at (-3.15, -0.4) {\small chord = bridge};
  \node[blue!70!black] at (-3.15, -0.7) {\small dual edge};
  \draw[->, blue!70!black, thin] (-2.5, -0.4) -- (-0.15, 0);
\end{tikzpicture}
\caption{Case 2 ($R$ = annulus) with $O$ a barbell.
$B_{\mathrm{out}}$ is the outer hexagon (red); $O$ has two
triangles $\{a_1, a_2, a_3\}$ and $\{b_1, b_2, b_3\}$ joined by
the bridge $a_3\text{--}b_1$ (all light red).  The annulus is
triangulated by $14$ annular triangles: $6$ ``outer-cap''
triangles (one per outer edge), $6$ ``inner-cap'' triangles (one
per non-bridge edge of $O$), and $2$ ``bridge-cap'' triangles
$\{u_0, a_3, b_1\}$ and $\{u_3, a_3, b_1\}$ adjacent to the
bridge.  Each blue dot sits at the centroid of an annular
triangle; blue edges connect dual vertices whose triangles share
an interior annular edge (spoke or bridge).  The two bridge-cap
vertices have $\Gamma$-degree $3$ (their triangles have no
boundary edge) and are joined by the dashed blue \emph{chord}
corresponding to the bridge; the remaining $13$ edges form the
Hamilton cycle that wraps around the annulus.  All $14$ vertices
lie on the outer face of the cycle-with-chord embedding, so
$\Gamma \cong \Theta(1, 7, 7)$ is outerplanar.}
\label{fig:inner-dual-annulus-case}
\end{figure}

\begin{remark}
\label{rem:hamilton-cycle-spoke-only}
In the \emph{spoke-only} case (Definition~\ref{def:tire-graph} with
$O$ $2$-connected and $E_{\mathrm{ann}}$ consisting only of spokes),
every annular face has exactly one boundary edge, every
$d_f$ has $\Gamma$-degree $2$, and the construction of the
Theorem~\ref{thm:inner-dual-outerplanar} proof reduces to the
classical Hamilton cycle $\Gamma \cong C_{n+m}$ with zero chords.
\end{remark}

\begin{remark}
\label{rem:bridge-case-theta}
When $O$ has a bridge $e_{\mathrm{br}} \in E(O)$ whose two
incident faces are annular triangles, $e_{\mathrm{br}}$ contributes
an interior annular edge in $\Gamma$ rather than two leaves in
$D(T)$ (see Definition~1.7 of \cite{bauerfeld-nested-tire-duals}).
The two bridge-incident annular triangles have $\Gamma$-degree $3$;
the resulting $\Gamma$ has the structure of a Hamilton cycle of
length $n + m_\partial$ plus a single chord (length $1$).  This
corresponds to the theta graph $\Theta(1, b, c)$ identified
empirically in \cite{bauerfeld-nested-tire-duals}, which has no
$K_{2,3}$ subdivision (since one of the three paths has length $1$
and so contributes no degree-$2$ branch vertex), hence is
outerplanar as predicted.
\end{remark}

\begin{theorem}[Tait correspondence: 4-colorings of a tire vs 3-edge-colorings of its inner dual]
\label{thm:tait-tire}
Let $T = (B_{\mathrm{out}}, O, E_{\mathrm{ann}})$ be a tire graph
(viewed as an annular triangulation of its tire tread $R$) and let
$\Gamma$ be its inner dual
(Theorem~\ref{thm:inner-dual-outerplanar}).  Then
\[
   \#\bigl\{\text{proper $4$-vertex-colorings of } T\bigr\} \big/ |S_4|
   \;=\;
   \#\bigl\{\text{proper $3$-edge-colorings of } \Gamma\bigr\} \big/ |S_3|.
\]
That is, the number of $4$-vertex-colorings of $T$ up to permutation
of the colour set $\{0, 1, 2, 3\}$ equals the number of
$3$-edge-colorings of $\Gamma$ up to permutation of the colour set
$\{1, 2, 3\}$.
\end{theorem}

\begin{proof}
The argument is the classical Tait correspondence
\cite{tait-original} adapted to the annular triangulation $T$.
Encode the four colours of a proper $4$-vertex-coloring $c \colon
V(T) \to \mathbb{Z}_2 \times \mathbb{Z}_2$.  For each interior
annular edge $e$ of $T$ (whose two incident faces both lie in
$F_{\mathrm{ann}}$, contributing a $\Gamma$-edge $e^*$), set
\[
    \chi^*(e^*) \;:=\; c(u) + c(v) \in \mathbb{Z}_2 \times \mathbb{Z}_2,
    \qquad \text{where } u, v \text{ are the endpoints of } e.
\]
Since $c(u) \ne c(v)$, we have $\chi^*(e^*) \ne 00$, so $\chi^*$
takes values in $\{01, 10, 11\}$, which we identify with the
$3$-edge-coloring palette $\{1, 2, 3\}$.

\emph{Properness.}  At each $\Gamma$-vertex $d_f$ corresponding to
an annular triangle $f = \{u, v, w\}$, the three incident
$\Gamma$-edges (one per cycle-edge of $f$) carry colours
$c(u) + c(v),\; c(v) + c(w),\; c(u) + c(w)$.  These three elements
of $\mathbb{Z}_2 \times \mathbb{Z}_2$ sum to $0$ and are pairwise
distinct (their pairwise differences are
$c(u) - c(w),\; c(v) - c(u),\; c(w) - c(v)$, each nonzero), so
they form a permutation of $\{01, 10, 11\}$ --- a proper edge
colouring at $d_f$.

\emph{Surjectivity onto cosets.}  Given a proper $3$-edge-coloring
$\chi^*$ of $\Gamma$, the equation $c(u) + c(v) = \chi^*(e^*)$
admits exactly $|\mathbb{Z}_2 \times \mathbb{Z}_2| = 4$ solutions
$c \colon V(T) \to \mathbb{Z}_2 \times \mathbb{Z}_2$ (a global
translation is the only freedom).  Hence the map $c \mapsto
\chi^*$ is $4$-to-$1$.

\emph{Count.}  Therefore $\#\{\text{4-colorings of } T\} = 4 \cdot
\#\{\text{3-edge-colorings of } \Gamma\}$.  Dividing by $|S_4|
= 24$ on the left and $|S_3| = 6$ on the right (since $S_4$ acts
faithfully on the $4$-colorings and $S_3$ on the $3$-edge-colorings,
and the $4$-to-$1$ map respects the $S_4/S_3 \cong S_3$ quotient
via the natural surjection $S_4 \twoheadrightarrow S_3$) gives the
stated equality.
\end{proof}

\begin{remark}
\label{rem:count-general-outerplanar}
Theorem~\ref{thm:tait-tire} reduces the $4$-colouring count of a
tire to the $3$-edge-coloring count of its outerplanar inner dual
$\Gamma$.  For the cycle case $\Gamma \cong C_n$ (the spoke-only
case of Remark~\ref{rem:hamilton-cycle-spoke-only}), the cycle
chromatic polynomial at $k = 3$ gives
$2^n + 2 (-1)^n$.  For an inner dual with one or more non-crossing
chords, the count depends on the chord structure, not just on the
pair (number of vertices, number of chords): two outerplanar graphs
with the same $n$ and number of chords can have different proper
$3$-edge-coloring counts depending on how the chords are arranged
(nested, sequential, sharing vertices, etc.).  Every such count
can nevertheless be computed in linear time by tree-decomposition
methods, since outerplanar graphs have treewidth at most $2$ and
the edge-chromatic polynomial admits a deletion--contraction
recursion that respects the cycle-plus-chord structure.
\end{remark}

\begin{theorem}[Tire treads form a rooted tree under face containment]
\label{thm:tread-tree}
Let $G$ be a maximal planar graph with planar embedding $\Pi_G$
and let $S \subseteq V(G)$ be a single-vertex level source
$\{v_0\}$ lying on the outer face of $\Pi_G$.  The collection
$\mathcal{R}(G, S)$ of tire treads
(Theorem~\ref{thm:tread-partition}) carries a canonical rooted
tree structure $\mathcal{T}(G, S)$ defined as follows.

\begin{itemize}
\item \emph{Root.}  The depth-$0$ tire tread $T_0$ --- the unique
      tire produced by Lemma~\ref{lem:tire-component} at $d = 0$,
      with degenerate outer boundary $B_{\mathrm{out}} = \{v_0\}$
      and inner outerplanar graph $O^{(0)} = G[L_1]$ --- is the
      root.
\item \emph{Parent.}  For each tire tread $T_c$ at depth $d \ge 1$,
      its outer boundary $B_{\mathrm{out}}^{(c)}$ is a cycle in
      $L_d$.  The \emph{parent} of $T_c$ is the unique tire tread
      $T_p$ at depth $d - 1$ whose inner outerplanar graph
      $O^{(p)}$ has $B_{\mathrm{out}}^{(c)}$ as the boundary cycle
      of one of its bounded faces.  Equivalently, $R_c$ lies
      inside this bounded face of $O^{(p)}$ (which is itself the
      region of the plane cut off by $B_{\mathrm{out}}^{(c)}$ on
      the side away from $S$).
\item \emph{Children.}  The children of a tire tread $T_p$ are in
      bijection with those bounded faces of $O^{(p)}$ whose
      interiors contain at least one vertex of $G$ at level
      $\ge d + 2$ --- equivalently, with the connected components
      of $G'_{d+1}$ whose tires have outer boundary cycle equal to
      a bounded face of $O^{(p)}$.
\end{itemize}

Every tire tread except $T_0$ has exactly one parent; a tire
tread may have zero, one, or several children.
\end{theorem}

\begin{proof}
\emph{Root is well-defined.}  At $d = 0$ with single-vertex source
$S = \{v_0\}$, the dual subgraph $G'_0$ is connected (every face
of $G$ incident to $v_0$ has dual depth $0$, and they form a
single fan around $v_0$).  By
Lemma~\ref{lem:tire-component}, the unique component of $G'_0$
gives the depth-$0$ tire $T_0$ described above.

\emph{Existence of parent.}  Fix a tire tread $T_c$ at depth $d
\ge 1$ arising from a connected component $C'_c$ of $G'_d$.  Its
outer boundary $B_{\mathrm{out}}^{(c)} = G[V_{C'_c} \cap L_d]$ is
a simple cycle in $L_d$ (Lemma~\ref{lem:tire-component}; the
source-side boundary of a tire is always a simple cycle, by
Proposition~\ref{prop:no-level-d-pinch}).  The faces of $G$
immediately outside $B_{\mathrm{out}}^{(c)}$ on the side facing $S$
have depth $d - 1$ (one of their three vertices lies in $L_{d-1}$,
two in $L_d$).  Let $C'_p$ be the connected component of
$G'_{d-1}$ containing the dual vertex of any such face.

\emph{Uniqueness of parent.}  $B_{\mathrm{out}}^{(c)}$ is a single
simple cycle in $G$, with a well-defined ``$S$-side'' (the side
of the cycle closer to $v_0$ in $\Pi_G$).  The depth-$(d-1)$
faces lying on this side form a single contiguous arc around
$B_{\mathrm{out}}^{(c)}$ in the dual --- they are all $G'$-adjacent
in sequence (each pair of consecutive arc faces shares an edge in
$B_{\mathrm{out}}^{(c)}$).  Hence they all lie in the same
connected component $C'_p$ of $G'_{d-1}$, which is therefore
unique.

\emph{$B_{\mathrm{out}}^{(c)}$ bounds a face of $O^{(p)}$.}  The
parent tire $T_p$ has $V(O^{(p)}) = V_{C'_p} \cap L_d \supseteq
V(B_{\mathrm{out}}^{(c)})$.  The cycle $B_{\mathrm{out}}^{(c)}$ is
a subgraph of $O^{(p)}$ that bounds a face of $O^{(p)}$ in the
inherited embedding: the cycle traces around a depth-$\ge d+1$
region (containing $R_c$ and any descendants of $T_c$), which is
exactly a bounded face of $O^{(p)}$.

\emph{Children description.}  The bounded faces of $O^{(p)}$ are
in bijection with the connected components of $G'_d$ whose
faces lie inside those bounded regions (= one component per
bounded face, by an argument analogous to the existence-and-
uniqueness step above, applied one level deeper).

\emph{Tree property.}  Every non-root $T_c$ has a unique parent at
strictly smaller depth.  Iterating the parent map strictly
decreases depth, terminating at $T_0$.  No cycles can form
(depth is monotone).  Hence $\mathcal{T}(G, S)$ is a rooted tree.
\end{proof}

\begin{remark}
\label{rem:tree-multiple-children}
A parent tire $T_p$ has multiple children precisely when its
inner outerplanar graph $O^{(p)}$ has multiple bounded faces with
non-trivial interiors (= containing depth-$\ge d+2$ vertices of
$G$).  This happens, for instance, when $O^{(p)}$ has chords or
cut-vertices that subdivide its inner region, or when $O^{(p)}$
has multiple connected components in $G[L_{d+1}] \cap V_{C'_p}$.
By contrast, if $O^{(p)}$ is a simple cycle (the spoke-only case
of Remark~\ref{rem:hamilton-cycle-spoke-only}) with a non-empty
interior, $T_p$ has exactly one child.
\end{remark}

\begin{remark}
\label{rem:tree-coloring-factorisation}
Combining Theorem~\ref{thm:tread-partition} (treads partition
the bounded faces of $G$) with
Theorem~\ref{thm:tread-tree} (treads form a rooted tree), any
proper coloring problem on $G$'s bounded faces factors through:
\begin{itemize}
\item local coloring problems on each tread (the inner dual of
      each tread is outerplanar by
      Theorem~\ref{thm:inner-dual-outerplanar}), plus
\item consistency constraints along parent-child interfaces (the
      cycle $B_{\mathrm{out}}^{(c)}$ shared between a child and the
      face of its parent's $O^{(p)}$).
\end{itemize}
This is the structural setup underlying the chain-pigeonhole
program for tire treads.
\end{remark}

\begin{definition}[Iso of rooted trees of tire treads; combinatorial, $O$-preserved]
\label{def:tree-iso-O-preserved}
Let $\mathcal{T}_1, \mathcal{T}_2$ be rooted trees of tire treads.  A
\emph{combinatorial, $O$-preserved iso} from $\mathcal{T}_1$ to
$\mathcal{T}_2$ is a pair $(\varphi, \{\varphi_T\}_{T \in \mathcal{T}_1})$
satisfying:
\begin{itemize}
\item $\varphi : \mathcal{T}_1 \to \mathcal{T}_2$ is a rooted-tree iso
      (root to root, parent edges to parent edges);
\item for each tread $T \in \mathcal{T}_1$, $\varphi_T : O^{(T)} \to
      O^{(\varphi(T))}$ is an iso of plane outerplanar graphs --- in
      particular, the set of bounded faces of $O^{(T)}$ is sent
      bijectively to that of $O^{(\varphi(T))}$, with cyclic structure
      of each face preserved;
\item the child--face correspondence commutes with $\varphi$: if $T_c$
      is the child of $T$ at the bounded face $f$ of $O^{(T)}$, then
      $\varphi(T_c)$ is the child of $\varphi(T)$ at the bounded face
      $\varphi_T(f)$ of $O^{(\varphi(T))}$.
\end{itemize}
The outer boundaries $B_{\mathrm{out}}^{(T)}$ are \emph{not} required to
correspond.  In particular, the root tread's outer boundary may be
degenerate (a single vertex) in $\mathcal{T}_1$ and a simple cycle in
$\mathcal{T}_2$; this is essential because the root tread of a
\emph{sub-tree} of $\mathcal{T}(\tilde G, \tilde S)$ inherits a
non-degenerate $B_{\mathrm{out}}$ from its parent, even when it is iso
to a tree arising from a single-vertex level source.
\end{definition}

\begin{conjecture}[Universal nesting of tire-tread trees]
\label{conj:universal-nesting}
Let $\mathcal{T}_2 = \mathcal{T}(G_2, S_2)$ be a tree of tire treads
arising from a maximal planar $G_2$ with single-vertex level source
$S_2$.  Let $T \in \mathcal{T}_2$ be a tread at depth $d$, and let $f$
be a non-trivial bounded face of $O^{(T)}$ (i.e.\ a face whose interior
contains depth-$\ge d+2$ vertices of $G_2$).  Let $\mathcal{T}_1 =
\mathcal{T}(G_1, S_1)$ be any other tree of tire treads.

Then there exists a maximal planar graph $\tilde G$ with single-vertex
level source $\tilde S$ such that:
\begin{itemize}
\item[(N1)] $\mathcal{T}(\tilde G, \tilde S)$ contains, as a rooted
            sub-tree, an iso copy (in the sense of
            Definition~\ref{def:tree-iso-O-preserved}) of the
            truncation $\mathcal{T}_2 \setminus \mathrm{Desc}(T, f)$
            obtained from $\mathcal{T}_2$ by deleting the descendant
            sub-tree of $T$ at face $f$;
\item[(N2)] the sub-tree of $\mathcal{T}(\tilde G, \tilde S)$ rooted at
            the (new) child of $T$'s image at (the image of) $f$ is iso,
            in the sense of Definition~\ref{def:tree-iso-O-preserved},
            to $\mathcal{T}_1$.
\end{itemize}

Informally: trees of tire treads are closed under face-slot insertion,
where the slot at face $f$ in $\mathcal{T}_2$ is filled by the entirety
of $\mathcal{T}_1$.  The class of trees of tire treads is
\emph{closed under composition} by face-slot insertion.
\end{conjecture}

\begin{conjecture}[Seam realizability; technical core of nesting]
\label{conj:seam-realizability}
Let $\mathcal{T}_1 = \mathcal{T}(G_1, S_1)$ be a tree of tire treads.
For every integer $k \ge 3$ there exists a planar graph $H_k$, embedded
in a closed disk $D \subset \mathbb{R}^2$ with $\partial D$ a $k$-cycle,
such that:
\begin{itemize}
\item[(S1)] $\partial H_k = \partial D$, as a cyclic sequence of $k$
            vertices;
\item[(S2)] every bounded face of $H_k$ is a triangle;
\item[(S3)] BFS in $H_k$ from the cycle $\partial H_k$ assigns levels to
            $V(H_k) \setminus V(\partial H_k)$, and the resulting rooted
            tree of tire treads --- with the depth-$0$ tread taking
            outer boundary $\partial H_k$ in place of a single-vertex
            source --- is iso, in the sense of
            Definition~\ref{def:tree-iso-O-preserved}, to
            $\mathcal{T}_1$.
\end{itemize}
The construction $\mathcal{T}(H_k, \partial H_k)$ is the natural
extension of Theorem~\ref{thm:tread-tree} from single-vertex sources to
cycle sources: the depth-$0$ tread has non-degenerate
$B_{\mathrm{out}} = \partial H_k$ and the rest of the construction is
unchanged.
\end{conjecture}

\begin{figure}[h]
\centering
\includegraphics[width=0.95\textwidth]{fig_seam_construction.png}
\caption{Seam realizability for a small example.  $(a)$ A stacked-ring
triangulation $G_1$ with single-vertex source $S_1 = \{0\}$ and concentric
levels $L_1, L_2, L_3$; its tree of tire treads is the chain $T_0 \to T_1
\to T_2$ with $O^{(T_d)} = G_1[L_{d+1}]$ a $3$-cycle on each tread.
$(b)$ The apex-removal seam construction
$H_5 = (G_1 \setminus S_1) \cup A_5$, re-embedded so that the former
fan-face around $S_1$ becomes the outer face (with $L_1$ now the
outermost $G_1$-derived ring and $L_3$ innermost), and with an annular
triangulation $A_5$ (orange) attaching to a fresh $5$-cycle $\partial H_5$
(red).  Vertex labels show $\mathrm{BFS}_{\partial H_5}$ levels in $H_5$:
they agree with $\ell_{G_1}$ on $V(G_1) \setminus \{S_1\}$, so
$\mathcal{T}(H_5, \partial H_5)$ is iso (combinatorial, $O$-preserved) to
$\mathcal{T}(G_1, S_1)$.}
\label{fig:seam-construction}
\end{figure}

\begin{remark}[Nesting reduces to seam realizability]
\label{rem:seam-reduces-nesting}
Conjecture~\ref{conj:universal-nesting} follows from
Conjecture~\ref{conj:seam-realizability} by a direct gluing argument
within the framework of this paper.  Briefly: given a disk realization
$H_k$ of $\mathcal{T}_1$ with $k = |C_f|$, where $C_f$ is the cycle
bounding $f$ in $O^{(T)}$, excise from $G_2$ all vertices and edges
strictly inside $f$, then glue $H_k$ into the resulting hole by
identifying $\partial H_k$ with $C_f$.  The verification that the glued
graph $\tilde G$ is maximal planar, retains $\tilde S = S_2$ as a
single-vertex level source, and realizes the claimed nesting --- the
levels of $\tilde G$ from $S_2$ inside $f$ being just BFS-from-$C_f$ in
$H_k$ shifted by $d + 1$ --- is mechanical from
Theorems~\ref{thm:tread-partition},
\ref{thm:inner-dual-outerplanar},
\ref{thm:tread-tree} and the parent--child interface description of
Remark~\ref{rem:tree-coloring-factorisation}.

The substantive content of universal nesting thus sits entirely in
Conjecture~\ref{conj:seam-realizability}: given an arbitrary tree of
tire treads, can it be realized as the BFS-from-boundary tree of treads
of a triangulated planar disk, for every boundary length $k \ge 3$?
\end{remark}

\begin{remark}[Motivation and open questions]
\label{rem:nesting-motivation}
The conjectured closure under nesting carries two structural
implications for the Four Colour Theorem programme:
\begin{itemize}
\item \emph{Compositional colourability.}  If colourability of
      $\tilde G$ in (N1)--(N2) can be decided from the colourability
      of $G_1$ and $G_2$ alone (via the parent--child consistency
      constraints of Remark~\ref{rem:tree-coloring-factorisation}),
      then $4$-colourability propagates through nesting.  A minimum
      $4$CT counterexample (if it exists) would have to be
      \emph{irreducible} under such nesting --- it could not be
      decomposed into strictly smaller trees of tire treads whose
      colourings combine to a colouring of the whole.
\item \emph{Universality.}  Universal nesting positions trees of
      tire treads as a kind of ``term algebra'' for the structural
      decomposition of plane triangulations.  Coloring arguments
      can then be formulated inductively on this term algebra,
      with the chain-pigeonhole step
      (Remark~\ref{rem:tree-coloring-factorisation}) supplying the
      composition rule.
\end{itemize}

\medskip

Open questions:
\begin{itemize}
\item \emph{Candidate seam construction.}  A natural candidate for
      $H_k$ in Conjecture~\ref{conj:seam-realizability} is the
      \emph{apex-removal} construction:
      $H_k = (G_1 \setminus S_1) \cup A_k$, where $A_k$ is a
      triangulated annulus from the cycle $L_1^{(G_1)}$ to a fresh
      $k$-cycle that serves as $\partial H_k$; the embedding is chosen
      so the former fan-face around $S_1$ in $G_1$ becomes the outer
      face.  Showing that $\mathcal{T}(H_k, \partial H_k)$ is iso
      (combinatorial, $O$-preserved) to $\mathcal{T}_1$ amounts to
      verifying that BFS distances from $\partial H_k$ in $H_k$
      reproduce $\ell_{G_1}(\cdot)$ on $V(G_1) \setminus \{S_1\}$ ---
      which follows from the observation that every shortest path in
      $H_k$ from a non-boundary vertex to $\partial H_k$ passes through
      $L_1^{(G_1)}$.
\item \emph{$6$-connectivity preservation.}  Does nesting respect
      Birkhoff's internally $6$-connected condition for minimum $4$CT
      counterexamples?  The gluing seam $C_f \sim \partial H_k$ is
      exactly the low-connectivity site, so even when $G_1, G_2$ are
      internally $6$-connected the resulting $\tilde G$ is generically
      not, absent further hypotheses on $(G_1, G_2, f, k)$.  Identifying
      sufficient conditions for $6$-connected-preserving nesting is the
      relevant subproblem for the $4$CT application.
\item \emph{Stronger iso notions.}
      Definition~\ref{def:tree-iso-O-preserved} allows
      $B_{\mathrm{out}}^{(T)}$'s to differ.  A strictly stronger
      version of Conjecture~\ref{conj:universal-nesting} would require
      $B_{\mathrm{out}}$'s to correspond as cycles, but this is
      generically false: the cycle $C_f$ has fixed length $k$
      determined by $G_2$, while the depth-$1$ cycle $L_1^{(G_1)}$ has
      length $\deg_{G_1}(S_1)$ determined by $G_1$.  The combinatorial,
      $O$-preserved version of Definition~\ref{def:tree-iso-O-preserved}
      is exactly the notion that allows the seam to absorb this length
      mismatch.
\end{itemize}
\end{remark}


\begin{thebibliography}{9}

\bibitem{tait-original}
P.~G.~Tait,
\emph{Remarks on the colouring of maps},
Proc.\ Roy.\ Soc.\ Edinburgh \textbf{10} (1880), 729.

\bibitem{bauerfeld-depth}
E.~Bauerfeld,
\emph{Plane Depth},
manuscript (math-research repository), 2026.

\bibitem{bauerfeld-nested-tire-duals}
E.~Bauerfeld,
\emph{Coloring Nested Tire Dual Graphs},
manuscript (math-research repository), 2026.

\end{thebibliography}

\end{document}
