Introduce flip neighborhood and contradiction target Thm 4.5
Defines the flip neighborhood N(G) and recasts the colored edge flip class as a transitive closure rather than a single-step set, then states Theorem 4.5: no colored flip class of a flip-neighbor of a minimum-order 5-chromatic G contains G itself. The proof is one inductive line from the definition; the theorem is intended as the contradiction target for a future argument that some other condition would force G into such a class.
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\@writefile{toc}{\contentsline {section}{\tocsection {}{3}{Flip-symmetric maximal planar graphs}}{1}{}\protected@file@percent }
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\@writefile{lof}{\contentsline {figure}{\numberline {1}{\ignorespaces An edge flip replaces the diagonal $uv$ of the quadrilateral $uwvx$ with the diagonal $wx$.}}{2}{}\protected@file@percent }
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\newlabel{def:flip-symmetric}{{3.1}{2}}
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\newlabel{def:colored-flip-class}{{3.2}{2}}
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\newlabel{def:flip-neighborhood}{{3.2}{2}}
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\@writefile{toc}{\contentsline {section}{\tocsection {}{4}{A minimal four-colorable counterexample}}{2}{}\protected@file@percent }
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\newlabel{def:edge-deletion}{{4.1}{2}}
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\newlabel{lem:edge-deletion-4colorable}{{4.2}{2}}
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@@ -15,6 +16,7 @@
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\newlabel{tocindent2}{0pt}
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\newlabel{tocindent3}{0pt}
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\newlabel{thm:min-five-chromatic-not-flip-symmetric}{{4.4}{3}}
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\newlabel{thm:no-colored-class-contains-G}{{4.5}{3}}
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\@writefile{lof}{\contentsline {figure}{\numberline {2}{\ignorespaces Case\nonbreakingspace 2 of the proof of Theorem\nonbreakingspace 4.4\hbox {}: $u, v$ share color $a$ and $w, x$ share color $c$. The $\{a, b\}$-Kempe path $P$ from $u$ to $v$ separates $w$ from $x$ in the plane, so no $\{c, d\}$-path between $w$ and $x$ can avoid crossing $P$; since the color sets $\{a, b\}$ and $\{c, d\}$ are disjoint, no such path exists.}}{4}{}\protected@file@percent }
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\newlabel{fig:flip-proof-case-two}{{2}{4}}
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@@ -172,19 +172,30 @@ $G^{\mathrm{flip}(uv)} \cong G$. We write $\mathcal{F}$ for the class
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of flip-symmetric maximal planar graphs.
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\end{definition}
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\begin{definition}[Flip neighborhood]\label{def:flip-neighborhood}
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Let $G$ be a maximal planar graph. The \emph{flip neighborhood} of
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$G$ is the set
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\[
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\mathcal{N}(G) \;=\; \bigl\{\, G^{\mathrm{flip}(uv)} : uv \in E(G)
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\text{ and the flip at } uv \text{ is admissible} \,\bigr\}
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\]
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of maximal planar graphs obtainable from $G$ by a single admissible
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edge flip.
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\end{definition}
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\begin{definition}[Colored edge flip class]\label{def:colored-flip-class}
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Let $G$ be a maximal planar graph and let $\varphi$ be a proper
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$4$-coloring of $G$. The \emph{colored edge flip class} of
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$(G, \varphi)$ is the set
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\[
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\mathcal{C}(G, \varphi) \;=\; \bigl\{\, G^{\mathrm{flip}(uv)} :
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uv \in E(G),\ \text{the flip at } uv \text{ is admissible, and}\
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\varphi(w) \neq \varphi(x) \,\bigr\},
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\]
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where $w, x$ are the third vertices of the two triangular faces of
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$G$ containing $uv$. Equivalently, $\mathcal{C}(G, \varphi)$ is the
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set of graphs obtained from $G$ by an admissible edge flip under
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which $\varphi$ remains a proper $4$-coloring.
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$(G, \varphi)$ is the set $\mathcal{C}(G, \varphi)$ of maximal planar
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graphs reachable from $G$ by some (possibly empty) sequence of
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admissible edge flips, each of which leaves $\varphi$ a proper
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$4$-coloring of the resulting graph. Explicitly,
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$H \in \mathcal{C}(G, \varphi)$ iff there exist graphs
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$G = G_0, G_1, \ldots, G_k = H$ such that for each $0 \leq i < k$,
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$G_{i+1} = G_i^{\mathrm{flip}(u_i v_i)}$ for some
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$u_i v_i \in E(G_i)$ whose flip is admissible in $G_i$ and whose
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opposite vertices $w_i, x_i$ satisfy
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$\varphi(w_i) \neq \varphi(x_i)$.
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\end{definition}
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\section{A minimal four-colorable counterexample}
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@@ -250,8 +261,7 @@ applied to $\varphi'$.
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\begin{theorem}\label{thm:min-five-chromatic-not-flip-symmetric}
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Let $G$ be a minimum-order maximal planar graph with $\chi(G) \geq 5$.
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Then for every edge $e \in E(G)$, the graph induced by an edge flip
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of $e$ is $4$-colorable.
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Then every $H \in \mathcal{N}(G)$ is $4$-colorable.
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\end{theorem}
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\begin{proof}
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@@ -324,6 +334,27 @@ exists.}
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\label{fig:flip-proof-case-two}
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\end{figure}
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\begin{theorem}\label{thm:no-colored-class-contains-G}
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Let $G$ be a minimum-order maximal planar graph with $\chi(G) \geq 5$.
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Then for every $H \in \mathcal{N}(G)$ and every proper $4$-coloring
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$\varphi$ of $H$,
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\[
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G \;\notin\; \mathcal{C}(H, \varphi).
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\]
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\end{theorem}
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\begin{proof}
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Suppose, for contradiction, that $G \in \mathcal{C}(H, \varphi)$ for
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some $H \in \mathcal{N}(G)$ and some proper $4$-coloring $\varphi$ of
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$H$. By Definition~\ref{def:colored-flip-class}, there exists a
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sequence of maximal planar graphs $H = H_0, H_1, \ldots, H_k = G$ in
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which each $H_{i+1}$ is obtained from $H_i$ by an admissible edge
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flip that leaves $\varphi$ a proper $4$-coloring of $H_{i+1}$. By
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induction on $i$, $\varphi$ is a proper $4$-coloring of every $H_i$;
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in particular, $\varphi$ is a proper $4$-coloring of $H_k = G$. But
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$\chi(G) \geq 5$ admits no such coloring, a contradiction.
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\end{proof}
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\end{document}
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%-----------------------------------------------------------------------
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