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Records a proof skeleton and the +single lemma it reduces to; nothing here is proved. Companion to +\texttt{boundary\_restriction\_structure.tex} and \texttt{paper.tex}.} + +\section*{Setup: the move} + +Let $G$ be a minimal counterexample to the Four Colour Theorem (a smallest +triangulation with no proper $4$-vertex-colouring). By Euler $G$ has a +vertex $v$ of degree $\le 5$; degrees $\le 4$ are classically reducible, so +take $\deg(v)=5$. The link of $v$ is a pentagon $a,x,b,y,z$ (consecutive +neighbours adjacent). For a non-adjacent (diagonal) pair, say $a,b$, the +\emph{double contraction} contracts both edges $(a,v)$ and $(b,v)$, +identifying $a=v=b$ into a single vertex. There are $5$ diagonal pairs. + +The double contraction is a proper minor with two fewer vertices, hence a +smaller planar graph $G'$; by minimality $G'$ is $4$-colourable. It stays a +simple triangulation iff each contracted edge is non-separating (its +endpoints have exactly two common neighbours). + +\section*{Kempe chains as Heawood face-chains} + +Work in the cubic dual $G'$ (vertices $=$ faces of $G$, so the Heawood +$\pm1$ labels sit on dual vertices). Tait-colour the edges with the three +nonzero Klein-4 elements $\{a,b,c\}$ (edge colour $=$ colour-difference +across the primal edge). Then: + +\begin{obs}[Kempe chain $=$ Tait cycle $=$ flip-set] +A primal Kempe chain (component of two vertex-colour classes of $G$) +corresponds to a connected component of the two-edge-colour subgraph +$\{a,b\}$ of the dual. That subgraph is $2$-regular, so it is a disjoint +union of cycles; each such cycle alternates $a,b,a,b,\dots$ and is therefore +\textbf{always even}. The cycle is literally a cyclic chain of faces of $G$. +A Kempe swap reverses the rotational order $(a,b,c)$ at exactly the vertices +on the cycle, so the set of faces whose Heawood sign \emph{flips} under the +swap is precisely the Kempe chain. +\end{obs} + +\begin{obs}[Sign alternation $=$ same-side rule] +Even-ness does \emph{not} force the Heawood signs to alternate along the +cycle. At each cycle vertex the third (off-cycle, colour-$c$) edge points +either inside or outside the region bounded by the cycle. For consecutive +vertices $A,B$: +\[ + \text{same side (both in / both out)} \Rightarrow \text{signs flip;} + \qquad + \text{opposite side} \Rightarrow \text{signs repeat.} +\] +(Local chirality computation: the four in/out cases give +$+,-$ / $-,+$ / $+,+$ / $-,-$.) Hence the Heawood sign sequence around a +Tait cycle is determined, up to one global sign, by the in/out pattern of +the third edges; it is fully alternating iff there are zero side-switches, +i.e.\ all third edges lie on one side. +\end{obs} + +\section*{The transport hypothesis (L1)} + +\begin{conj}[Chain transport, claimed for all triangulations] +The double contraction preserves the relevant (crossing) Kempe/Heawood +chain structure as it propagates up the nested tire decomposition: the two +chains anchored at the pentagon survive, consistently, through every tire +interface above $v$. +\end{conj} + +\section*{The reductio (two nested contradictions)} + +We are \emph{not} claiming $G'$ is uncolourable, and we are not claiming +intertwined Kempe chains are impossible (they occur, e.g.\ in the Errera +graph). The reducible object is \emph{forced} intertwining. + +\paragraph{Inner.} Suppose intertwining is the \emph{only} way to colour +$G'$ (every colouring is intertwined at $v$). By transport the two crossing +chains pass through every interface above; if uncrossing is excluded at each +interface, the tire restriction relations $R_{\mathsf K}$ collapse to their +``crossed-only'' sub-relations. Contract each interface \textbf{along the +chain} to obtain a strictly smaller triangulation $H''$. If the crossed-only +sub-relations admit no compatible gluing on $H''$, then $H''$ is +uncolourable and smaller than $G$ --- contradicting minimality. + +\paragraph{Outer.} Therefore $G'$ admits a non-intertwined colouring; it +uncrosses at $v$, frees a colour for $v$, and lifts to a colouring of $G$ --- +contradicting that $G$ is a counterexample. \hfill$\square$ (modulo the +lemma below) + +\section*{What it all reduces to} + +\begin{claim}[the only open content] +Under the crossed-only collapse, the surviving Heawood boundary sequences at +a (forced-short) tire interface have empty pointwise-negation gluing on +$H''$ --- not merely small, genuinely empty. +\end{claim} + +This is where the $2^{n-2}$ constraint floor and the mod-3 side-pattern must +do real work: the floor guarantees that \emph{smallness alone} never empties +a relation, so the emptiness must come from the chains pinning the +sub-relation down, not from short interface length. + +\section*{Open points / to pin next} + +\begin{itemize} +\item \textbf{Define $H''$ precisely.} Current candidate: contract each tire + interface along the transported chain (the global analogue of the local + double contraction). Confirm this is a definite, strictly smaller + triangulation. +\item \textbf{Lift is not automatic.} ``Non-intertwined'' must be defined as + ``admits a swap collapsing the pentagon $a,x,b,y,z$ to $\le 3$ colours,'' + or freeing $v$ fails even after uncrossing (with $c(a)=c(b)$ the pentagon + can still show $4$ colours). +\item \textbf{Errera oracle.} Run the whole construction against the Errera + graph (and Fritsch / Kittell). These are colourable but Kempe-intertwined; + the hypothesis ``every colouring intertwined'' must \emph{fail} for them, + so the construction must decline. \emph{Why} it declines is exactly the + mechanism the Claim must deny in the counterexample case. +\item \textbf{Prove or break L1} as a statement about all triangulations. +\end{itemize} + +\end{document}