didericis
037d987c7d
The previous statement "Heawood is constant on K through merged" was
strictly stronger than what the proof actually established without
Conjecture 5.3. Restate the lemma in the contrapositive direction:
If h_phi is constant on V(K), then no edge e in E(K) admits a face
F of G'^hat and edges e_1, e_2 on dF realising the clause-(3) arc
of Conjecture 5.1 at the endpoints of e.
Proof structure is mostly preserved (same F_R/F_L geometry, same case
split on phi(e) in {a, b}, same reading-off of cyclic colour orders).
The hypothesis "h_phi(v_0) != h_phi(v_1)" becomes "h_phi(v_0) =
h_phi(v_1)", which flips the conclusion: the same-coloured non-e
edges at v_0, v_1 land on opposite faces of e instead of the same
face. No dependency on Conjecture 5.3 or Theorem 4.X.
Redraw the figure to match the new lemma: both vertices labelled
h_phi = +1, both showing CW order (a, b, c), and the same-colour pair
(b-edges in Case A, a-edges in Case B) drawn on opposite sides of e.
Co-Authored-By: Claude Opus 4.7 <noreply@anthropic.com>
58 lines
5.6 KiB
TeX
58 lines
5.6 KiB
TeX
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\citation{AH77a}
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\citation{AHK77}
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\@writefile{lof}{\contentsline {figure}{\numberline {1}{\ignorespaces The four steps of Definition\nonbreakingspace 3.2\hbox {}, illustrated on $G' = $ the dodecahedron (dual of the icosahedron) with $F_v$ the inner pentagon and $i = 0$. Top left: delete the five boundary vertices of $F_v$, leaving five degree-$2$ vertices on a new face $F$. Top right: order them clockwise as $A_0,\dots ,A_4$. Bottom left: add $v_n$ joined to $A_0, A_1, A_2$. Bottom right: add the chord $A_3 A_4$, giving the cubic plane graph $\setbox \z@ \hbox {\mathsurround \z@ $\textstyle G$}\mathaccent "0362{G}'_{v,0}$.}}{5}{}\protected@file@percent }
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\@writefile{lof}{\contentsline {figure}{\numberline {3}{\ignorespaces Edge suppression (Definition\nonbreakingspace 4.1\hbox {}). Left: a fragment of a cubic plane graph with the suppressed edge $e = uv$ highlighted in red. Middle: deleting $e$ leaves $u$ and $v$ of degree\nonbreakingspace $2$. Right: smoothing $u$ and $v$ replaces each pair of incident edges by a single new edge, removing $u, v$ and giving a cubic plane graph again.}}{9}{}\protected@file@percent }
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\@writefile{lof}{\contentsline {figure}{\numberline {4}{\ignorespaces The recolouring used in the proof of Theorem\nonbreakingspace 4.2\hbox {}. Left: the $4$-face $f$ of $H$ under $\varphi $, with the forced colours $\varphi (e_0) = a$, $\varphi (e_1) = b$, $\varphi (e_2) = \varphi (e_3) = c$, $\varphi (w_0) = \varphi (w_1) = b$, and $\varphi (w_2) = \varphi (w_3) = a$. Right: the suppressed graph $H'$ under $\varphi '$. The smoothed-in edges $e_2', e_3'$ inherit the colour $b$ from $w_0, w_1$, and $e_1$ is recoloured from $b$ to $c$; every edge outside the face neighbourhood keeps its $\varphi $-colour (dotted in red: the five edges of $H$ removed by the suppression).}}{10}{}\protected@file@percent }
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