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face_monochromatic_pairs: reframe Lemma 5.2 as a non-existence result

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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>
This commit is contained in:
didericis
2026-05-24 22:31:10 -04:00
parent d99f8e23b3
commit 037d987c7d
6 changed files with 148 additions and 159 deletions
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papers/face_monochromatic_pairs/experiments/draw_lemma_kempe_heawood.py
+79 -89
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@@ -1,17 +1,17 @@
"""Two-panel illustration of the proof of Lemma 5.2
(Heawood constant on Kempe cycles through merged).
"""Two-panel illustration of the proof of Lemma 5.2 (a Heawood-constant
Kempe cycle does not admit the clause-(3) arc of Conjecture 5.1).
Each panel shows two consecutive vertices v_0, v_1 on the {a, b}-Kempe
cycle K, joined by an edge e, with h(v_0) = +1 (CW colour order (a, b, c))
and h(v_1) = -1 (CW colour order (a, c, b)).
cycle K, joined by an edge e, with h(v_0) = h(v_1) = +1: i.e., both
have the same clockwise colour order (a, b, c). The would-be witness
edges (b-edges in Case A, a-edges in Case B) lie on opposite sides of
e, so no face of the graph contains both of them.
Left panel (Case A): phi(e) = a. The two b-edges at v_0, v_1 both lie on
the same face F = F_R (right side of e); they form
the witness (e_1, e_2).
Left panel (Case A): phi(e) = a. The b-edges at v_0, v_1 are on
opposite sides of e (one south, one north).
Right panel (Case B): phi(e) = b. The two a-edges at v_0, v_1 both lie
on the same face F = F_L (left side of e); they
form the witness (e_1, e_2).
Right panel (Case B): phi(e) = b. The a-edges at v_0, v_1 are on
opposite sides of e.
Produces fig_lemma_kempe_heawood.png.
"""
@@ -25,12 +25,12 @@ OUT_DIR = os.path.dirname(os.path.dirname(os.path.abspath(__file__)))
DARK = '#374151'
GRAY = '#9ca3af'
# Colour code matching earlier figures: a=red/orange, b=blue, c=green.
COL_A = '#ea580c' # 'a'
COL_B = '#2563eb' # 'b'
COL_C = '#16a34a' # 'c'
FACE_FILL = '#fef3c7'
FACE_FILL_R = '#fef3c7' # F_R shading (south)
FACE_FILL_L = '#dbeafe' # F_L shading (north)
V0 = (-1.6, 0.0)
V1 = ( 1.6, 0.0)
@@ -62,7 +62,7 @@ def label_text(ax, p, text, color=DARK, fontsize=12, dx=0, dy=0,
edgecolor='none', alpha=0.85))
def label_edge_midpoint(ax, p, q, text, color, fontsize=11, offset=(0, 0)):
def label_edge_midpoint(ax, p, q, text, color, fontsize=10, offset=(0, 0)):
mid = ((p[0] + q[0]) / 2 + offset[0],
(p[1] + q[1]) / 2 + offset[1])
ax.text(mid[0], mid[1], text, ha='center', va='center',
@@ -71,132 +71,122 @@ def label_edge_midpoint(ax, p, q, text, color, fontsize=11, offset=(0, 0)):
edgecolor='none', alpha=0.9))
def shade_face(ax, pts, color=FACE_FILL, alpha=0.7):
def shade_face(ax, pts, color, alpha=0.55):
poly = Polygon(pts, facecolor=color, edgecolor='none',
alpha=alpha, zorder=1)
ax.add_patch(poly)
def panel_case_A(ax):
# phi(e) = a. v_0 has CW order (a, b, c) starting from e at 0 deg:
# e (a) at 0 deg, b-edge at 300 deg (southeast), c-edge at 120 deg
# (northwest).
# v_1 has CW order (a, c, b) starting from e at 180 deg:
# e (a) at 180 deg, c-edge at 60 deg (northeast), b-edge at 240 deg
# (south-southwest).
e_color = COL_A
# Other endpoints (stubs) of the non-e edges.
b0 = edge_at(V0, -60) # b-edge at v_0, southeast
c0 = edge_at(V0, 120) # c-edge at v_0, northwest
c1 = edge_at(V1, 60) # c-edge at v_1, northeast
b1 = edge_at(V1, 240) # b-edge at v_1, southwest
# Same Heawood: v_0 and v_1 both have CW order (a, b, c) with e = a.
# v_0: e at 0 deg (east), b at 300 deg (south), c at 60 deg (north).
# v_1: e at 180 deg (west), b at 90 deg (north), c at 270 deg (south).
# The b-edges land on opposite sides of e (south of v_0, north of v_1).
b0 = edge_at(V0, 300) # south of v_0
c0 = edge_at(V0, 60) # north of v_0
b1 = edge_at(V1, 90) # north of v_1
c1 = edge_at(V1, 270) # south of v_1
# Shade F_R = south face: vertices roughly (b0, V0, V1, b1) plus a
# closing polygon below.
shade_face(ax, [V0, V1, b1, (b1[0] + 0.2, b1[1] - 0.6),
(b0[0] - 0.2, b0[1] - 0.6), b0])
label_text(ax, ((V0[0] + V1[0]) / 2, -1.6), 'face $F$', color=DARK,
fontsize=12, weight='bold')
# Shade both F_R (south) and F_L (north) lightly.
shade_face(ax, [V0, V1, c1, (c1[0] + 0.3, c1[1] - 0.6),
(b0[0] - 0.3, b0[1] - 0.6), b0], color=FACE_FILL_R)
shade_face(ax, [V0, c0, (c0[0] - 0.3, c0[1] + 0.6),
(b1[0] + 0.3, b1[1] + 0.6), b1, V1], color=FACE_FILL_L)
label_text(ax, ((V0[0] + V1[0]) / 2, -1.7), '$F_R$', color=DARK,
fontsize=11, weight='bold')
label_text(ax, ((V0[0] + V1[0]) / 2, 1.7), '$F_L$', color=DARK,
fontsize=11, weight='bold')
# Edges
draw_edge(ax, V0, V1, e_color) # e (color a)
draw_edge(ax, V0, b0, COL_B) # b-edge at v_0
draw_edge(ax, V0, c0, COL_C) # c-edge at v_0
draw_edge(ax, V1, c1, COL_C) # c-edge at v_1
draw_edge(ax, V1, b1, COL_B) # b-edge at v_1
draw_edge(ax, V0, V1, COL_A)
draw_edge(ax, V0, b0, COL_B)
draw_edge(ax, V0, c0, COL_C)
draw_edge(ax, V1, b1, COL_B)
draw_edge(ax, V1, c1, COL_C)
# Vertices
draw_vertex(ax, V0, DARK)
draw_vertex(ax, V1, DARK)
draw_vertex(ax, V0, DARK); draw_vertex(ax, V1, DARK)
draw_stub(ax, b0); draw_stub(ax, c0)
draw_stub(ax, c1); draw_stub(ax, b1)
draw_stub(ax, b1); draw_stub(ax, c1)
# Labels
label_text(ax, V0, '$v_0$', dy=0.28, fontsize=12)
label_text(ax, (V0[0] - 0.05, V0[1] - 0.28), '$h_\\varphi\\!=\\!+1$',
color=DARK, fontsize=9)
label_text(ax, V1, '$v_1$', dy=0.28, fontsize=12)
label_text(ax, (V1[0] + 0.05, V1[1] - 0.28), '$h_\\varphi\\!=\\!-1$',
label_text(ax, (V1[0] + 0.05, V1[1] - 0.28), '$h_\\varphi\\!=\\!+1$',
color=DARK, fontsize=9)
label_edge_midpoint(ax, V0, V1, '$e\\!=\\!a$', color=COL_A,
offset=(0, 0.16))
label_edge_midpoint(ax, V0, b0, '$e_1\\!=\\!b$', color=COL_B,
offset=(-0.05, 0.05))
offset=(0, 0.18))
label_edge_midpoint(ax, V0, b0, '$b$', color=COL_B,
offset=(-0.15, 0))
label_edge_midpoint(ax, V0, c0, '$c$', color=COL_C,
offset=(0.05, 0))
offset=(-0.15, 0))
label_edge_midpoint(ax, V1, b1, '$b$', color=COL_B,
offset=(0.15, 0))
label_edge_midpoint(ax, V1, c1, '$c$', color=COL_C,
offset=(-0.05, 0))
label_edge_midpoint(ax, V1, b1, '$e_2\\!=\\!b$', color=COL_B,
offset=(0.05, 0.05))
offset=(0.15, 0))
ax.set_title('Case A: $\\varphi(e) = a$. The two $b$-edges'
' at $v_0, v_1$ lie on $\\partial F$',
fontsize=11, color=DARK, pad=10, fontweight='bold')
ax.set_title('Case A: $\\varphi(e) = a$. The two $b$-edges are on\n'
'opposite sides of $e$ -- no common face',
fontsize=11, color=DARK, pad=8, fontweight='bold')
def panel_case_B(ax):
# phi(e) = b. v_0 has CW order (a, b, c) with b = e at 0 deg:
# a-edge at 60 deg (northeast), e (b) at 0 deg, c-edge at 300 deg
# (southeast).
# v_1 has CW order (a, c, b) with b = e at 180 deg:
# a-edge at 60 deg (northeast), c-edge at 300 deg (southeast), e
# (b) at 180 deg.
e_color = COL_B
a0 = edge_at(V0, 60) # a-edge at v_0, northeast
c0 = edge_at(V0, -60) # c-edge at v_0, southeast
a1 = edge_at(V1, 120) # a-edge at v_1, northwest
c1 = edge_at(V1, 240) # c-edge at v_1, southwest
# Same Heawood: v_0 and v_1 both have CW order (a, b, c) with e = b.
# v_0: a at 60 deg (north), e (b) at 0 deg (east), c at 300 deg (south).
# v_1: a at 270 deg (south), e (b) at 180 deg (west), c at 90 deg (north).
# The a-edges land on opposite sides of e (north of v_0, south of v_1).
a0 = edge_at(V0, 60) # north of v_0
c0 = edge_at(V0, 300) # south of v_0
a1 = edge_at(V1, 270) # south of v_1
c1 = edge_at(V1, 90) # north of v_1
# Shade F_L = north face: a0, V0, V1, a1, plus a closing polygon above.
shade_face(ax, [V0, a0, (a0[0] - 0.2, a0[1] + 0.6),
(a1[0] + 0.2, a1[1] + 0.6), a1, V1])
label_text(ax, ((V0[0] + V1[0]) / 2, 1.6), 'face $F$', color=DARK,
fontsize=12, weight='bold')
shade_face(ax, [V0, V1, a1, (a1[0] + 0.3, a1[1] - 0.6),
(c0[0] - 0.3, c0[1] - 0.6), c0], color=FACE_FILL_R)
shade_face(ax, [V0, a0, (a0[0] - 0.3, a0[1] + 0.6),
(c1[0] + 0.3, c1[1] + 0.6), c1, V1], color=FACE_FILL_L)
label_text(ax, ((V0[0] + V1[0]) / 2, -1.7), '$F_R$', color=DARK,
fontsize=11, weight='bold')
label_text(ax, ((V0[0] + V1[0]) / 2, 1.7), '$F_L$', color=DARK,
fontsize=11, weight='bold')
# Edges
draw_edge(ax, V0, V1, e_color)
draw_edge(ax, V0, V1, COL_B)
draw_edge(ax, V0, a0, COL_A)
draw_edge(ax, V0, c0, COL_C)
draw_edge(ax, V1, a1, COL_A)
draw_edge(ax, V1, c1, COL_C)
# Vertices
draw_vertex(ax, V0, DARK)
draw_vertex(ax, V1, DARK)
draw_vertex(ax, V0, DARK); draw_vertex(ax, V1, DARK)
draw_stub(ax, a0); draw_stub(ax, c0)
draw_stub(ax, a1); draw_stub(ax, c1)
# Labels
label_text(ax, V0, '$v_0$', dy=0.28, fontsize=12)
label_text(ax, (V0[0] - 0.05, V0[1] - 0.28), '$h_\\varphi\\!=\\!+1$',
color=DARK, fontsize=9)
label_text(ax, V1, '$v_1$', dy=0.28, fontsize=12)
label_text(ax, (V1[0] + 0.05, V1[1] - 0.28), '$h_\\varphi\\!=\\!-1$',
label_text(ax, (V1[0] + 0.05, V1[1] - 0.28), '$h_\\varphi\\!=\\!+1$',
color=DARK, fontsize=9)
label_edge_midpoint(ax, V0, V1, '$e\\!=\\!b$', color=COL_B,
offset=(0, -0.18))
label_edge_midpoint(ax, V0, a0, '$e_1\\!=\\!a$', color=COL_A,
offset=(-0.05, 0))
label_edge_midpoint(ax, V0, a0, '$a$', color=COL_A,
offset=(-0.15, 0))
label_edge_midpoint(ax, V0, c0, '$c$', color=COL_C,
offset=(0.05, 0))
label_edge_midpoint(ax, V1, a1, '$e_2\\!=\\!a$', color=COL_A,
offset=(0.05, 0))
offset=(-0.15, 0))
label_edge_midpoint(ax, V1, a1, '$a$', color=COL_A,
offset=(0.15, 0))
label_edge_midpoint(ax, V1, c1, '$c$', color=COL_C,
offset=(-0.05, 0))
offset=(0.15, 0))
ax.set_title('Case B: $\\varphi(e) = b$. The two $a$-edges'
' at $v_0, v_1$ lie on $\\partial F$',
fontsize=11, color=DARK, pad=10, fontweight='bold')
ax.set_title('Case B: $\\varphi(e) = b$. The two $a$-edges are on\n'
'opposite sides of $e$ -- no common face',
fontsize=11, color=DARK, pad=8, fontweight='bold')
def main():
plt.rcParams['text.usetex'] = False # keep matplotlib defaults
fig, axes = plt.subplots(1, 2, figsize=(13, 5.5))
fig, axes = plt.subplots(1, 2, figsize=(13, 5.8))
for ax in axes:
ax.set_xlim(-3.5, 3.5)
ax.set_ylim(-2.4, 2.4)
ax.set_ylim(-2.5, 2.5)
ax.set_aspect('equal')
ax.axis('off')
panel_case_A(axes[0])
papers/face_monochromatic_pairs/fig_lemma_kempe_heawood.png
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papers/face_monochromatic_pairs/paper.aux
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@@ -38,9 +38,9 @@
\newlabel{conj:face-monochromatic-pair-on-merged-kempe-cycle}{{5.1}{10}}
\newlabel{lem:kempe-heawood-constant}{{5.2}{11}}
\newlabel{rem:conj-3-6-empirical}{{5.3}{11}}
\@writefile{lof}{\contentsline {figure}{\numberline {5}{\ignorespaces The two cases in the proof of Lemma\nonbreakingspace 5.2\hbox {}. Vertices $v_0, v_1$ are consecutive on the $\{a, b\}$-Kempe cycle $K$, joined by an edge $e$, with $h_\varphi (v_0) = +1$ (clockwise colour order $(a, b, c)$) and $h_\varphi (v_1) = -1$ (clockwise order $(a, c, b)$). \emph {Left (Case\nonbreakingspace A):} when $\varphi (e) = a$, the two $b$-edges at $v_0, v_1$ lie on the same face $F$, with $e$ as the unique $\partial F$-edge between them. \emph {Right (Case\nonbreakingspace B):} when $\varphi (e) = b$, the two $a$-edges at $v_0, v_1$ lie on the opposite face $F$ instead, again with $e$ between them on one arc. In either case $(F, e_1, e_2)$ witnesses clauses (1)--(3) of Conjecture\nonbreakingspace 5.1\hbox {}.}}{12}{}\protected@file@percent }
\@writefile{lof}{\contentsline {figure}{\numberline {5}{\ignorespaces The two cases in the proof of Lemma\nonbreakingspace 5.2\hbox {}. Vertices $v_0, v_1$ are consecutive on the $\{a, b\}$-Kempe cycle $K$, joined by an edge $e$, with the lemma's hypothesis $h_\varphi (v_0) = h_\varphi (v_1) = +1$ --- so both vertices share the clockwise colour order $(a, b, c)$. \emph {Left (Case\nonbreakingspace A):} when $\varphi (e) = a$, the colour-$b$ edge at $v_0$ lies south of $e$ (on $\partial F_R$) and the colour-$b$ edge at $v_1$ lies north of $e$ (on $\partial F_L$); the two would-be witness edges are on opposite faces, so no face of $\setbox \z@ \hbox {\mathsurround \z@ $\textstyle G$}\mathaccent "0362{G}'_{v,i}$ contains both. \emph {Right (Case\nonbreakingspace B):} when $\varphi (e) = b$, the colour-$a$ edges at $v_0, v_1$ are likewise on opposite sides of $e$. In either case the clause-$(3)$ arc of Conjecture\nonbreakingspace 5.1\hbox {} cannot be realised at $e$.}}{12}{}\protected@file@percent }
\newlabel{fig:lemma-kempe-heawood}{{5}{12}}
\newlabel{conj:face-monochromatic-pair-strengthened}{{5.4}{13}}
\newlabel{conj:face-monochromatic-pair-strengthened}{{5.4}{12}}
\newlabel{rem:conj-3-8-empirical}{{5.5}{13}}
\newlabel{rem:implication-4ct}{{5.6}{13}}
\bibcite{Heawood1898}{1}
papers/face_monochromatic_pairs/paper.log
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\OT1/cmr/m/it/10 Remark \OT1/cmr/m/n/10 5.5\OT1/cmr/m/it/10 . \OT1/cmr/m/n/10 T
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[]
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papers/face_monochromatic_pairs/paper.tex
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@@ -662,67 +662,64 @@ merged edge, such that:
\end{enumerate}
\end{conjecture}
\begin{lemma}[Heawood number is constant on the Kempe cycles through the merged edge]
\begin{lemma}[A Heawood-constant Kempe cycle does not admit the clause-(3) arc]
\label{lem:kempe-heawood-constant}
Let $G$ be a minimal counterexample to the Four Colour Theorem, fix a
reduced dual $\widehat{G}'_{v,i}$ of $G' = \mathrm{dual}(G)$, and let
$\varphi$ be a proper $3$-edge-colouring of $\widehat{G}'_{v,i}$. Set
$a := \varphi(\mathrm{merged})$. Then for each
$b \in \{1, 2, 3\} \setminus \{a\}$, every vertex of the
$\{a, b\}$-Kempe cycle of $\varphi$ through the merged edge has the same
Heawood number $h_\varphi$.
$a := \varphi(\mathrm{merged})$ and let $K$ be the $\{a, b\}$-Kempe
cycle of $\varphi$ through the merged edge for some
$b \in \{1, 2, 3\} \setminus \{a\}$. If $h_\varphi$ is constant on
$V(K)$, then no edge $e \in E(K)$ admits a face $F$ of
$\widehat{G}'_{v,i}$ and two non-incident edges
$e_1, e_2 \in \partial F$ such that
$\varphi(e_1) = \varphi(e_2)$ and $e$ is the unique edge of
$\partial F$ between $e_1$ and $e_2$ along one of the two arcs of
$\partial F$ --- that is, no edge of $K$ admits the clause-(3) arc of
Conjecture~\ref{conj:face-monochromatic-pair-on-merged-kempe-cycle}
with $e_1, e_2$ at its two endpoints.
\end{lemma}
\begin{proof}
Fix $b \in \{1, 2, 3\} \setminus \{a\}$, let $K$ be the $\{a, b\}$-Kempe
cycle of $\varphi$ through the merged edge, and let $c$ be the third
colour. Suppose for contradiction that $h_\varphi$ is not constant on
$V(K)$. Since $K$ is a closed cycle, there exist consecutive vertices
$v_0, v_1 \in V(K)$, joined by an edge $e \in E(K)$, with
$h_\varphi(v_0) \neq h_\varphi(v_1)$. After possibly swapping
$v_0, v_1$, take $h_\varphi(v_0) = +1$ and $h_\varphi(v_1) = -1$. By
Definition~\ref{def:heawood-number}, the clockwise cyclic colour order
at $v_0$ is $(a, b, c)$ (an even cyclic permutation), and at $v_1$ it is
$(a, c, b)$ (an odd one).
Let $c$ be the third colour. Fix any edge $e \in E(K)$ joining
$v_0, v_1 \in V(K)$. By hypothesis $h_\varphi(v_0) = h_\varphi(v_1)$;
after possibly relabelling we may take
$h_\varphi(v_0) = h_\varphi(v_1) = +1$, so by
Definition~\ref{def:heawood-number} the clockwise cyclic colour order
at $v_0$ and at $v_1$ is the same even cyclic class $(a, b, c)$.
Let $F_R, F_L$ be the two faces of $\widehat{G}'_{v,i}$ on the two sides
of $e$, with $F_R$ on the right side as one walks from $v_0$ to $v_1$.
For a vertex $v \in \{v_0, v_1\}$, the non-$e$ edge of $\partial F_R$ at
$v$ is the next-clockwise edge from $e$ around $v_0$ (since at $v_0$ the
right side coincides with the clockwise next edge from $e$) and the
next-counter-clockwise edge from $e$ around $v_1$ (since at $v_1$ the
orientation of $e$ is reversed, so the right side coincides with the
counter-clockwise next edge from $e$).
Let $F_R, F_L$ be the two faces of $\widehat{G}'_{v,i}$ on the two
sides of $e$, with $F_R$ on the right side as one walks from $v_0$ to
$v_1$. For a vertex $v \in \{v_0, v_1\}$, the non-$e$ edge of
$\partial F_R$ at $v$ is the next-clockwise edge from $e$ around $v_0$
(since at $v_0$ the right side coincides with the clockwise next edge
from $e$) and the next-counter-clockwise edge from $e$ around $v_1$
(since at $v_1$ the orientation of $e$ is reversed, so the right side
coincides with the counter-clockwise next edge from $e$).
\emph{Case~A: $\varphi(e) = a$.} The CW order $(a, b, c)$ at $v_0$ makes
the next-CW edge from $e$ the colour-$b$ edge at $v_0$; the CW order
$(a, c, b)$ at $v_1$ makes the next-CCW edge from $e$ the colour-$b$
edge at $v_1$. Let $e_1, e_2$ be these colour-$b$ edges at $v_0$ and
$v_1$ respectively. Then $e_1, e_2 \in \partial F_R$, they are
non-incident (their endpoints other than $v_0, v_1$ are distinct
vertices), and $e$ is the unique edge of $\partial F_R$ lying between
$e_1$ and $e_2$ along one of the two arcs of $\partial F_R$. Both
$e_1, e_2$ lie on $K$ (the colour-$b$ edge at any $K$-vertex is a
$K$-edge), so $e_1, e_2$, and the merged edge are on a common
$\{a, b\}$-Kempe cycle, and $\varphi(e_1) = \varphi(e_2) = b \neq a$
means neither equals the merged edge.
\emph{Case~A: $\varphi(e) = a$.} In the CW order $(a, b, c)$ at $v_0$
the next-CW edge from $e$ has colour $b$; in the same CW order
$(a, b, c)$ at $v_1$ the next-CCW edge from $e$ has colour $c$ (since
CCW-next from $a$ in cyclic order $(a, b, c)$ is $c$). Hence the
non-$e$ edge of $\partial F_R$ at $v_0$ has colour $b$, while the
non-$e$ edge of $\partial F_R$ at $v_1$ has colour $c$ --- these
differ. Symmetrically, the non-$e$ edges of $\partial F_L$ at $v_0$
and $v_1$ have colours $c$ and $b$ respectively, again different.
Hence the colour-$b$ edges at $v_0$ and $v_1$ lie on opposite faces
of $e$, and the same for the colour-$c$ edges; no face of
$\widehat{G}'_{v,i}$ contains two same-coloured non-$e$ edges at
$\{v_0, v_1\}$.
\emph{Case~B: $\varphi(e) = b$.} By the symmetric reasoning applied to
$F_L$, the colour-$a$ edges at $v_0$ and $v_1$ both lie on
$\partial F_L$, with $e$ as the unique edge of $\partial F_L$ between
them on one arc; both lie on $K$, and if neither $v_0$ nor $v_1$ is an
endpoint of the merged edge (which can be arranged by choosing the
differing-Heawood pair $(v_0, v_1)$ appropriately on $K$), neither
colour-$a$ edge equals the merged edge.
\emph{Case~B: $\varphi(e) = b$.} By the analogous reasoning, the
non-$e$ edges of $\partial F_R$ at $v_0$ and $v_1$ have colours $c$
and $a$ respectively, and those of $\partial F_L$ have colours $a$
and $c$. The colour-$a$ edges at $v_0, v_1$ thus lie on opposite
faces of $e$, and so do the colour-$c$ edges.
Either way, the cyclic colour orders at $v_0, v_1$ force a face $F$ of
$\widehat{G}'_{v,i}$ and two non-incident edges $e_1, e_2 \in \partial
F$, both of the same colour and both on $K$ together with the merged
edge, such that $e_1$ and $e_2$ lie on a common arc of $\partial F$ with
exactly one edge of $\partial F$ between them.
The triple $(F, e_1, e_2)$ then satisfies clauses~(1)--(3) of
Conjecture~\ref{conj:face-monochromatic-pair-on-merged-kempe-cycle}.
In either case, no face $F$ of $\widehat{G}'_{v,i}$ has two same-coloured
non-$e$ edges at $\{v_0, v_1\}$ on $\partial F$, so the clause-(3) arc
(with $e$ as the unique $\partial F$-edge between $e_1$ and $e_2$ at the
endpoints of $e$) cannot be realised.
\end{proof}
\begin{figure}[h]
@@ -731,15 +728,17 @@ Conjecture~\ref{conj:face-monochromatic-pair-on-merged-kempe-cycle}.
\caption{The two cases in the proof of
Lemma~\ref{lem:kempe-heawood-constant}. Vertices $v_0, v_1$ are
consecutive on the $\{a, b\}$-Kempe cycle $K$, joined by an edge $e$,
with $h_\varphi(v_0) = +1$ (clockwise colour order $(a, b, c)$) and
$h_\varphi(v_1) = -1$ (clockwise order $(a, c, b)$). \emph{Left
(Case~A):} when $\varphi(e) = a$, the two $b$-edges at $v_0, v_1$ lie on
the same face $F$, with $e$ as the unique $\partial F$-edge between
them. \emph{Right (Case~B):} when $\varphi(e) = b$, the two $a$-edges at
$v_0, v_1$ lie on the opposite face $F$ instead, again with $e$ between
them on one arc. In either case $(F, e_1, e_2)$ witnesses clauses
(1)--(3) of
Conjecture~\ref{conj:face-monochromatic-pair-on-merged-kempe-cycle}.}
with the lemma's hypothesis $h_\varphi(v_0) = h_\varphi(v_1) = +1$ ---
so both vertices share the clockwise colour order $(a, b, c)$.
\emph{Left (Case~A):} when $\varphi(e) = a$, the colour-$b$ edge at
$v_0$ lies south of $e$ (on $\partial F_R$) and the colour-$b$ edge at
$v_1$ lies north of $e$ (on $\partial F_L$); the two would-be witness
edges are on opposite faces, so no face of $\widehat{G}'_{v,i}$ contains
both. \emph{Right (Case~B):} when $\varphi(e) = b$, the colour-$a$
edges at $v_0, v_1$ are likewise on opposite sides of $e$. In either
case the clause-$(3)$ arc of
Conjecture~\ref{conj:face-monochromatic-pair-on-merged-kempe-cycle}
cannot be realised at $e$.}
\label{fig:lemma-kempe-heawood}
\end{figure}