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@@ -35,6 +35,8 @@
% Remove any commented or uncommented macros you do not use.
\documentclass{amsart}
\usepackage{tikz}
\usetikzlibrary{hobby}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
@@ -75,7 +77,7 @@
\date{}
\dedicatory{Dedicated to all who value more than machines}
\dedicatory{Dedicated to those who doubt the machine}
\begin{abstract}
\end{abstract}
@@ -130,106 +132,212 @@ In both cases we obtain a valid 4-coloring of $G$, a contradiction.
\section{Resolution of Degree 5 Case}
\subsection{At Least 12 Vertices of Degree 5}
\begin{figure}[h]
\centering
\begin{tikzpicture}[every node/.style={circle, draw, fill=black, inner sep=2pt, minimum size=5pt}]
\node[label=below:{$v_0$}] (v0) at (0,0) {};
\node[label=above:{$a_0$}] (a0) at (90:1.8) {};
\node[label=left:{$a_1$}] (a1) at (162:1.8) {};
\node[label=left:{$a_2$}] (a2) at (234:1.8) {};
\node[label=right:{$a_3$}] (a3) at (306:1.8) {};
\node[label=right:{$v_1$}] (v1) at (18:1.8) {};
\draw (a0) -- (a1) -- (a2) -- (a3) -- (v1) -- (a0);
\draw (v0) -- (a0);
\draw (v0) -- (a1);
\draw (v0) -- (a2);
\draw (v0) -- (a3);
\draw (v0) -- (v1);
\draw[dotted] (a0) -- ++(75:0.7) (a0) -- ++(105:0.7);
\draw[dotted] (a1) -- ++(147:0.7) (a1) -- ++(177:0.7);
\draw[dotted] (a2) -- ++(219:0.7) (a2) -- ++(249:0.7);
\draw[dotted] (a3) -- ++(291:0.7) (a3) -- ++(321:0.7);
\draw[dotted] (v1) -- ++(3:0.7) (v1) -- ++(33:0.7);
\end{tikzpicture}
\caption{The neighborhood of a degree-5 vertex $v_0$: a 5-cycle on $a_0,a_1,a_2,a_3,v_1$ with $v_0$ adjacent to each.}
\end{figure}
Since we have already handled every vertex of degree $\leq 4$, we may assume without loss of generality that the minimal counterexample $G$ has minimum degree 5. We now show that $G$ must contain at least 12 vertices of degree exactly 5.
Let $G$ be a minimum example of a maximal planar graph requiring 5 colors, and let $v_0$ be a graph in $G$ of degree 5, and let $v_1$ be a neighbor of $v_0$.
\begin{lemma}
If $G$ is a planar graph with minimum degree 5, then $G$ contains at least 12 vertices of degree exactly 5.
Let $G^*$ be the subgraph of $G$ obtained by deleting the edge $\{v_0, v_1\}$, and let $G'$ be the graph obtained by merging vertices $a_0$ and $a_3$ in $G^*$. Then there is at least one proper 4-coloring of $G*$ which assigns $v_0$ and $v_1$ the same color and $a_0$ and $a_3$ the same color.
\end{lemma}
\begin{proof}
For each $k \geq 5$, let $n_k$ denote the number of vertices of degree exactly $k$ in $G$. Since every vertex has degree $\geq 5$,
\[
V = \sum_{k \geq 5} n_k, \qquad 2E = \sum_{k \geq 5} k\, n_k.
\]
Using $E \leq 3V - 6$, we obtain $2E \leq 6V - 12$, so
\[
\sum_{k \geq 5} k\, n_k \leq 6\sum_{k \geq 5} n_k - 12.
\]
Moving all terms to the right-hand side gives
\[
12 \leq 6\sum_{k \geq 5} n_k - \sum_{k \geq 5} k\, n_k = \sum_{k \geq 5} (6 - k)\, n_k.
\]
We now expand this sum by separating the $k = 5$, $k = 6$, and $k \geq 7$ terms:
\[
\sum_{k \geq 5}(6-k)\,n_k
= \underbrace{(6-5)\,n_5}_{=\,n_5}
+ \underbrace{(6-6)\,n_6}_{=\,0}
+ \sum_{k \geq 7}(6-k)\,n_k
= n_5 - \sum_{k \geq 7}(k-6)\,n_k.
\]
Therefore $12 \leq n_5 - \sum_{k \geq 7}(k-6)\,n_k$, which rearranges to
\[
n_5 \geq 12 + \sum_{k \geq 7}(k-6)\,n_k.
\]
Since $k - 6 \geq 1 > 0$ and $n_k \geq 0$ for all $k \geq 7$, each term in the sum is non-negative, so $\sum_{k \geq 7}(k-6)\,n_k \geq 0$ and thus $n_5 \geq 12$.
Since $G'$ has fewer vertices than $G$, it admits a proper 4-coloring $\phi$ by minimality. Any such $\phi$ must assign $v_0$ and $v_1$ the same color: if they received distinct colors, $\phi$ would induce a proper 4-coloring of $G^*$ in which $v_0 \neq v_1$, and re-inserting edge $\{v_0,v_1\}$ would yield a proper 4-coloring of $G$, contradicting minimality. The induced coloring $\Phi$ of $G^*$ assigns $a_0$ and $a_3$ the common color $\phi(a_{03})$ by construction, so $\Phi$ is the required coloring.
\end{proof}
\subsection{Two Non-Adjacent Vertices of Degree 5}
\begin{figure}[h]
\centering
\begin{tikzpicture}[every node/.style={circle, draw, fill=black, inner sep=2pt, minimum size=5pt}]
\node[label=below:{$v_0$}] (v0) at (0,0) {};
\node[label=above:{$a_0$}] (a0) at (90:1.8) {};
\node[label=left:{$a_1$}] (a1) at (162:1.8) {};
\node[label=left:{$a_2$}] (a2) at (234:1.8) {};
\node[label=right:{$a_3$}] (a3) at (306:1.8) {};
\node[label=right:{$v_1$}] (v1) at (18:1.8) {};
\draw (a0) -- (a1) -- (a2) -- (a3) -- (v1) -- (a0);
\draw (v0) -- (a0);
\draw (v0) -- (a1);
\draw (v0) -- (a2);
\draw (v0) -- (a3);
\draw[dotted] (a0) -- ++(75:0.7) (a0) -- ++(105:0.7);
\draw[dotted] (a1) -- ++(147:0.7) (a1) -- ++(177:0.7);
\draw[dotted] (a2) -- ++(219:0.7) (a2) -- ++(249:0.7);
\draw[dotted] (a3) -- ++(291:0.7) (a3) -- ++(321:0.7);
\draw[dotted] (v1) -- ++(3:0.7) (v1) -- ++(33:0.7);
\end{tikzpicture}
\caption{The neighborhood of $v_0$ and $v_1$ in $G^*$: $G$ with the edge $\{v_0, v_1\}$ deleted, so the spoke to $v_1$ is absent.}
\end{figure}
\begin{figure}[h]
\centering
\begin{tikzpicture}[every node/.style={circle, draw, fill=black, inner sep=2pt, minimum size=5pt}]
\node[label=below:{$v_0$}] (v0) at (0,0) {};
\node[label=above:{$a_{03}$}] (a03) at (18:0.9) {};
\node[label=left:{$a_1$}] (a1) at (162:1.8) {};
\node[label=left:{$a_2$}] (a2) at (234:1.8) {};
\node[label=right:{$v_1$}] (v1) at (18:1.8) {};
\draw (a03) -- (a1) -- (a2) -- (a03);
\draw (a03) -- (v1);
\draw (v0) -- (a03);
\draw (v0) -- (a1);
\draw (v0) -- (a2);
\draw[dotted] (a03) -- ++(75:0.7) (a03) -- ++(105:0.7);
\draw[dotted] (a03) -- ++(255:0.7) (a03) -- ++(285:0.7);
\draw[dotted] (a1) -- ++(147:0.7) (a1) -- ++(177:0.7);
\draw[dotted] (a2) -- ++(219:0.7) (a2) -- ++(249:0.7);
\draw[dotted] (v1) -- ++(3:0.7) (v1) -- ++(33:0.7);
\end{tikzpicture}
\caption{The neighborhood of $v_0$ and $v_1$ in $G'$: $G^*$ with $a_0$ and $a_3$ merged into $a_{03}$.}
\end{figure}
\begin{lemma}
If $G$ is a planar graph with minimum degree 5, then $G$ contains two non-adjacent vertices of degree exactly 5.
Let $\Phi$ be a proper 4-coloring of $G*$ induced by a proper 4-coloring of $G'$, let $\alpha = \Phi(v_0) = \Phi(v_1)$, and let $\beta = \Phi(a_{0}) = \Phi(a_{3})$. Then for each of the two remaining colors $\gamma$ and $\delta$, there exists a Kempe chain in $G'$ connecting $v_0$ to $v_1$ using colors $\{\alpha, \gamma\}$ and $\{\alpha, \delta\}$ respectively.
\end{lemma}
\begin{proof}
By the previous lemma, $G$ contains at least 12 vertices of degree exactly 5. Let $S$ denote the set of all degree-5 vertices, so $|S| \geq 12$. Suppose for contradiction that every two vertices in $S$ are adjacent, i.e., $S$ induces a clique in $G$. Then every vertex $v \in S$ is adjacent to all other $|S| - 1 \geq 11$ vertices of $S$. But $\deg(v) = 5$, so $v$ has at most 5 neighbors in total. Since $11 > 5$, this is a contradiction. Therefore $S$ does not form a clique, and there exist two vertices $v_1, v_2 \in S$ that are non-adjacent.
We show $v_0$ and $v_1$ lie in the same $\{\alpha, \gamma\}$-Kempe chain; the argument for $\{\alpha, \delta\}$ is identical.
By the previous lemma, $\Phi$ must exist. Suppose for contradiction that $v_0$ and $v_1$ lie in \emph{different} $\{\alpha, \gamma\}$-Kempe chains. Swapping colors $\alpha$ and $\gamma$ throughout $v_0$'s chain yields a new proper 4-coloring $\Phi'$ of $G*$ in which $\Phi'(v_0) = \gamma$ while $\Phi'(v_1) = \alpha$. Since $\gamma \neq \alpha$, the coloring $\Phi'$ assigns $v_0$ and $v_1$ distinct colors, which allows us to properly color $G$. Therefore $v_0$ and $v_1$ lie in the same $\{\alpha, \gamma\}$-Kempe chain, and by the same argument they lie in the same $\{\alpha, \delta\}$-Kempe chain.
\end{proof}
\subsection{The Reduced Subgraph $G'$}
By the previous lemma, we may fix two non-adjacent vertices $v_0, v_1 \in G$ each of degree 5. Define
\[
G' = G - \{v_0, v_1\},
\]
as the subgraph of $G$ obtained by deleting $v_0$ and $v_1$ together with all edges incident to either vertex. Since $G'$ has strictly fewer vertices than $G$, the minimality of $G$ guarantees that $G'$ admits a proper 4-coloring $\phi : V(G') \to \{1,2,3,4\}$. Because $v_0$ and $v_1$ are non-adjacent in $G$, neither is a neighbor of the other, so each retains all 5 of its neighbors in $G'$. Let $N(v_i)$ denote the set of neighbors of $v_i$ in $G$; then $N(v_0)$ and $N(v_1)$ are each sets of 5 vertices, all of which lie in $G'$ and are assigned colors by $\phi$. Note also that $N(v_0)$ and $N(v_1)$ may overlap. Let $\Phi$ be the set of all possible 4-colorings of $G'$.
\begin{figure}[h]
\centering
\begin{tikzpicture}
\coordinate (v0pos) at (0,0);
\coordinate (a03pos) at (18:0.9);
\coordinate (a1pos) at (162:1.8);
\coordinate (a2pos) at (234:1.8);
\coordinate (v1pos) at (18:1.8);
% Kempe chain highlights beneath nodes
\draw[line width=4pt, green!50, dashed] (a1pos) to[bend left=40] (v1pos);
\draw[line width=4pt, green!80, dashed] (a1pos) to (v0pos);
\draw[line width=4pt, yellow!80, dashed] (a2pos) to[bend right=40] (v1pos);
\draw[line width=4pt, yellow!80, dashed] (a2pos) to (v0pos);
% Regular edges
\draw (a03pos) -- (a1pos) -- (a2pos) -- (a03pos);
\draw (a03pos) -- (v1pos);
\draw (v0pos) -- (a03pos);
\draw (v0pos) -- (a1pos);
\draw (v0pos) -- (a2pos);
% Dotted lines
\draw[dotted] (a03pos) -- ++(75:0.7) (a03pos) -- ++(105:0.7);
\draw[dotted] (a03pos) -- ++(255:0.7) (a03pos) -- ++(285:0.7);
\draw[dotted] (a1pos) -- ++(147:0.7) (a1pos) -- ++(177:0.7);
\draw[dotted] (a2pos) -- ++(219:0.7) (a2pos) -- ++(249:0.7);
\draw[dotted] (v1pos) -- ++(3:0.7) (v1pos) -- ++(33:0.7);
% Colored nodes
\node[circle, draw, fill=red!60, inner sep=2pt, minimum size=10pt, label=below:{$v_0$}] at (v0pos) {};
\node[circle, draw, fill=blue!60, inner sep=2pt, minimum size=10pt, label=above:{$a_{03}$}] at (a03pos) {};
\node[circle, draw, fill=green!60, inner sep=2pt, minimum size=10pt, label=left:{$a_1$}] at (a1pos) {};
\node[circle, draw, fill=yellow!80, inner sep=2pt, minimum size=10pt, label=left:{$a_2$}] at (a2pos) {};
\node[circle, draw, fill=red!60, inner sep=2pt, minimum size=10pt, label=right:{$v_1$}] at (v1pos) {};
\end{tikzpicture}
\caption{A proper 4-coloring of $G'$ with $\phi(v_0)=\phi(v_1)=\text{red}$, $\phi(a_{03})=\text{blue}$, $\phi(a_1)=\text{green}$, $\phi(a_2)=\text{yellow}$. By the previous lemma, $v_0$ and $v_1$ must lie in both the $\{\text{red},\text{green}\}$-Kempe chain (dashed green, passing through $a_1$) and the $\{\text{red},\text{yellow}\}$-Kempe chain (dashed yellow, passing through $a_2$).}
\end{figure}
\subsection{Not All Colorings in $\Phi$ Saturate Both Neighborhoods}
\begin{figure}[h]
\centering
\begin{tikzpicture}[every node/.style={circle, draw, inner sep=2pt, minimum size=10pt}]
\coordinate (v0pos) at (0,0);
\coordinate (a0pos) at (90:1.8);
\coordinate (a1pos) at (162:1.8);
\coordinate (a2pos) at (234:1.8);
\coordinate (a3pos) at (306:1.8);
\coordinate (v1pos) at (18:1.8);
\coordinate (green_mid) at (0, 2.4);
\coordinate (yellow_mid) at (306:2.3);
% Kempe chain highlights beneath nodes
\draw[line width=4pt, green!50, dashed] (a1pos) to[curve through={(green_mid)}] (v1pos);
\draw[line width=4pt, green!80, dashed] (a1pos) to (v0pos);
\draw[line width=4pt, yellow!80, dashed] (a2pos) to[curve through={(yellow_mid)}] (v1pos);
\draw[line width=4pt, yellow!80, dashed] (a2pos) to (v0pos);
\node[fill=red!60, label=below:{$v_0$}] (v0) at (v0pos) {};
\node[fill=blue!60, label=above:{$a_0$}] (a0) at (a0pos) {};
\node[fill=green!60, label=left:{$a_1$}] (a1) at (a1pos) {};
\node[fill=yellow!80, label=left:{$a_2$}] (a2) at (a2pos) {};
\node[fill=blue!60, label=right:{$a_3$}] (a3) at (a3pos) {};
\node[fill=red!60, label=right:{$v_1$}] (v1) at (v1pos) {};
\draw (a0) -- (a1) -- (a2) -- (a3) -- (v1) -- (a0);
\draw (v0) -- (a0);
\draw (v0) -- (a1);
\draw (v0) -- (a2);
\draw (v0) -- (a3);
\draw[dotted] (a0) -- ++(75:0.7) (a0) -- ++(105:0.7);
\draw[dotted] (a1) -- ++(147:0.7) (a1) -- ++(177:0.7);
\draw[dotted] (a2) -- ++(219:0.7) (a2) -- ++(249:0.7);
\draw[dotted] (a3) -- ++(291:0.7) (a3) -- ++(321:0.7);
\draw[dotted] (v1) -- ++(3:0.7) (v1) -- ++(33:0.7);
\end{tikzpicture}
\caption{The same coloring lifted to $G^*$: $\phi(v_0)=\phi(v_1)=\text{red}$, $\phi(a_0)=\phi(a_3)=\text{blue}$, $\phi(a_1)=\text{green}$, $\phi(a_2)=\text{yellow}$. The $\{\text{red},\text{green}\}$- and $\{\text{red},\text{yellow}\}$-Kempe chains connecting $v_0$ to $v_1$ pass through $a_1$ and $a_2$ respectively.}
\end{figure}
\begin{lemma}
It is not possible that every $\phi \in \Phi$ satisfies both $|\phi(N(v_0))| = 4$ and $|\phi(N(v_1))| = 4$.
Let $\Phi$ be a proper 4-coloring of $G*$ induced by a proper 4-coloring of $G'$, let $\alpha = \Phi(v_0) = \Phi(v_1)$, let $\beta = \Phi(a_{0}) = \Phi(a_{3})$, let $\delta = \Phi(a_{2})$, and let $\gamma = \Phi(a_{1})$. Then the coloring $\Phi'$ obtained by doing a Kempe switch on the $\{\beta, \delta\}$-Kempe chain incident to $a_0$ must assign $\alpha = \Phi'(v_0) = \Phi'(v_1)$, $\beta = \Phi'(a_3)$, $\delta = \Phi'(a_0) = \Phi'(a_2)$, and $\gamma = \Phi'(a_1)$ (IE the coloring for all vertices except $a_0$ remains the same)
\end{lemma}
\begin{proof}
We show the stronger fact that there exists $\phi \in \Phi$ with $|\phi(N(v_0))| \leq 3$.
Since $v_1$ is a single vertex, $G - \{v_1\}$ is a planar graph on $V - 1$ vertices. Because $V - 1 < V$ and $G$ is a minimal counterexample, $G - \{v_1\}$ is 4-colorable. Fix a proper 4-coloring $\psi$ of $G - \{v_1\}$. This coloring assigns a color to every vertex of $G'$ and also to $v_0$.
Let $\phi = \psi|_{V(G')}$ be the restriction of $\psi$ to $G'$. Since $\psi$ is a proper coloring of $G - \{v_1\} \supseteq G'$, the restriction $\phi$ is a proper 4-coloring of $G'$, so $\phi \in \Phi$.
Because $\psi$ is a proper coloring of $G - \{v_1\}$ and $v_0 \in V(G - \{v_1\})$, the color $\psi(v_0)$ differs from the color of every neighbor of $v_0$. In particular, $\psi(v_0) \notin \phi(N(v_0))$. Since $\psi(v_0) \in \{1,2,3,4\}$ and $\psi(v_0)$ does not appear in $\phi(N(v_0))$, it follows that $|\phi(N(v_0))| \leq 3$. Therefore not every coloring in $\Phi$ uses all 4 colors on $N(v_0)$, and in particular it is impossible that every $\phi \in \Phi$ saturates both $N(v_0)$ and $N(v_1)$ with all 4 colors.
The kempe switch will assign $\Phi'(a_0) = \delta$ by definition. By the previous lemma, there must be a $\{\alpha, \gamma\}$ kempe chain incident to $v_0$, $v_1$, and $a_1$. Therefore the $\{\beta, \delta\}$ component incident to $a_2$ and $a_3$ cannot be connected to the $\{\beta, \delta\}$ component incident to $a_0$, so the kempe switch we do to obtain $\Phi'$ will not recolor vertices $a_2$ and $a_3$. Since the $\{\beta, \delta\}$ kempe switch also maintains vertices assigned $\alpha$ and $\gamma$ in $\Phi$, $\Phi'$ will not recolor vertices $a_1$, $v_0$, and $v_1$. Therefore $\alpha = \Phi'(v_0) = \Phi'(v_1)$, $\beta = \Phi'(a_3)$, $\delta = \Phi'(a_0) = \Phi'(a_2)$, and $\gamma = \Phi'(a_1)$.
\end{proof}
\section{Merged Subgraphs of $G'$}
\begin{figure}[h]
\centering
\begin{tikzpicture}[every node/.style={circle, draw, inner sep=2pt, minimum size=10pt}]
\coordinate (v0pos) at (0,0);
\coordinate (a0pos) at (90:1.8);
\coordinate (a1pos) at (162:1.8);
\coordinate (a2pos) at (234:1.8);
\coordinate (a3pos) at (306:1.8);
\coordinate (v1pos) at (18:1.8);
Let $N(v_0) = \{a_0, a_1, a_2, a_3, a_4\}$ be the five neighbors of $v_0$ in $G'$. For each pair of non-adjacent vertices $a_i, a_j \in N(v_0)$ (i.e., $\{a_i, a_j\} \notin E(G')$), define the graph $G'_{ij}$ to be the result of identifying $a_i$ and $a_j$ in $G'$: formally, $G'_{ij}$ is obtained from $G'$ by removing $a_i$ and $a_j$ and adding a new vertex $a_{ij}$ adjacent to every vertex in $N_{G'}(a_i) \cup N_{G'}(a_j)$. Define the set of all such merged subgraphs by
\[
\mathcal{M} = \bigl\{\, G'_{ij} : a_i, a_j \in N(v_0),\ \{a_i, a_j\} \notin E(G') \,\bigr\}.
\]
\coordinate (green_mid) at (0, 2.4);
\coordinate (yellow_mid) at (306:2.3);
% Kempe chain highlights beneath nodes
\draw[line width=4pt, green!50, dashed] (a1pos) to[curve through={(green_mid)}] (v1pos);
\draw[line width=4pt, green!80, dashed] (a1pos) to (v0pos);
% \draw[line width=4pt, yellow!80, dashed] (a2pos) to[curve through={(yellow_mid)}] (v1pos);
% \draw[line width=4pt, yellow!80, dashed] (a2pos) to (v0pos);
\subsection{Colorings of Merged Subgraphs Extend to $G'$}
\node[fill=red!60, label=below:{$v_0$}] (v0) at (v0pos) {};
\node[fill=yellow!60, label=above:{$a_0$}] (a0) at (a0pos) {};
\node[fill=green!60, label=left:{$a_1$}] (a1) at (a1pos) {};
\node[fill=yellow!80, label=left:{$a_2$}] (a2) at (a2pos) {};
\node[fill=blue!60, label=right:{$a_3$}] (a3) at (a3pos) {};
\node[fill=red!60, label=right:{$v_1$}] (v1) at (v1pos) {};
For each $G'_{ij} \in \mathcal{M}$, let $\Phi(G'_{ij})$ denote the set of proper 4-colorings of $G'_{ij}$. Each such coloring can be extended to a coloring of $G'$ by assigning the color of the merged vertex $a_{ij}$ back to both $a_i$ and $a_j$. Formally, for $\psi \in \Phi(G'_{ij})$, define $\widetilde{\psi} : V(G') \to \{1,2,3,4\}$ by
\[
\widetilde{\psi}(v) = \begin{cases} \psi(a_{ij}) & \text{if } v = a_i \text{ or } v = a_j, \\ \psi(v) & \text{otherwise.} \end{cases}
\]
Let $\widetilde{\Phi}_{ij} = \{\,\widetilde{\psi} : \psi \in \Phi(G'_{ij})\,\}$ denote the set of all such extensions.
\begin{lemma}
For each $G'_{ij} \in \mathcal{M}$, we have $\widetilde{\Phi}_{ij} \subseteq \Phi$.
\end{lemma}
\begin{proof}
Let $\psi \in \Phi(G'_{ij})$ and let $\widetilde{\psi}$ be its extension to $G'$. We verify that $\widetilde{\psi}$ is a proper coloring of $G'$ by checking every edge $\{u, w\} \in E(G')$.
\textbf{Case 1: neither $u$ nor $w$ is $a_i$ or $a_j$.} Then $\{u, w\}$ is also an edge of $G'_{ij}$, and $\widetilde{\psi}(u) = \psi(u) \neq \psi(w) = \widetilde{\psi}(w)$.
\textbf{Case 2: $u \in \{a_i, a_j\}$ and $w \notin \{a_i, a_j\}$ (or vice versa).} Then $w \in N_{G'}(a_i) \cup N_{G'}(a_j)$, so $w$ is adjacent to $a_{ij}$ in $G'_{ij}$. Since $\psi$ is proper, $\psi(w) \neq \psi(a_{ij}) = \widetilde{\psi}(u)$.
\textbf{Case 3: $u = a_i$ and $w = a_j$.} This case cannot occur, since $a_i$ and $a_j$ are non-adjacent in $G'$ by the definition of $\mathcal{M}$.
In all cases the endpoints of every edge receive distinct colors, so $\widetilde{\psi} \in \Phi$.
\end{proof}
\draw (a0) -- (a1) -- (a2) -- (a3) -- (v1) -- (a0);
\draw (v0) -- (a0);
\draw (v0) -- (a1);
\draw (v0) -- (a2);
\draw (v0) -- (a3);
\draw[dotted] (a0) -- ++(75:0.7) (a0) -- ++(105:0.7);
\draw[dotted] (a1) -- ++(147:0.7) (a1) -- ++(177:0.7);
\draw[dotted] (a2) -- ++(219:0.7) (a2) -- ++(249:0.7);
\draw[dotted] (a3) -- ++(291:0.7) (a3) -- ++(321:0.7);
\draw[dotted] (v1) -- ++(3:0.7) (v1) -- ++(33:0.7);
\end{tikzpicture}
\end{figure}
\end{document}