Every basic bull-free graph is 2-narrow

We argue by induction on $\mathrm{\mid }V(G)\mathrm{\mid }$. Let $g$ be a good function on $G$. If $\mathrm{\mid }V(G)\mathrm{\mid }\le 1$, then ${\sum }_{v}g(v{)}^2\le 1$ because every one-vertex graph is perfect and therefore the good-function condition already gives $g(v)\le 1$. Assume now $\mathrm{\mid }V(G)\mathrm{\mid }\ge 2$, and choose $u\in V(G)$ with $g(u)$ maximal. Since every two-vertex induced subgraph is perfect, the good-function inequality implies $g(u)+g(v)\le 1$ for every $v\mathrm{\ne }u$, so if $g(u)=1$ then all other weights are $0$ and the desired inequality is immediate. Thus we may assume $g(u)<1$. By [L1], [L2], and [L3], after replacing $G$ by its complement if necessary we may assume that $G[N(u)]$ is perfect; this replacement preserves basicness, good functions, and the two-narrow inequality. Put $N=N(u)$ and $M=V(G)\setminus N[u]$. Any composite witness inside $G[M]$ would also be a composite witness inside $G$, so $G[M]$ is basic; by induction it is two-narrow.

For every perfect induced subgraph $P$ of $G[M]$, the graph $G[P\cup \{u\}]$ is perfect because $u$ is anticomplete to $P$ and adjoining an isolated vertex preserves the equalities $\chi =\omega$ on every induced subgraph. Hence the function $f(v)=g(v)\mathrm{/}(1-g(u))$ on $M$ is good on $G[M]$, so induction and [F1] give ${\sum }_{v\in M}g(v{)}^2\le (1-g(u){)}^2$. Also $G[N\cup \{u\}]$ is perfect because $u$ is complete to $N$ and adjoining a universal vertex raises both $\chi$ and $\omega$ by $1$. Thus the good-function inequality gives ${\sum }_{v\in N}g(v)\le 1-g(u)$. Since $g(u)$ is maximal, $g(v{)}^2\le g(u)g(v)$ for every $v\in N$, and therefore ${\sum }_{v\in N}g(v{)}^2\le g(u)(1-g(u))$.

Combining the contributions of $u$, $M$, and $N$ gives ${\sum }_{v\in V(G)}g(v{)}^2\le g(u{)}^2+(1-g(u){)}^2+g(u)(1-g(u))=1-g(u)+g(u{)}^2\le 1$. Since $g$ was an arbitrary good function, [F1] shows that $G$ is two-narrow.