The C5-free graphs satisfy a polynomial kappa bound

Choose $ϵ$ with $0<ϵ<1\mathrm{/}400$. Apply [L1] with $H=C_5$ and this $ϵ$ to obtain ${\delta }_0>0$. Set $\delta :=\min \{{\delta }_0,1\}\mathrm{/}2$. Then every nonempty $C_5$-free graph has a set $X$ with $\mathrm{\mid }X\mathrm{\mid }\ge \delta \mathrm{\mid }V(G)\mathrm{\mid }$ such that one of $G[X]$ or $\bar{G}[X]$ has maximum degree at most $ϵ\delta \mathrm{\mid }V(G)\mathrm{\mid }$. Let ${\tau }_0$ be the constant from [L2] for this pair $(\delta ,ϵ)$. Because $400ϵ<1$, choose $\tau \in (0,{\tau }_0]\cap (0,1\mathrm{/}2)$ so small that $(400ϵ{)}^{2-1\mathrm{/}\tau }>400ϵ\mathrm{/}\delta$.

Suppose for contradiction that some nonempty $C_5$-free graph satisfies $\kappa (G)<\mathrm{\mid }V(G){\mathrm{\mid }}^{\tau }$. Choose such a graph of minimum order. Then [L4] makes it $\tau$-critical.

Apply the last sentence of step 1.1 to this minimal counterexample. There is a set $X\subseteq V(G)$ with $\mathrm{\mid }X\mathrm{\mid }\ge \delta \mathrm{\mid }V(G)\mathrm{\mid }$ such that one of $G[X]$ or $\bar{G}[X]$ has maximum degree at most $ϵ\delta \mathrm{\mid }V(G)\mathrm{\mid }$. If the low-degree graph is $\bar{G}[X]$, replace $G$ by its complement. This preserves the order, preserves $\kappa$ because complement swaps cliques and stable sets, preserves $\tau$-criticality because induced subgraphs and complements commute, and preserves $C_5$-freeness because $C_5\cong \overline{C_5}$. So after this replacement we may assume that $G[X]$ itself has maximum degree at most $ϵ\delta \mathrm{\mid }V(G)\mathrm{\mid }$.

Apply [L2] to the $\tau$-critical graph $G$ and the set $X$. We obtain a rooted stable-tooth comb $(v,((a_i,B_i):1\le i\le t))$ in $G[X]$ such that $t\ge 1\mathrm{/}(400ϵ)$ and $\mathrm{\mid }{B}_i\mathrm{\mid }\ge \delta \mathrm{\mid }V(G)\mathrm{\mid }\mathrm{/}(400ϵt^2)$ for each $i$. If some block $B_i$ meets another block $B_j$ by an edge, then [L3] gives an induced $C_5$ in $G$, impossible. Therefore the blocks $B_1,\dots ,B_t$ are pairwise anticomplete.

Each $B_i$ is a proper induced subgraph of $G$, so $\tau$-criticality and [L5] give $\mathrm{\mid }{B}_i{\mathrm{\mid }}^{\tau }\le \kappa (B_i)=\alpha (B_i)\omega (B_i)\le \alpha (B_i)\omega (G)$. Hence $\alpha (B_i)\ge \mathrm{\mid }{B}_i{\mathrm{\mid }}^{\tau }\mathrm{/}\omega (G)\ge {(\delta \mathrm{\mid }V(G)\mathrm{\mid }\mathrm{/}(400ϵt^2))}^{\tau }\mathrm{/}\omega (G)$.

Because the blocks are pairwise anticomplete, stable sets chosen inside different $B_i$ may be united. Thus $\alpha (G)\ge {\sum }_{i=1}^t\alpha (B_i)\ge t{(\delta \mathrm{\mid }V(G)\mathrm{\mid }\mathrm{/}(400ϵt^2))}^{\tau }\mathrm{/}\omega (G)$. Multiplying by $\omega (G)$ and using [L5], $\kappa (G)\ge t{(\delta \mathrm{\mid }V(G)\mathrm{\mid }\mathrm{/}(400ϵt^2))}^{\tau }$. Since step 1.2 assumes $\kappa (G)<\mathrm{\mid }V(G){\mathrm{\mid }}^{\tau }$, cancelling $\mathrm{\mid }V(G){\mathrm{\mid }}^{\tau }$ yields $400ϵ\mathrm{/}\delta >t^{1\mathrm{/}\tau -2}$.

Step 3.1 gives $t\ge 1\mathrm{/}(400ϵ)$, and step 1.1 has $\tau <1\mathrm{/}2$, so $1\mathrm{/}\tau -2>0$. Therefore $t^{1\mathrm{/}\tau -2}\ge (1\mathrm{/}(400ϵ){)}^{1\mathrm{/}\tau -2}=(400ϵ{)}^{2-1\mathrm{/}\tau }$. Combining with step 5.1 gives $400ϵ\mathrm{/}\delta >(400ϵ{)}^{2-1\mathrm{/}\tau }$, contrary to step 1.1. This contradiction proves that no counterexample exists, so every nonempty $C_5$-free graph satisfies $\kappa (G)\ge \mathrm{\mid }V(G){\mathrm{\mid }}^{\tau }$.