Substituting two $\alpha$-narrow graphs yields another $\alpha$-narrow graph

Let $g$ be a good function on $G$. Let ${\mathcal{P}}_2$ be the family of perfect induced subgraphs of $H_2$, and let $K={\max }_{P\in {\mathcal{P}}_2}{\sum }_{x\in V(P)}g(x)$. If $K=0$, then every one-vertex induced subgraph of $H_2$ has weight $0$, so $g$ vanishes on $V(H_2)$. Choose any vertex $x\in V(H_2)$, define $g_1$ on $H_1$ by copying $g$ outside $v$ and setting $g_1(v)=0$, and note from [F2] that every perfect induced subgraph of $H_1$ corresponds either to the same perfect induced subgraph of $G$ or to one obtained by replacing $v$ with $x$. Hence $g_1$ is good on $H_1$, so [F1] gives ${\sum }_{y\in V(H_1)}{g}_1(y{)}^{\alpha }\le 1$. Because $g$ vanishes on $H_2$, this is exactly ${\sum }_{z\in V(G)}g(z{)}^{\alpha }\le 1$.

Assume now that $K>0$. Define $g_1$ on $H_1$ by copying $g$ outside $v$ and setting $g_1(v)=K$. If $Q$ is a perfect induced subgraph of $H_1$ not containing $v$, then it appears unchanged in $G$ and has total $g_1$-weight at most $1$. If $v\in V(Q)$, choose $P\in {\mathcal{P}}_2$ with ${\sum }_{x\in V(P)}g(x)=K$; then [L1] makes the substitution $Q[v\to P]$ a perfect induced subgraph of $G$, so ${\sum }_{y\in V(Q)}{g}_1(y)={\sum }_{y\in V(Q)\setminus \{v\}}g(y)+K={\sum }_{z\in V(Q[v\to P])}g(z)\le 1$. Thus $g_1$ is good on $H_1$. Likewise $g_2:=g\mathrm{/}K$ is good on $H_2$ by the definition of $K$. Applying [F1] to $g_1$ and $g_2$ gives ${\sum }_{y\in V(H_1)\setminus \{v\}}g(y{)}^{\alpha }+K^{\alpha }\le 1$ and ${\sum }_{x\in V(H_2)}g(x{)}^{\alpha }\le K^{\alpha }$.

In the case $K>0$, step 1.2 yields ${\sum }_{z\in V(G)}g(z{)}^{\alpha }={\sum }_{y\in V(H_1)\setminus \{v\}}g(y{)}^{\alpha }+{\sum }_{x\in V(H_2)}g(x{)}^{\alpha }\le 1$. Together with step 1.1, this proves that every good function on $G$ has $\alpha$-power sum at most $1$. Therefore [F1] shows that $G$ is $\alpha$-narrow.