The five-cycle is 2-narrow but not 1-narrow

Deleting any vertex of $C_5$ leaves a four-vertex path. Every induced subgraph of that path is a disjoint union of paths. If it has an edge, alternating colours on each path give chromatic number at most $2$, and the edge gives clique number at least $2$, so both numbers are $2$; if it is nonempty and has no edge, both are $1$; and if it is empty, both are $0$. Thus every vertex deletion is perfect by [F3]. Since $C_5$ has five vertices, [L1] proves that $C_5$ is two-narrow.

Define $g(v)=1\mathrm{/}4$ for every vertex of $C_5$. Every proper induced subgraph lies in some vertex deletion, so it is perfect by step 1.1. Since [L2] says the whole $C_5$ is not perfect, every perfect induced subgraph is proper and has at most four vertices. Thus [F2] makes $g$ a good function: its total on each perfect induced subgraph is at most $4\cdot (1\mathrm{/}4)=1$. But ${\sum }_{v\in V(C_5)}g(v)=5\mathrm{/}4>1$, so [F1] shows that $C_5$ is not one-narrow.

Thus $C_5$ is two-narrow by step 1.1 but not one-narrow by step 2.1.