I'm reading the following rotation lemma on graphs. There's this statement in the proof: "If $y \in V(P)$, rotate $P'$ along the edge $\{v,y\}$", which I don't see how it could happened when applied to the example I've drawn.
Rotations: Suppose $P - v_0v_1...v_t$ is a path in $G$. Suppose $\left\{v_{i}, v_{t}\right\} \in E(G), 1 \leq i \leq t-2$. Rotation along $\left\{v_{i}, v_{t}\right\}: P^{\prime}=v_{0} v_{1} \ldots v_{i} v_{t} v_{t-1} \ldots v_{i+1}$ is also a path of length $t$.
Lemma 5.3.7. Let $P=v_{0} v_{1} \ldots v_{t}$ be a longest path in a graph $G,$ and let $R$ be the set of endpoints reachable from $v_{0}$ by sequences of rotations. Then $N_{G}(R) \subseteq N_{P}(R)$.
Proof
- After rotating along $\left\{v_{i}, v_{t}\right\},$ only $v_{i}, v_{t}$ get new neighbours on the path.
- Let $v \in R$. Rotate to path $P^{\prime}$ with $v$ as an endpoint.
- Let $y \in N(v) \backslash R$
- If $y \notin V(P)$, extend $P^{\prime}$ to $y \Rightarrow$ longer path than $P$
- If $y \in V(P)$, rotate $P^{\prime}$ along the edge $\{v, y\}$
- $\Rightarrow$ a neighbour $x$ of $y$ on $P^{\prime}$ is an endpoint of the new path, so $x \in R$
- If $x$ also a neighbour of $y$ on $P$, then $y \in N_{P}(R)$
- Otherwise must have rotated along an edge incident to $y \Rightarrow y \in N_{P}(R)$
If you rotate a path along the last edge, this does nothing. You get the original path back.
In the example you drew, rotating $P'$ along the edge $\{v,y\}$ gives the path $P'$ back again, and it's still true that the endpoint of the resulting path (namely, $v$) is a neighbor of $y$.