I need help understanding the solution to this problem. This problem has been answered here, however, my doubt is not addressed.
Problem: Let $G$ be a connected Eulerian graph with at least $3$ vertices. A vertex $′v′$ in $G$ is extendible if every trail beginning at $′v′$ can be extended to form an Eulerian Circuit.
Prove following statement: A vertex $v\in V(G)$ is extendible if and only if $G-v$ is a forest.
Solution :
Necessity: We prove the contrapositive. If $G − v$ is not a forest, then $G − v$ has a cycle $C$ . In $G − E(C)$ , every vertex has even degree, so the component of $G − E(C)$ containing $v$ has an Eulerian circuit. This circuit starts and ends at $v$ and exhausts all edges of $G$ incident to $v$, so it cannot be extended to reach $C$ and complete an Eulerian circuit of $G$.
Sufficiency: If $G −v$ is a forest, then every cycle of $G$ contains $v$ . Given a trail $T$ starting at $v$, extend it arbitrarily at the end until it can be extended no farther. Because every vertex has even degree, the process can end only at $v$. The resulting closed trail $T'$ must use every edge incident to $v$, else it could extend farther. Since $T'$ is closed, every vertex in $G − E(T' )$ has even degree. If $G − E(T)$ has any edges, then minimum degree at least two in a component of $G − E(T)$ yields a cycle in $G − E(T')$; this cycle avoids $v$, since $T'$ exhausted the edges incident to $v$. Since we have assumed that $G − v$ has no cycles, we conclude that $G − E(T')$ has no edges, so $T'$ is an Eulerian circuit that extends $T$.
Please explain the necessity part, especially the highlighted part.