Why no cut-vertices or cut edges in a graph where eccentricity is same for all vertices

Question

I need help to prove the following statement.

There are no cut-vertices or cut-edges(bridges) in a graph where eccentricity is same for all vertices. I am getting that if the graph contains a cut-edge, then the eccentricity of end vertices of the bridge may contradict the fact that all vertices have same eccentricity. How to prove it theoretically. Thanks for help

Answer 1

You can have a cut-edge if the eccentricity of each vertex is $1$: o---o.

Let $e>1$ be the common eccentricity of the vertices of $G$, and suppose that $\{u,v\}$ is a cut-edge. Let $w$ be a vertex of $G$ such that $d(u,w)=e$. Let $G'$ be the graph that remains when the edge $\{u,v\}$ is removed. If $w$ is in the same component of $G'$ as $u$, then any path from $v$ to $w$ must pass through $u$, and $d(v,w)=1+d(u,w)=1+e$, contradicting the hypothesis that the eccentricity of $v$ is $e$. Thus, $w$ is in the same component of $G'$ as $v$, and $d(v,w)=d(u,w)-1=e-1$. Similarly, if $z$ is a vertex such that $d(v,z)=e$, then $z$ is in the same component of $G'$ as $u$, and $d(u,z)=e-1$. What does this tell you about $d(w,z)$?