Proving that $\kappa(G)-1 \leq \kappa(G-e) \leq \kappa(G)$

Question

$\kappa(G)$ is defined to be the minimum number of vertices that must be deleted to disconnect G. $\kappa(G-e) \leq \kappa(G)$ is true by definition since a graph with one less edges can't have more vertices that needs to be deleted in order to get a disconnected graph. However, I am struggling to see why $\kappa(G)-1 \leq \kappa(G-e)$. Would I need to divide the case into when e is a bridge and when e is not a bridge? or something similar?

Thanks.

Answer 1

Suppose that $S$ is an optimal cut set in $G - e$. Then if $G - S$ is connected, it must be that $e$ is a bridge. Then if we let $S'$ be $S$ along with one of the vertices incident to $e$, $G - S'$ is disconnected.

Technical details:

$G - e$ is not complete, so $G$ has at least $\kappa(G - e) + 2$ vertices.

Hence, $\kappa(G) \leq |S'| = \kappa(G - e) + 1$.