From Graph Theory with Applications by Bondy and Murty:
7.1.2. A graph $G$ is $\alpha$-critical if $\alpha(G-e)>\alpha(G)$ for any edge $e$ in $G$. Show that if $G$ is $\alpha$-critical, then $G$ has no cut vertex.
Here, $\alpha(G)$ is the number of vertices in a maximum independent set of $G$.
I need to show that any two edges in $G$ lie on a common cycle.
It's not hard to find that this implies that $\beta(G-e)<\beta(G)$ for any edge $e$, but I am not sure if that is useful, where $\beta(G)$ is the number of vertices in a minimal covering of $G$.