I've been reading up on Brooks' Theorem as stated in Graphs and Digraphs by Chartrand et al. As shown in the above snippet. What has stumped me is the highlighted sentence. Since $H$ is $k$-critical, it implies that it is $2$-connected, then how does it happen that we consider the case that $H$ is $3$-connected?
What confuses me is that since $H$ is $2$-connected, then the deletion of $2$ or more vertices from $H$ makes $H$ to be disconnected or a trivial graph. However, considering the case that $H$ is $3$-connected, then the deletion of fewer than $3$ vertices does not cause $H$ to be disconnected but does that not contradict our assumption?
I really need clarification on this! Thank you in advance.
A $2$-connected graph is a graph $G$ with $\kappa(G) \ge2$. A $3$-connected graph is a graph $G$ with $\kappa(G) \ge 3$. So it's possible that a $2$-connected graph is also $3$-connected.
To phrase the argument in the proof in terms of the connectivity $\kappa$: we start out knowing that $\kappa(H) \ge 2$, because $H$ is a $\chi$-critical graph. So there are two cases to consider: either $\kappa(H) = 2$, or else $\kappa(H) \ge 3$.
We know that $\kappa(H) \ge 2$, because we have excluded the case where $H$ has a cut vertex $v$. So here, we are doing casework based on whether $H$ has a cut $\{v,w\}$ for some two vertices $v$ and $w$.