Show that a k-critical graph is connected. Furthermore, show that it does not have a vertex whose removal disconnects the graph (such a vertex is known as a cut vertex).
I have managed to proove , I think, the first part Let's assume G is not connected. Since χ(G) = k, (If G1,G2,...,Gr are the components of a disconnected graph G, then χ(G) = max χ(Gi) 1≤i≤r ) then there is a component G1 of G such that χ(G1)=k.If v is any vertex of G which is not in G1,then G1 isa component of the subgraph G − v. Therefore, χ (G − v) = χ (G1 ) = k. This contradicts the fact that G is k-critical. Hence G is connected.
But I can't manage to prove the second part..ie that there is not cut vertex. Anyone can help?