# Elementary question on Graph Theory, Edge Contraction and Separability [closed]

Let $$G = (V,E)$$ be a 2-connected, simple graph. Apparently, if $$G/e$$ is separable then there exist two 2-connected graphs $$G_1$$ and $$G_2$$ such that $$G_1 \cup G_2 = G$$ and $$G_1 \cap G_2 = K_2$$. I wonder does the contracted edge in $$G$$ (and the resulting vertex $$G/e$$) have to be the vertex of articulation of $$G/e$$? It's hard to visualise how this works out.

• What is $K_2$?? Dec 20 '19 at 22:30
• The complete graph with 2 vertices. Dec 20 '19 at 23:29
• The title mentions edge contraction, so presumably $G/e$ means the graph obtained from $G$ by removing edge $e$ and identifying the two vertices it previously connected. If I understand the meaning of $G/e$ separable, it should be connected but not $2$-connected. Does the statement beginning "Apparently..." ask for confirmation of the claim, or does it serve as additional assumptions? Dec 21 '19 at 19:55
• That is true - $G/e$ being separable means that it is 1-connected but not 2-connected (that is the definition of seperable). I used the word apprently as I have not seen the proof yet so it is a claim is true (made in a research paper) but for me to say it is true, will have to see how, which is why I am here :) Dec 21 '19 at 22:18

• In particular, this example shows that there do not necessarily exist two 2-connected graphs $G_1$ and $G_2$ as in the question. Dec 20 '19 at 18:45
• Actually, I must apologise that there was a lemma that also implied that given some constraints are satisfied, the Graph $G$ is also two-connected. Just made the edit. Dec 20 '19 at 20:42