Let $ G$ be a bridgeless cubic (simple) graph, and let $ M$ be a perfect matching in $ G$ . $ G-M$ will necessarily be a set of circuits. For example, if we delete a perfect matching from $ K_{3,3}$ we end up with one circuit. If we delete a perfect matching from the Petersen graph we end up with two circuits. And in general, one could end up with many. The following question comes to mind.

QuestionLet $ G$ be a cubic bridgeless (simple) graph which is cyclically edge-5-connected, and let $ M$ be a perfect matching in $ G$ . Is there a subset $ K$ of edges in $ M$ such that $ G-K$ has exactly two components, such that either none of the two components are circuits, or both are circuits?

For example, any perfect matching in the Petersen graph is an example of such an edge cut.

A graph $ G$ is *cyclically edge-5-connected* if no set of fewer than 5 edges is cycle separating. A set $ K$ of edges is *cycle separating* if $ G-K$ is disconnected and at least two of its components contain circuits.