On the Kuratowski graph planarity criterion

📝 Abstract

This paper is purely expositional. The statement of the Kuratowski graph planarity criterion is simple and well-known. However, its classical proof is not easy. In this paper we present the Makarychev proof (with further simplifications by Prasolov, Telishev, Zaslavski and the author) which is possibly the simplest. In the Rusian version before the proof we present all the necessary definitions, and afterwards we state some close results on graphs and more general spaces. The paper is accessible for students familiar with the notion of a graph, and could be an interesting easy reading for mature mathematicians.

💡 Analysis

This paper is purely expositional. The statement of the Kuratowski graph planarity criterion is simple and well-known. However, its classical proof is not easy. In this paper we present the Makarychev proof (with further simplifications by Prasolov, Telishev, Zaslavski and the author) which is possibly the simplest. In the Rusian version before the proof we present all the necessary definitions, and afterwards we state some close results on graphs and more general spaces. The paper is accessible for students familiar with the notion of a graph, and could be an interesting easy reading for mature mathematicians.

📄 Content

arXiv:0802.3820v3 [math.GT] 18 Jul 2012 A SHORT PROOF OF THE KURATOWSKI GRAPH PLANARITY CRITERION 1 A. Skopenkov 2 Abstract. This paper is purely expositional. The statement of the Kuratowski graph planarity criterion is simple and well-known. However, its classical proof is not easy. In this paper we present the Makarychev proof (with further simplifications by Prasolov, Telishev, Zaslavski and the author) which is possibly the simplest. The paper is accessible for students familiar with the notion of a graph, and could be an interesting easy reading for mature mathematicians. A graph is called planar if it can be drawn in the plane without self-intersections. The Kuratowski Theorem. A graph is planar if and only if it does not contain a subgraph homeomorphic to K5 or to K3,3 (fig. 1). K5 K3,3 Figure 1: The Kuratowski graphs For definition of homeomorphic graphs, as well as for a short proof of the ‘only if’ part of the Kuratowski theorem see [Pr04]. For results related to this theorem see either the Russian version of this text or [Cl34, Cl37, Ep66, GHV79, HJ64, Ku00, MS67, MA41, RS90, RS99, Sa91, Sk95, Sk08, Sk, Wh33]. Here we present a simple proof of the ‘if’ part of the Kuratowski Theorem based on [Ma97], cf. [Th81], with further simplifications by Prasolov, Telishev, Zaslavski and the author. Deletion of an edge: G −e, contraction of an edge: G/e, and deletion of a vertex: G −x Clearly, it suffices to prove the Kuratowski Theorem for graphs without loops and multiple edges. So we consider only such graphs. By contraction of an edge we would understand contraction of an edge with replacement of each obtained edge of multiplicity greater than 1 by an edge of multiplicity 1. 1This note is based on the author’s lectures at the Kirov Region Summer School, St Petersburg Summer School, the Moscow Olympic School, mathematical circles at Kolmogorov College and at Moscow Center for Continuous Mathematical Education. I would like to acknowledge B. Mohar, D. Permyakov, V. Volkov, M. Vyalyi and T. Shaihieva for useful discussions. 2http://www.mccme.ru/˜ skopenkov. Supported by Simons-IUM Foundation. 1 We prove the ‘if’ part of the Kuratowski Theorem in the following equivalent form. Proposition. If a connected graph G is not isomorphic to K5 or to K3,3, and for each edge e of G both graphs G −e and G/e are planar, then G is planar. x y x y x y Figure 2: ’Uncontraction of an edge’ in the Kuratowski graphs Proof that Proposition implies the ‘if’ part of the Kuratowski Theorem. The statement ‘graph G contains a subgraph homeomorphic to the graph H’ will be abbreviated to ’G ⊃H’. The ‘if part’ of the Kuratowski Theorem is proved by induction on the number of edges in the graph. By Proposition the inductive step follows because G ⊃K5 or G ⊃K3,3 if either G −e ⊃K5 or G −e ⊃K3,3 or G/e ⊃K5 or G/e ⊃K3,3 for some edge e of G. ‘For G −e’ the italicized assertion is obvious. If G/e ⊃K3,3 then G ⊃K3,3, and if G/e ⊃K5 then G ⊃K5 or G ⊃K3,3 (fig. 2). QED Lemma on the Kuratowski Graphs. For each graph G the following three conditions are equivalent: (1) For each edge xy of G the graph G −x −y does not contain θ-graphs, and from each vertex of G −x −y at least two edges are issuing. (2) For each edge xy of the graph G the graph G −x −y is a cycle (containing n ≥3 vertices). (3) G is isomorphic either to K5 or to K3,3. The implications (3) ⇒(2) ⇒(1) in the Lemma on the Kuratowski Graphs are clear and are not used in the proof of the Kuratowski theorem. Figure 3: A ’tree’ of cycles Proof of the implication (1) ⇒(2) in the Lemma on the Kuratowski Graphs. Condition (1) implies that K −x −y is a disjoint union of ‘trees’ whose ‘vertices’ are cycles. (fig. 3; formally, each block of K is a cycle). Therefore K −x −y contains a ‘hanging’ cycle, i.e. a cycle C having only one common vertex v with the remaining graph. This cycle C has at most two other vertices p and q. Since K −x −y does not contain isolated vertices, from each vertex of 2 K there issue at least three edges. Hence each of these vertices p and q is joined either with x or with y. Therefore in the union of C and the edges of K joining vertices x, y, p, q we can find a θ-subgraph. Hence by (1) each edge of K −x −y has an end on C. Since by (1) the graph K −x −y does not contain hanging vertices, this graph coincides with C. QED Proof of the implication (2) ⇒(3) in the Lemma on the Kuratowski Graphs. If n = 3 then for each two vertices b and c of the cycle K −x −y the graph K −b −c is a cycle, hence the remaining vertex of the cycle K −x −y is joined (by an edge of K) both to x and to y. Hence K = K5. If n ≥4, then take any four consecutive vertices a, b, c, d of the cycle K −x −y. Since K −b −c is a cycle, in the graph K one of the vertices a or d is joined (by an edge) to x (and not joined to y), the other is joined to y (and not joined to x), whereas the vertices of the cycle K −x −y different from a, b, c, d (which do not exist when n = 4) are not joined neither to x nor to y. F

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