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A well known theorem of Kuratowski in 1932 states that a graph is planar if, and only if, it does not contain a subdivision of K 5 or K 3,3 . Wagner proved in 1937 that if a graph other than K 5 does not contain any subdivision of K 3,3 then it is planar or it admits a cut of size at most 2. Kelmans and, independently, Seymour conjectured in the 1970s that if a graph does not contain any subdivision of K 5 then it is planar or it admits a cut of size at most 4. In this paper, we give a proof of the Kelmans-Seymour conjecture. We also discuss several related results and problems. IntroductionFor a graph G, we use T G to denote a subdivision of G, and the vertices in T G that correspond to the vertices of G are said to be its branch vertices. Thus, T K 5 denotes a subdivision of K 5 , and the vertices in a T K 5 of degree four are its branch vertices.The well known result of Kuratowski [18] states that a graph is planar if, and only if, it does not contain T K 5 or T K 3,3 . A simple application of Euler's formula for planar graphs shows that, for n ≥ 3, if an n-vertex graph has at least 3n − 5 edges then it must be nonplanar and, hence, contains T K 5 or T K 3,3 . Dirac [5] conjectured that for n ≥ 3, if an n-vertex graph has at least 3n − 5 edges then it must contain T K 5 . This conjecture was also reported by Erdős and Hajnal [7]. Kézdy and McGuiness [15] showed that a minimal counterexample to Dirac's conjecture must be 5-connected and contains K − 4 (obtained from the complete graph K 4 by deleting an edge). After some partial results in [28,30,32,33], Dirac's conjecture was proved by Mader [22], where he also showed that every 5-connected n-vertex graph with at least 3n − 6 edges contains T K 5 or K − 4 . Seymour [25] (also see [22,33]) and, independently, Kelmans [14] conjectured that every 5-connected nonplanar graph contains T K 5 . Thus, the Kelmans-Seymour conjecture implies Mader's theorem. This conjecture is also related to several interesting problems, which we will discuss in Section 7.The authors [9-11] produced lemmas needed for proving this Kelmans-Seymour conjecture, and we are now ready to prove it in this paper. Theorem 1.1 Every 5-connected non-planar graph contains T K 5 . The starting point of our work is the following result of Ma and Yu [20, 21]: Every 5-connected nonplanar graph containing K − 4 has a T K 5 . This result, combined with the result of Kézdy and McGuiness [15] on minimal counterexamples to Dirac's conjecture, gives an alternative proof of Mader's theorem. Also using this result, Aigner-Horev [1]proved that every 5-connected nonplanar apex graph contains T K 5 . A simpler proof of Aigner-Horev's result using discharging argument was obtained by Ma, Thomas and Yu, and, independently, by Kawarabayashi, see [13].We now briefly describe the process for proving Theorem 1.1. For a more detailed version, we recommend the reader to read Section 6 first, which should also give motivation to some of the technical lemmas listed in Sections 2, 3, 4 and 5.Suppose G is a 5-connected non-...

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Seymour and, independently, Kelmans conjectured in the 1970s that every 5-connected nonplanar graph contains a subdivision of K 5 . This conjecture was proved by Ma and Yu for graphs containing K − 4 , and an important step in their proof is to deal with a 5separation in the graph with a planar side. In order to establish the Kelmans-Seymour conjecture for all graphs, we need to consider 5-separations and 6-separations with less restrictive structures. The goal of this paper is to deal with special 5-separations and 6separations, including those with an apex side. Results will be used in subsequent papers to prove the Kelmans-Seymour conjecture.Let K be a graph; we use T K to denote a subdivision of K and call the vertices of the T K corresponding to the vertices of K its branch vertices. Kuratowski's theorem states that a graph is planar iff it contains neither T K 3,3 nor T K 5 . Graphs containing no T K 3,3 can be constructed from planar graphs and copies of K 5 by pasting them along cliques of size at most two. The structure of graphs containing no T K 5 is not well understood. Kelmans [4] and, independently, Seymour [9] conjectured that 5-connected nonplanar graphs must contain T K 5 . Thus, if the Kelmans-Seymour conjecture is true then graphs containing no T K 5 is planar or admits a cut of size at most 4. Note that the requirement on connectivity is best possible (see, for example, K 4,4 ).Ma and Yu [7,8] proved the Kelmans-Seymour conjecture for graphs containing K − 4 , and Kawarabayashi, Ma and Yu [3] proved the Kelmans-Seymour conjecture for graphs containing K 2,3 . We refer the reader to [3,7,8] for problems and results (as well as references) related to the Kelmans-Seymour conjecture.It turns out that K − 4 is the right intermediate structure for studying the Kelmans-Seymour structure. By a result of Kawarabayashi [2], any 5-connected graph containing no K − 4 has an edge e that is contractible (i.e., G/e is also 5-connected). Thus, our strategy for proving the Kelmans-Seymour conjecture is to keep contracting edges incident with a special vertex to produce a smaller 5-connected graph. To avoid trivial components associated with 5-cuts or 6-cuts, we also contract triangles (but we give preference to edges).We now give a more detailed description of our strategy. For a graph G and a connected subgraph M of G (respectively, an edge e of G), we use G/M (respectively, G/e) to denote the graph obtained from G by contracting M (respectively, e). Let G be a 5-connected nonplanar graph containing no K − 4 . Then G contains an edge e such that G/e is 5-connected. If G/e is planar, we can apply a discharging argument. So assume G/e is not planar. Let M be a maximal connected subgraph of G such that G/M is 5-connected and nonplanar. Let z denote the vertex representing the contraction of M , and let H = G/M . Then one of the following holds.(a) H contains a K − 4 in which z is of degree 2.(b) H contains a K − 4 in which z is of degree 3.(c) H does not contain K − 4 , and there exists T ⊆ H such that z ∈ V (T...

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