Non-planar extensions of subdivisions of planar graphs

A graph G is almost 4-connected if it is simple, 3-connected, has at least five vertices, and V (G) cannot be partitioned into three sets A,B,C in such a way that |C| = 3, |A| ≥ 2, |B| ≥ 2, and no edge of G has one end in A and the other end in B. A graph K is a subdivision of a graph G if K is obtained from G by replacing its edges by internally disjoint nonzero length paths with the same ends, called segments. Let G,H be almost 4-connected graphs such that G is planar and has at least seven vertices, H is non-planar, and H has a subgraph isomorphic to a subdivision of G. If K is a subgraph of H, then a K-path in H is a path with at least one edge, both ends in K, and otherwise disjoint from K. We prove that there exists a subgraph K of H isomorphic to a subdivision of G such that one of the following conditions holds. (i) There exists a K-path in H such that no face boundary of K contains both ends of the path. (ii) There exist two disjoint K-paths with ends s1, t1 and s2, t2, respectively, such that the vertices s1, s2, t1, t2 belong to some face boundary of K in the order listed. Moreover, for i = 1, 2 the vertices si and ti do not belong to the same segment of K, and if two segments of K include all of s1, t1, s2, t2, then those segments are vertex-disjoint. This is a lemma to be used in other papers. In fact, we prove a more general theorem, where we relax the connectivity assumptions and do not assume that G is planar. Instead of face boundaries we work with a collection of cycles that cover every edge twice and have pairwise connected intersection. Finally, we prove a version of this result that applies when G\X is planar for some set X ⊆ V (G) of size at most k, but H\Y is non-planar for every set Y ⊆ V (H) of size at most k.