If $G$ is bipartite and does not have $K_{3,3}$ as a topological minor, then $G$ is planar.

Question

Prove or disprove: If $G$ is bipartite and does not have $K_{3,3}$ as a topological minor, then $G$ is planar.

Answer 1

This is not true, and a counterexample is the graph $K_5$ with each edge subdivided once.

The thought process is that this might not be true, so let's start with a small graph that is nonplanar but doesn't have $K_{3,3}$ as a topological minor. The graph $K_5$ is the natural choice here. Then we can "make $K_5$ bipartite" by subdividing each edge once (this is a nice idea that works for any graph). We can show pretty easily that the resulting graph is still nonplanar, and still doesn't have $K_{3,3}$ as a topological minor (because it has too few vertices of degree at least three).