Subdivision of nonplanar graph is nonplanar?

Question

If $G$ is a nonplanar graph, does it follow that every subdivision of $G$ is nonplanar?

Answer 1

Yes, because Kuratowski's Theorem states that a graph $G$ is nonplanar if and only if it contains a subgraph $H$ that is a subdivision of $K_{3,3}$ or $K_5$.

If $G$ is nonplanar, let $H$ be such a subgraph. A subdivision $G'$ of $G$ induces a subdivision $H'$ of $H$; since $H$ is a subdivision of either $K_{3,3}$ or $K_5$, so is $H'$, and so $G'$ is nonplanar.

There is an alternative (basically identical) proof using Wagner's theorem.

A third, more direct proof: If $G' = (V',E')$ is a subdivision of $G = (V,E)$, and $G'$ is planar, then draw $G'$ without intersecting edges. Now remove the vertices in $V' \setminus V$ but keep the edges. Now you have a drawing of $G$ without intersecting edges, so $G$ is planar.