equivalence between the minor and topological minor of $K_5$ or $K_{3,3}$

$\Leftarrow )$ Remark that if a graph G contains $K_5$ or $K_{3,3}$ as a topological minor (http://planetmath.org/subdivision) it is easy to see that it also contains respectively $K_5$ or $K_{3,3}$ as a minor.

$\Rightarrow )$ Same goes for $K_{3,3}$ as minor. If graph G contains a minor of $K_{3,3}$ as minor it also contains a topological minor (http://planetmath.org/subdivision) of $K_{3,3}$ in G.So it’s suffices to show that if graph G contains a $K_5$ as minor, it also contains a $K_5$ or $K_{3,3}$ as topological minor (http://planetmath.org/subdivision).

Suppose that $G⪰K_5$, let $K\subseteq G$ be minimal such that $K$ is a minor of $G$. Because $K$ is minimal every branch set of $K$ induces a tree in $K$ and every two branch sets have exactly one edge between them.For every branch tree $V_x$ we can add all four edges joining it to the other branches, this way we obtain a tree $T_x$.If each of the trees $T_x$ is a topological minor (http://planetmath.org/subdivision) of $K_{1,4}$, $K$ will be a topological minor (http://planetmath.org/subdivision) of $K_5$. This image should make clear how our $K$ looks like. The big circles are presenting different sets of vertices.

If one of the trees is not a $T{K}_{1,4}$ it has exactly two vertices of degree 3, which means that $K$ is a topological minor (http://planetmath.org/subdivision) of $K_{3,3}$. $K$ now looks like this. The big circles and triangles are presenting different sets of vertices.

Thus if $G⪰K_5$ then $G$ contains a topological minor (http://planetmath.org/subdivision) of $K_{3,3}$ or $K_5$. Which proofs this theorem.