Kuratowski's Theorem

Theorem
The Kuratowski-Wagner Theorem states that the following conditions on a graph $$\Gamma$$ are equivalent:

(a) $$\Gamma$$ is planar;

(b) $$\Gamma$$ contains no subdivision of $$K_5$$ or $$K_{3,3}$$.

Proof
The proof proceeds in two parts: (a)$$\Rightarrow$$(b) and (b)$$\Rightarrow$$(a).

Part 1: (a)$$\Rightarrow$$(b)

First we consider $$K_5$$. $$K_5$$ has 5 vertices and 10 edges, so by the Euler Polyhedron formula V-E+F=2, a planar embedding would have 7 faces. But each face has at least 3 edges, while each edge bounds at most two faces. If we count the incident edge-face pairs, the number of faces is at most (Edges)* 2/3 = 6 + 2/3, a contradiction. Hence $$K_5$$ is non-planar.

Now consider $$K_{3,3}$$. This graph has 6 vertices and 9 edges, and hence 5 faces if it is planar. Since this graph is bi-partite, each closed Euler trail has an even number of edges. Hence any faces must have at least 4 edges, and so the number of faces are at most (Edges)*2/4 = 4.5, a contradiction. Hence $$K_{3,3}$$ is non-planar.

Now consider a subdivision of either graph. A subdivision of a graph is a replacement of any edge $$[x,y]$$ with two edges, $$[x,z]$$ and $$[z,y]$$, where Z is a new vertex. The arguments as above continue to work as before, since both V and E have increased by one in the Euler formula, and so the expected value of F is unchanged.

Hence if a graph $$\Gamma$$ contains a subdivision of $$K_5$$ or $$K_{3,3}$$, then $$\Gamma$$ is not planar, and taking the contrapositive, we have

$$\Gamma$$ is a planar graph $$\Rightarrow \Gamma$$ does not contain a subdivision of $$K_5$$ or $$K_{3,3}$$.

Part 2: (a)$$\Leftarrow$$(b)