How can I prove that $K_{1,2,2,2}$ has a crossing number of 3?
I have an example of a it with 3 crossings, now I need to prove that it cannot have only 2 crossings.
My guess is that: Suppose $cr(K_{1,2,2,2})=2$, as the graph has 7 vertices then the two crossings must share a common vertex. Removing this vertex will give a planar graph - which should cause a contradiction. But removing the vertex in the vertex class with only 1 vertex gives $K_{2,2,2}$ which is planar, no contradiction...How do I remove this case to get a proper contradiction? Is there a simpler way to prove this?
Let $G = K_{1,2,2,2}$. Suppose $cr(G) = 2$, then by removing two edges you get a planar graph $H$. But then $e(H) \le 3 v(H) - 6$. That is:
$e(H) = e(G) - 2 = 16 <= 3 v(H) - 6 = 15$ which is a contradiction.