# Prove that every dual graph of a planar graph is planar

It seems obvious, but how to prove it properly? I tried Kuratowski, but got stuck at $K_{3,3}$

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$K_{3,3}$ is not planar, so would not need to be checked if you are to show "the dual graph of a planar graph is planar". – coffeemath Dec 14 '13 at 9:54
Well I need to prove that no dual contains $K_{3,3}$ or $K_5$ or their division as a subgraph. For $K_5$ it's easy - 5 faces that are neighbours to each other? That's impossible by four colors theorem. But how to prove that $K_{3,3}$ in dual is impossible as well? – Klobbbyyy Dec 14 '13 at 10:00
You don't really need to use Kuratowski to prove a graph is planar, provided you show it is planar by making a specific graph of it. I just put up an answer about how to do that from any given planar graph, that is to get its dual graph as a planar graph drawn on top of it. – coffeemath Dec 14 '13 at 10:05
The answer is wrong - check Eu Yu's comment. – Klobbbyyy Dec 14 '13 at 10:10
I agree my answer was wrong as it initially used line graph instead of dual. Fixed now. – coffeemath Dec 14 '13 at 10:17

The dual graph is not defined this way. You're talking about the line graph. Note that line graphs of planar graphs are not necessarily planar. The $5$ star $K_{1,5}$ has line graph $K_5$ for example. – EuYu Dec 14 '13 at 10:03
@MathN00b The thing that needs care is that the new edges don't cross each other. Initially put a mark on the interior of each edge. A given face may be mapped into a circle by some map $h$, then from the center $c$ of that circle one can draw radii to the points where the marks on the edges have moved under $h$. Then going back by $h^{-1}$ the radii become nonintersecting arcs going to the marked points on the edges of the given face. This done for all faces gives the copy of the dual graph. – coffeemath Dec 14 '13 at 11:24