Dual of an eulerian graph

Question

I was studying graph theory in Bollobas book's and I found a result given without proof that I wasn't able to understand. Here is my problem: "Le G be an eulerian graph, then its planar dual is a bipartite graph". Can anyone explain me why it is true. Thanks in advance for the help.

Answer 1

The edge set of a circuit in $G$ correspond to (inclusion wise) minimal cuts in $G^*$ and vice versa.

Now we have the following theorem:

Let $G$ be a graph, $G$ is eulerian if and only if every minimal cut has even cardinality.

Proof: $"\Longleftarrow"$ Let $v \in V(G)$ be a vertex then the cut $\delta(v)$ has even cardinality, thus $\text{deg}(v)=|\delta(v)|$ is even.

$"\Longrightarrow"$ Let $\delta(X)$ be a cut, $X \subset V$. As $G$ is eulerian we can partition its edge set into edge-disjoint closed walks $C_1,\ldots, C_k$. But note that every of those walks intersects $\delta(X)$ in an even number of edges.

Thus $G$ is eulerian iff every minimal cut has even cardinality. Every minimal cut in $G$ has even cardinality iff every circuit in $G^*$ has even cardinality.

It follows that $G$ is eulerian iff $G^*$ is bipartite.