A famous theorem on the embedding of plane graph.
Theorem (Whitney). A $3$-connected planar graph has a unique embedding, up to composition with a homeomorphism of $S^2$.
I saw a proof of the theorem on the Internet see here. It seems to be correct, but some details I have not been fully understood.
Before proving Theorem author gives the following lemma.
Lemma. Let $G$ be a planar graph and let $C ⊂ G$ be a cycle.
The cycle $C$ is the boundary of a face for every embedding of $G$
in $S^2$ if and only if $G − C$ is connected.
The full text of the theorem proof is placed below.
Proof. Say there are two embeddings of $G$ in $S^2$. Then some
cycle $C \subset G$ is the boundary of a face for one embedding, but
not the other. By the Lemma, $G − C$ has at least two
components. Look at an embedding where $C$ is a face.
The other components of G − C have to fit in the “gaps”. Here is a cut pair.
The proof of the theorem came to an abrupt end in beamer. I am surprised why the unique embedding is proved in the end. Does that mean it suffies to prove that cycle $C \subset G$ is the boundary of a face for any embedding of $G$ in $S^2$?
Why is that enough?
Best Answer
Removing those two vertices leaves us with two components. But we said as a prerequisite that the Graph is 3-connected ("G is still connected after removing any 2 vertices."). Contradiction.