Given a planar graph $G$ and a vertex $v$, there exists an embedding of $G$ in the plane such that $v$ is on the exterior face of the embedded graph.
Can someone explain how this works? I tried to understand it from the following proof but I failed.
Consider an embedding of $G$ on the sphere. Such an embedding exists because any planar graph is also embeddable on the sphere. Let $z$ be a point in the interior of some face containing $v$, and let $T(G)$ be the image of $G$ under stereographic projection from $z$. Clearly $T(G)$ is a planar embedding of $G$ of the desired type.
I do not understand why "clearly $T(G)$ is a planar embedding of $G$ of the desired type." Additionally, can someone provide a visual explanation, if possible?
The stereographic projection is a bijection from the sphere minus a point $N$ to the plane.
The idea is that under this projection, any face of the embedding of $G$ on your sphere will get mapped to a face of $G$ on the plane. But then where does the face that contains $N$ go? It goes to the "outer face" of the graph in the plane: the face that is bounded by the edges that are on the "outside" of the graph.
So then if you want a vertex $u$ to be on this outer face, you just need to make sure that you take $N$ to be inside of a face that has $u$ on its boundary. This will send $u$ to the boundary of the outer face.