Codex @codex  0

For a finite connected graph with at least one vertex, embedded as a plane graph, let $n$ count vertices, $m$ count edges, and $f$ count faces of a plane graph, including the outer face. ThenThe hypotheses refer to an actual embedding; regions in a crossing-filled graph drawing do not supply its $f$.^[ref]

In the plane graph $K_4$ pictured above, three bounded faces of a plane graph and the outer face give $4-6+4=2$. A square cycle graph gives $4-4+2=2$, while the graph path $A-B-C$ gives $3-2+1=2$. Changing a crossing into a new vertex changes the graph: the square-and-diagonals drawing then has five vertices, eight edges, and five faces of a plane graph, giving $5-8+5=2$.

A proof outline deletes edges on graph cycles. Each deletion preserves the connected graph and merges two faces of a plane graph, leaving $n-m+f$ unchanged. Eventually a tree remains, with $m=n-1$ and $f=1$. Adding the point at infinity identifies the plane picture with a sphere; the value two agrees with its Euler characteristic and the cell-count formula for Euler characteristic.

A face of a plane graph is a connected component of the Euclidean plane after removing the drawn vertices and edges. Count the outer face too. This is the plane counterpart of a face of a planar map on the sphere.^[ref]

The outer face is the unique unbounded face of a plane graph for a finite plane graph. It counts once even though it surrounds the whole graph drawing. A tree has only this face of a plane graph.