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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]

A planar graph admits at least one graph drawing in which edges meet only at common endpoints. This is a property of the abstract graph, not a requirement on every graph drawing. A plane graph specifies one such drawing.^[ref]

For example, the complete graph $K_4$ joins every pair of four vertices, giving six edges. Put its vertices at the corners of a square and its diagonals cross. Instead put $A,B,C$ at the corners of a triangle and $D$ inside it: the same six edges $AB,AC,BC,AD,BD,CD$ now meet only at their endpoints. The first graph drawing obscures planarity; the second proves it. In particular, its crossing number is zero.

The distinction matters whenever connections must fit on one surface without intersecting. A failed graph drawing may only need rearrangement, whereas nonplanarity by edge counts proves that every rearrangement fails.