Euler's formula $\mathrm{\mid }V\mathrm{\mid }-\mathrm{\mid }E\mathrm{\mid }+\mathrm{\mid }F\mathrm{\mid }=2$ for every connected plane graph

Choose a spanning tree $T$ by [L1]. It has all $\mathrm{\mid }V\mathrm{\mid }$ vertices, $\mathrm{\mid }V\mathrm{\mid }-1$ edges, and exactly one face by [L3]. Thus $\mathrm{\mid }V\mathrm{\mid }-\mathrm{\mid }E(T)\mathrm{\mid }+\mathrm{\mid }F(T)\mathrm{\mid }=\mathrm{\mid }V\mathrm{\mid }-(\mathrm{\mid }V\mathrm{\mid }-1)+1=2$.

We prove the exact face correspondence for deleting a cycle edge $e$ from a connected plane graph $H$. Let $X$ be its drawing, let $e^{\circ }$ be the edge arc without its endpoints, and put $\mathrm{\Omega }={\mathbb{R}}^2\setminus (X\setminus e^{\circ })$, the complement after deleting $e$. By [L2], precisely two distinct old faces $F_1,F_2$ meet the two local sides of $e^{\circ }$. Put $R=F_1\cup e^{\circ }\cup F_2$. This is connected: each $F_i$ is connected and has every point of the connected arc $e^{\circ }$ in its closure. It is open in $\mathrm{\Omega }$: for any point of $e^{\circ }$, a sufficiently small disk avoids $X\setminus e^{\circ }$, and its two sides of the local straight piece, or of the two adjacent pieces at a bend, lie in $F_1,F_2$ respectively. Openness at points of the faces is immediate. Every other old face $F^{\mathrm{\prime }}$ is connected and open in $\mathrm{\Omega }$; its frontier misses $e^{\circ }$ by [L2] and is contained in $X\setminus e^{\circ }$, so $F^{\mathrm{\prime }}$ is also closed in $\mathrm{\Omega }$. Now $\mathrm{\Omega }$ is the disjoint union of $R$ and all these other old faces. The complement of $R$ is open, so $R$ is closed in $\mathrm{\Omega }$ as well. These connected open-and-closed pieces are exactly the new faces: $F_1,F_2$ merge into $R$, while every other old face remains distinct. Consequently deleting $e$ decreases both the edge count and the face count by one, leaving the vertex count fixed.

Starting from $G$, delete the finitely many edges of $E(G)\setminus E(T)$ one by one. At every stage $T$ remains, so the current graph is connected. Each edge $e$ to be deleted joins vertices already linked by a path in $T$ and hence lies on a cycle of the current graph; step 1.2 applies. Each deletion lowers $\mathrm{\mid }E\mathrm{\mid }$ and $\mathrm{\mid }F\mathrm{\mid }$ by one without changing $\mathrm{\mid }V\mathrm{\mid }$, so the Euler expression stays constant throughout the finite induction.

The final drawing is $T$, whose Euler expression equals $2$ by step 1.1. The invariant from step 2.1 therefore gives $\mathrm{\mid }V(G)\mathrm{\mid }-\mathrm{\mid }E(G)\mathrm{\mid }+\mathrm{\mid }F(G)\mathrm{\mid }=2$ for the original connected plane graph.