For a plane graph with $c$ components, $\mathrm{\mid }V\mathrm{\mid }-\mathrm{\mid }E\mathrm{\mid }+\mathrm{\mid }F\mathrm{\mid }=1+c$, including the null graph

Insert the connected component drawings one at a time in their inherited positions. Before a component is inserted, its connected drawing lies in one face of the components already inserted. By [L3] choose a spanning tree; [L4] gives it $\mathrm{\mid }V(G_i)\mathrm{\mid }-1$ edges, and [L5] shows that inserting this tree does not split the containing face. Add the remaining edges in their inherited drawing order. Each joins vertices already connected by the tree, hence lies on a cycle, so [L6] shows that its insertion splits one face into two. There are $\mathrm{\mid }E(G_i)\mathrm{\mid }-\mathrm{\mid }V(G_i)\mathrm{\mid }+1=\mathrm{\mid }F(G_i)\mathrm{\mid }-1$ such edges, where the last equality is [L1]. Starting from the one face of the empty drawing therefore gives $\mathrm{\mid }F\mathrm{\mid }=1+{\sum }_i(\mathrm{\mid }F(G_i)\mathrm{\mid }-1)$. Vertices and edges add disjointly by [L2].

Therefore $\mathrm{\mid }V\mathrm{\mid }-\mathrm{\mid }E\mathrm{\mid }+\mathrm{\mid }F\mathrm{\mid }={\sum }_i(\mathrm{\mid }V(G_i)\mathrm{\mid }-\mathrm{\mid }E(G_i)\mathrm{\mid }+\mathrm{\mid }F(G_i)\mathrm{\mid })-c+1=2c-c+1=1+c$.

For the null graph, $\mathrm{\mid }V\mathrm{\mid }=\mathrm{\mid }E\mathrm{\mid }=c=0$ and $\mathrm{\mid }F\mathrm{\mid }=1$, so the same formula reads $1=1$.