Euler formula for connected plane multigraphs

For each edge $e$ choose two distinct interior points at one-third and two-thirds of its polygonal parametrization, using the finitely many edge parametrizations in the code. Subdivide $e$ at those points into three edges and declare the two points new vertices. If $e$ is a loop at $v$, its replacement is the path $v-x_e-y_e-v$ with three distinct vertices along the loop; it has no loop edge. New vertices are unique to their original edge, so replacement edges from different original edges cannot be parallel, and the three edges replacing one original edge have different endpoint pairs. Thus the subdivided graph $G^{\mathrm{\prime }}$ is simple. Subdivision preserves connectedness and leaves the point-set drawing unchanged, so its faces are exactly those of $G$. With $E$ original edges, $V^{\mathrm{\prime }}=V+2E$ and $E^{\mathrm{\prime }}=3E$, hence $V^{\mathrm{\prime }}-E^{\mathrm{\prime }}=V-E$. If $E=0$, nonempty connectedness forces $V=1$, and the same construction leaves the one-vertex simple plane graph unchanged. By [F1], $G^{\mathrm{\prime }}$ is a connected plane graph in the library's simple-graph convention in either case.

Apply [F2] to $G^{\mathrm{\prime }}$. Since $F^{\mathrm{\prime }}=F$ and $V^{\mathrm{\prime }}-E^{\mathrm{\prime }}=V-E$ by step 1.1, the equality $V^{\mathrm{\prime }}-E^{\mathrm{\prime }}+F^{\mathrm{\prime }}=2$ gives $V-E+F=2$.