  We prove that every locally Hamiltonian graph with $n\ge 3$ vertices and
possibly with multiple edges has at least $3n-6$ edges with equality if and
only if it triangulates the sphere. As a consequence, every edge-maximal
embedding of a graph $G$ graph on some 2-dimensional surface $\Sigma$ (not
necessarily compact) has at least $3n-6$ edges with equality if and only if $G$
also triangulates the sphere. If, in addition, $G$ is simple, then for each
vertex $v$, the cyclic ordering of the edges around $v$ on $\Sigma$ is the same
as the clockwise or anti-clockwise orientation around $v$ on the sphere. If $G$
contains no complete graph on 4 vertices and has at least 4 vertices, then the
face-boundaries are the same in the two embeddings.
