Lower bound on the number of faces of a polyhedron of genus g Is there a lower bound on the number of faces of a polyhedron of topological genus g?  
For example: it seems very reasonable that $g$ < $F$ 
i.e. the genus of a polehydron is less than the number of faces of the polyhedron, but i can't find a proof. 
To be clear what is meant by polyhedron let's use the definition from wikipedia:
"A polyhedron is a solid in three dimensions with flat polygonal faces, straight edges and sharp corners or vertices."
The genus can be calculated by $g = \frac{2-\chi}{2}$, where $\chi$ is the Euler characteristic of the polyhedron. 
 A: Example of polyhedron with 4096 faces and 4097 holes. 
P. McMullen, C. Schulz, and J.M. Wills. "Polyhedral manifolds in E3 with unusually large genus". Israel Journal of Mathematics., 46 (1983), no. 1-2, pages 127–144
https://link.springer.com/article/10.1007%2FBF02760627
A: The genus of an orientable surface represents the number of tori in a connected sum decomposition of the surface ($g=$ the number of "holes" in a closed surface). A polyhedron with $g>0$ is a toroidal polyhedron.
Let $F_i$ represent the initial number of faces of a polyhedron with $g=0$ that we will manipulate. A polyhedron must have a minimum of $4$ faces (tetrahedron).
$$ F_i \ge 4 $$
Let $F_h$ represent the number of faces added to our polyhedron when we add a hole to it. A polygon must have a minimum of $3$ edges (triangle), thus a minimum of $3$ faces are created by adding a hole in a polyhedron. Because $g=$ the number of holes,
$$ F_h \ge 3g $$
The sum of the initial faces and the faces added per hole gives the total faces:
$$ F = F_i + F_h $$
$$ F \ge 4 + 3g $$
$$ F > 3g $$
$$ F > g $$
