What is the least number of faces you need on a polyhedron that has Euler-characteristic $V-E+F = \chi = -2$ (i.e. 2 holes)?
So far I just found the solution of gluing two copies of the following toroidal shape together, which results in 16 faces, but is it possible to do it with even fewer?
(The faces should be simply connected (i.e. have no holes and one continuous border), and there should be no points or edges where the polynomial touches itself. )
Here is a quick 3d model:
(EDIT: This approach yields 15 faces if you flatten two faces two one, as I wrote in the comments, but this results in a non-convex face. So it would also be interesting what happens if we restrict this to only convex faces. Here is what it looks like with those flattened faces:
)
We can definitely do this with $12$ faces, $22$ vertices, and $36$ edges by gluing two copies of the Szilassi polyhedron along a face.
(The Szilassi polyhedron has two suitable faces to do this with. In the GIF animation on Wikipedia, you can see this clearly: they are the ones on which the polyhedron can lie flat.)
Here is a Mathematica-generated image of the result:
But I think we can do even better. This arXiv paper gives realizations of genus-$2$ polyhedra with $10$ vertices, which is best possible. I'm pretty sure this means that their duals are genus-$2$ polyhedra with $10$ faces, but I'm not sure how to figure out what their duals would look like.