Let $P$ be a polyhedron. Is it true that $P$ has always more/as many corners than facets? I haven't found a counterexample in $\mathbb R^2$ and $\mathbb R^3$ and intuitively I think the statement is true. Is there a proof?
$\begingroup$ Try using $V-E+F=2$? $\endgroup$– arasFeb 20 at 17:12
2$\begingroup$ Nope. see this picture for an polyhedron with $12$ faces and $9$ vertices. $\endgroup$– achille huiFeb 20 at 17:23
$\begingroup$ Is it also passible if $P$ has to be convex? @achillehui $\endgroup$– RobertFeb 20 at 18:57
2$\begingroup$ Yes, just push the "fan" on top of the polyhedron in previous comment towards the face of pyramid it rest on as much as possible. Similar construction/procedure for other combination of $f \ge v$ will give you a convex polyhedron. $\endgroup$– achille huiFeb 20 at 19:05
3$\begingroup$ “I haven't found a counterexample” is a little bit puzzling, since you have missed both the regular octahedron and the triangular dipyramid (which is just two tetrahedra stuck together). I would have expected both of those to be among the first ten polyhedra you checked. Can I suggest that you assemble a more comprehensive list of examples? If you have a conjecture about polyhedra, you can check it with the 5 platonic solids, a couple of prisms and antiprisms, maybe the other five deltahedra, some things like that. $\endgroup$– MJDFeb 20 at 23:48
In $\Bbb R^2$ the claim is trivial, since every polygon always has the same number of sides and vertices.
In $\Bbb R^3$, it's definitely false. For every polyhedron that has $F$ facets and $C$ corners, there is a so-called “dual” polyhedron that has $C$ facets and $F$ corners.
For example, the cube has 6 facets and 8 corners. Its dual is the regular octahedron, which has 8 facets but only 6 corners.
Did I misunderstand what you meant by "facets" and "corners"?