Relation between vertices, cells, and vertex-cell-incidences in 3D subdivisions Consider a planar subdivision, with F faces, V vertices, E edges, and I face-vertex incidences. For simplicity, assume a "non-degenerate" situation in which each vertex occurs on the boundary cycle of each face only once. Then I is the sum over all faces of the number of vertices per face, or equivalently, the sum over all vertices of the number of incident faces. We have I = 2E, and therefore Euler's formula gives us:

F + V = I/2 + 2.

All I actually care about is:

F + V &lt; I/2.

Does this also hold in for three-dimensional subdivisions, when F is the number of three-dimensional cells of the subdivision, and V and I are as defined above? 

Note that I am looking for an inequality that is independent of the number of one- and two-dimensional faces in the subdivision. When the cells are convex I can get something using quite a bit of geometry, but it seems to me that the requirement that the cells be convex is an artifact of my proof technique, and something more general should be possible and/or known. Does anybody have an idea where to look for the answer?
Doesn't $F + V \le 2I$ hold trivially for every system of incidences between things called vertices and things called faces? Under some reasonable assumptions, every vertex has at least one incidence, and every face has at least one incidence. If you add those incidences together you count each incidence at most twice.

To get from there to your desired $F + V \le I/2 + O(1)$ you'd need to use some actual geometry show that the vertices and faces each have at least four incidences on average. This is obvious for convex subdivisions but maybe needs more assumptions in general (e.g. consider a system of bubbles in 3d space, linked one to the next in a chain: each face has no vertex-face incidences, because there are no vertices).