Let $C$ be a $d$-dimensional *(abstract) polytopal complex*.
Most of what I say below could be asked in this general setting, but for a start, let's further restrict to *simple polytopal spheres*, that is, $C$ is homeomorphic to the $d$-sphere, and at each vertex meet exactly $d+1$ facets.

My goal is to formalize and then prove a certain kind of statement. For a warmup, here are two examples of what I am after:

- Assume that each vertex of $C$ is incident to $d+1$ facets, each one combinatorially equivalent to the $d$-cube. From that I can already deduce that $C$ is combinatorially equivalent to the (boundary of the) $(d+1)$-cube.
- Assume that $d=2$ and each vertex is incident to a 4-gonal cell and two 6-gonal cells. This already suffices to conclude that $C$ is combinatorially equivalent to the (boundary of the) permutahedron.

In general, I want to prove something like this:

If each vertex of a simple polytopal sphere $C$ looks the same locally (that is, is incident to the same type of facets), then $C$ is already uniquely determined.

So how to prove this?

Intuitively, this is easy. Take, for example, the second example from above (and see the image below): if you try to draw its edge graph, you will start with a single vertex (the white dot) surrounded by a 4-gon and two 6-gons. You then see that there are several vertices for which only a single face is missing (the black dots), and you add these. You go on until you realize that now all the vertices satisfy the constraint. You never made a choice, and so the result must be unique.

My question is the following:

Question:How can this proof be formalized?

There seem to be some subtleties, some are hinting at certain generalizations:

- You need to use that $C$ is a polytopal
*sphere*, otherwise the complex might not be unique. For example, there are infinitely many possibilities to realize a polytopal*torus*in which every vertex is incident to exactly three 6-gons. Maybe "sphere" can be replaced by*simply connected*, though. - We do not necessarily need
*simple*, but it is the "simplest" of several conditions that ensure that we never have to make a choice when placing the next facet. Here is a case where above reasoning fails because we dropped simplicity: if we want every vertex of a 2-dimensional complex to be surrounded by a 3-gon and three 4-gons, there are two solutions: this one and this one. - How do I know that there is always a vertex at which the next facet is determined?
- How can one be sure that the order in which I add new facets (which are forced on me) does not make a difference. Or does it?