I had a look at paragraph 1.1 from "Basic Topology" by M.A. Armstrong.
There a very nice proof outline presented there to the Euler's theorem.
$ V-E+F=2 $
In that proof outline (towards the end) there's this note:
"The proof breaks down here for a polyhedron such as that shown in Fig. 1.3, because the dual graph $\Gamma$ will contain loops."
But I don't quite get this note since... I just think the proof breaks much earlier. Because for the polyhedron in Fig 1.3, there's simply no tree $T$ as constructed in the proof. We have 2 trees instead of a single tree $T$. Is this observation correct?
I mean... in fact the polyhedra from Figs 1.2 and 1.3 do not satisfy at all the condition (a) of the theorem, that the graph formed by the vertices and edges of the polyhedron needs to be connected. Their respective graphs are composed of 2 separate connected components and hence are not connected graphs.
Not to talk that they don't satisfy also condition (b) which the text explicitly mentions.
So the whole construction of the proof breaks much earlier, I think.
I just don't see how you can start doing the proof's construction if condition (a) is not met.
Basically I realized that my confusion comes from not being sure about the following. Let's look at this prism which has a smaller prism cut through the middle. In that proof they start by building a graph G of all vertices and edges of the polyhedron. Then they build a spanning tree T of G. Then they build the dual graph $\Gamma$ which is defined as: vertices - the faces of the polyhedron, edges - two faces are connected if they are neighbors and the neighboring side is not an edge in T.
My question is: should I consider that edges AG, BH, CI and the analogical 3 edges from the top face exist? Seems this influences significantly the Euler's characteristic of the polyhedron?! Hm... In other words: I understand this unfilled prism does not satisfy condition (b) of the theorem, but does it satisfy (a) or not?
If I consider they exist, then I can understand the note in the book. We build G, G is a connected graph, and then we build a spanning tree T of G (which is in green on my drawing).
But if these 6 edges do not exist then well... the whole graph G has no spanning tree because it's not connected.
So in this context of topology, do we consider these vertices as connected (A with G, B with H, C with I) ?