I am looking back at the history of the formula "$V-E+F=2$" for any polyhedron in Euclidean 3-space ($V=$number of vertices; $E=$ number of edges, $F=$ number of faces). Cauchy gave following proof of this theorem.
Let P=the polyhedral, with V vertices, E edges, and F faces. To calculate V-E+F, remove one face, which will reduce $V-E+F$ by 1. Then, make any triangulation; this will not change $V-E+F$; then remove one by one faces whose one edge is also an edge of polyhedron; this will also not alter $V-E+F$. At last, we have one triangle, for which $V_1-E_1+F_1=1$; adding one both sides, LHS will be $V-E+F$ of original polyhedron, which is $2$, the RHS.
This is the only proof I could saw now. I heard that Descartes, and Euler has also given proof of this theorem.
Question 1. How they (Descartes and Euler) proved it?
Question Instead of removing one face (in Cauchy's proof), can we remove one vertex, or one edge? More precisely, if P is a polyhedron, then by removing on vertex or one edge, can we "topologically" embed (deform) it into the plane?
Euler proved it by constructing two trees out of the given Polyhedron (by the way $V+F-E = 2$ is valid for certain polyhedrons only).
One of the tree is the tree $T$ with vertices as the vertices from the polyhedron and edges (only a subset of course that are required to get the tree have all the vertices from the polyhedron) as edges from the polyhedron. The other tree $T^\prime$ is the dual of the tree $T$ - where the $T^\prime$ vertices are the faces of the polyhedron and the edges are the remainder of the edges in the polyhedron which are not in the tree $T$.
Now, for a tree, the difference in the number of vertices to the number of edges is 1. So, evaluating this for $T$ and $T^\prime$ and adding them together, we get the desired formula.