Формула Єйлера для многогранників

Euler's formula (for polyhedra)

Let $V$ be the number of vertices of a polyhedron, $F$ the number of faces of that polyhedron, and $E$ be the number of edges. The quantity $\chi = V-E+F$ is called the Euler characteristic (of a polyhedron). In the case of convex polyhedra, $\chi=2$.

Consider, for example, a tetrahedron (which is the simplest solid). It has 4 faces, $\frac{1}{2}(4)(3) = 6$ edges, and $\frac{1}{3}(4)(3) = 4$ vertices. Thus we have $V-E+F = 4 - 6 + 4 = 2$.

Euler's formula holds for all Platonic solids (tetrahedron, cube, octahedron, dodecahedron, and icosahedron). Since a cube and an octahedron are dual polyhedra (each is formed by connecting the centers of the faces of the other), their $V$ and $F$ values are equal to the $F$ an $V$ values of the other. (The same is true for the dodecahedron and icosahedron).

Cauchy's proof of the formula

In 1811, Cauchy published a proof of this formula. His proof is as follows (in reproducing this proof, I have not proved all of the steps).

Consider any convex polygon and remove any face from it. Then $\chi$ decreases by $1$ because $F$ is decreased by $1$, and $V$ and $E$ don't change. Now by moving the vertices, it is possible to deform this figure into a 2-dimensional figure. Now, if there is a face with $4$ or more vertices, draw a diagonal connecting any two nonadjacent vertices in that face (this increases $F$ and $E$ by 1 each, so $\chi$ is preserved). Repeat this process until all faces are triangular. Now, perform one of the following two operations:

(1) Remove a triangle with two edges adjacent to the exterior of the figure. This decreases $V$ by $1$, $E$ by $2$ and $F$ by $1$, so $\chi$ is preserved.

(2) Remove a triangle with one edge adjacent to the exterior of the figure. This decreases $F$ by $1$ and $E$ by $1$, so $\chi$ is again preserved.

You can repeat the process until you get to the last triangle. Now for a triangle, $V=3$, $E=3$ and $F=1$, so its $\chi = 1$. Thus since both operations preserve $\chi$, we proved that for the original deformed planar figure $\chi = 1$, so for the polyhedron, $\chi = 2$.

The value of $\chi$ for figures that aren't convex polyhedra

For any nondegenerate polygon (with at least 3 sides) in the plane, the value of $\chi$ is $1$ ($V=E$ and $F=1$). For a circle, you have $V=0$, $E=1$ and $F=1$, so $\chi=0$, and for a line segment, $V=2$, $E=1$ and $F=0$, so $\chi=1$. For a sphere, $V=E=0$ and $F=2$. A Mobius strip and a Klein bottle each have $\chi=0$, and $n$ non-intersecting figures each with $\chi=k$ have a total $\chi=nk$.