$v-e+f=2-2g$ as a topological invariant

Question

Euler's theorem was expanded to encompass polyhedrons homeomorphic to not only spheres but also $g$-holed toruses. I've tried to understand proofs about how $2-2g$ is a topological invariant but have always had trouble with the use of planar graphs.

Can anyone explain how $2-2g$ is a topological invariant without using the concept of planar graphs?

Answer 1

$2g-2$ is a topological invariant of any closed orientable surface $S$ of genus $g$, namely its Euler characteristic. There are several ways to see that the Euler characteristic is a topological invariant, see also this question, or this one.

Answer 2

Consider using the building tiles like the ones in the image below; you will be using planar graphs, but you will (hopefully) think about them differently, and maybe that helps:

If you use these to build a closed surface (i.e. every single edge has exactly two tiles meeting along it), you can count

• How many corners where two or more edges meet, $v$
• How many edges there are where two tiles meet, $e$
• How many tiles you used, $f$
• How many holes your surface has, $g$

and you will find that $v-e+f = 2-2g$

For instance, if you use triangular tiles to make an icosahedron, you have

• 12 corners
• 30 edges
• 20 tiles
• 0 holes

and we can confirm that $12-30+20 = 2+2\cdot0$.

You can also use them to make a figure like this:

You will get

• 32 corners
• 64 edges
• 32 tiles
• 1 hole

and we can confirm that $32 - 64 + 32 = 2-2\cdot 1$.

In general, you can use tiles like this which are bent and crooked (as long as the tiles themselves do not have any holes) to make any closed surface, and the result stays the same.