Euler characteristic of a polygon with a hole

Question

My doubt is about how you obtain the Euler characteristic of this polygon with hole.

It has $6$ vertices, $6$ edges and $1$ face, so $6+1-6=1$.

But if you triangulated it using $3$ parallelograms, you get $6$ vertices, $3$ faces, $9$ edges, so it is zero.

How is it possible? (Euler characteristic is an invariant.)

Thanks in advance!

Answer 1

Triangulations means using TRIANGLES, not any polygons.

Answer 2

If you polygon $P$ have $h$ holes, then

$$ \chi(P)=1-h $$

Answer 3

The Euler characteristic is invariant as long as the faces are simply connected.

In the original figure, the face is not simply connected. To make it simply connected you draw one edge, any edge, from the inner triangles to the outer one, and then your calculated Euler characteristic drops from $1$ to $0$ with just that edge.

If you now draw more connecting edges, like connecting every pair of corners in your second figure,the cat is out of the bag; the face(s) is/are already reduced to being simply connected and the Euler characteristic never changes again from $0$.