# Euler characteristic of a polygon with a hole

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.)

• If you're aware that the Euler characteristic is a homotopy invariant, then you know what the answer should be: the polygon with a hole is homotopy equivalent to $S^1$, whose Euler characteristic is $0$. However, as Tsemo points out in his answer, both of your computations are invalid because they don't use a triangulation, so it's more-or-less a coincidence that your second computation got the right value. Mar 6, 2020 at 14:39
• The second computation is fine, you don’t need to restrict yourself to triangles. OP, in the first computation the “face” you’re using doesn’t count as a face. Aug 13, 2020 at 2:22
• Generally the rule is that faces themselves can't have any "holes" in them. A sufficient condition is that faces always be convex polygons. Aug 13, 2020 at 3:40

Triangulations means using TRIANGLES, not any polygons.

• I thought any contractible polygon was valid. Mar 6, 2020 at 19:57
• Additionally, note that in lower case also it can be understood :) Mar 6, 2020 at 20:17

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 $$0$$.

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

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