Euler characteristic of this polyhedron?

Question

I´m trying to obtain the Euler characteristic of this polyhedron $P$, that is homeomorphic to the torus $T$ (I think):

So it should be $\mathcal{X}(P)=\mathcal{X}(T)=0$.

But we get $V=16, F=10, E=24$, so $\mathcal{X}(P)=2$.

However, if we consider a triangulation as this two cases:

it is $\mathcal{X}(P)=0$, because $V=C=16$ and $E=32$, and $V=16, F=32, E=48$, respectively.

So, what is it wrong?

Thanks for the support!

Answer 1

The problem is that your first polygon setup uses non-simple polygons. add single edge to each donut shaped face and it will work better.

Cauchy's basic proof for Euler's formula for characteristic requires that we triangulate the polygons and takes advantage of a particular property of this triangulation: adding a single edge via splitting a polygon must also add a single face. This is true of simple polygons, yes, but complex polygons like the donut shape don't have this property: adding an edge may not split the polygon into two.