# Euler characteristic of this polyhedron?

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!

• The leftmost image in the second figure a quadrangulation, not a triangulation! – Pedro Tamaroff Feb 18 at 10:39
• Ok, I really notice that a triangulation can be done for any convex polygon. – LH8 Feb 18 at 10:43