TITLE: The Euler characteristic & a cube with holes?
QUESTION [6 upvotes]: Possible Duplicate:
Is the Euler characteristic $\chi =2$ for the prism with a hole? 

In attempting to lead a bunch of high school students to an understanding of the Gauss-Bonnet theorem, I've run into a problem determining the Euler characteristic of a cube with a square hole through it.  The problem is that the surface is like that of a torus, and so ought to have $\chi = 0$, but a simple counting of $V$, $F$, and $E$ yields $\chi = 2$.
I have a wide range of abilities in my class, so I need to motivate this theorem without the use of calculus.  My hope was that by examining polyhedra of different types, they would come to "discover" a connection between the Euler characteristic, the total angle defect, and the number of holes through the surface, but now I'm stuck trying to come up with a way to rig the counting to make $\chi$ come out right.
I would love some insight on  

how to make the count come out to $\chi = 0$
how to justify such a counting method so that it does not seem arbitrary to my students.

REPLY [5 votes]: You just have to count carefully.
           

$V=16$: 8 on the hole, 8 on the outside box.
$F=16$: 4 groups of 4.
$E=32$: 12 on top, 12 on bottom, 8 vertical.
$$V-E+F=16-32+16=0 \;.$$ 
In terms of the Gauss-Bonnet formula, there are 8 vertices with curvature $\pi/2$ (the outside corners),
and 8 vertices with curvature $-\pi/2$ (the hole vertices), yielding a total curvature
of $0 = 2 \pi \chi$.
For justification, you might look at Discrete and Computational Geometry, from which the figure above is taken, Chapter 6.