Knot theory: Genus of a surface Use Euler characteristic to determine the genus of the surface in Figure 4.24 in picture below.
I am stuck with this question 4.10 from Colin Adams, the Knot Book.
 A: I know it isn't using the Euler characteristic, but I couldn't help it. Consider the following 'proof by picture':
Start by rounding everything out, so that your original picture looks like a ball with holes drilled out of it. Then just follow the pictures.


This is an ambient isotopy of the manifold, so the genus is preserved, and it follows that the genus of your original surface is 2.
A: I believe that you should first break up those faces a bit more, to make them polygonal. That mean, add an edge between any vertex on an inner square to the outer square in a given plane. There are 3 big faces, plus 12 big faces that have been cut in 4, plus 4 faces from the very small rod, 4 from the medium rod, 7 medium faces in the inner cube and 12 more trapezoidal faces after breaking up the remaining bad faces. 
Now count the edges. I believe there should be 84 of them, but it might be a good idea to recount them while drawing the diagram. 
Finally, no vertices were added in the polygonization, so there are still 40 of them. As the comments say, $\chi=V-E+F=40-84+42=-2=2-2g$, so the genus is 2. 
A: This question came up today in my class.  The isotopy above is probably the easiest way to see that $g = 2$ but one can also count as wiiza did - (we did this together!)  However, in the spirit of the section - there's a shorter way (shown to me by a student) by just subdividing so that the faces have disks as boundaries.  It's a bit of a head twister, but here's a photo: 
