Nerves of good covers for projective plane and torus in Bott-Tu I am completely stuck on two examples in Bott-Tu, p. 105 and 106. There, two nerves for open covers of the projective plane and the torus are given. My problem is that I don't visualize these covers, and consecutively, I don't see what could go wrong in any arbitrary such picture (with care on identifications). 
For instance, in the picture of nerve for projective plane, do the 0 and 1 open subset intersect? It might sound silly, but a nerve is a simplicial complex, while here they are linked by an arc...Also, are there any 'filled' triangles in this picture?
Last but not least, it seems like the cohomology computations in this paragraph can be done without visualizing anything (just by counting intersections etc). Is this a good perspective?
Thanks for any help.
 A: I am going to go over the projective plane and give you an idea how it looks like and give an intuition why the nerve has 6 vertices.   

To visualize the real projective plane,  we begin with an octahedral. We remove two triangles ABD and BEC. Next we construct inside the   octahedral a surface I have called Surface 1. It is made from three triangles AOD, DOB and DBC. If you follow the top arrows, you see it sort of wraps around from A to C. Note the two dashed triangles  AOD and DOB are our additions and not part of the octahedral we started with but DBC is part of it. We do the same thing on the bottom but on the opposite side and we get Surface 0. (again follow the arrows below 0). Note the two surfaces meet exactly along line OB.  We have in fact at this point constructed something which is precisely equivalent to Möbius strip. Now if we fill the triangle ABC (Surface 2), we get the projective plane. Projective plane is indeed a Möbius strip to whose boundary one attaches a disk (or a triangle in our case,  topologically equivalent). You might wonder how we go from the strip to this construction. Imagine the Möbius strip as a flexible surface and its boundary like a wire. Grab it and move it apart until you get an "8" shape, then untwist the 8 to get a circle. After all if topologically the boundary  of the Möbius strip is a circle, you should be able to deform it into a circle. If you do this, the flexible strip will have  this  "wrap-around" and cross its own boundary!! This is the surface  ODBE crossing ABC. So when attach the disk, it looks like ODBE has to cut through Surface 2 but it really does not. This is because you cannot immerse the projective plane in 3D. 
Now note that we have four triangles behind ADCE on the back of our octahedral (the dashed lines should guide you to where they join on the back). You can think of the backside to be topologically speaking  a hemisphere. This is the "simple" side of the strip where there is no twisting going on. Recall how you make the Mobius strip from a piece of paper by twisting the end and gluing it. In our figure the "twisted" side where all the action is resides on the front of the picture. 
Now in Bott and Tu:
http://www.maths.ed.ac.uk/~aar/papers/botttu.pdf
take a look at page 101, Figure 9.3. Here you see the hemisphere on the bottom is covered by 3 open sets, sort of like three sepals wrapping around a rosebud. Now take a look at the nerve in the question, Figure 9.6. The covers 3,4,5 are in fact such a cover for the back our of octahedral after we deform it as I just mentioned into a hemisphere.
How about the nerves 0, 1, 2? These are roughly speaking the surfaces 0,1,2 in our figure, 2 is surface 2  etc. Our figure has 2-symmetry 0 and 1 being anti-symmetric on top and bottom. And Figure 9.6 is symmetric with respect to 0,1,2. So you have sort of take the Surfaces 0,1,2 and divide them with overlaps into three "overlapping" parts. It is hard to show. But, I hope you get the idea. 
We cannot use less than 3 covers on the back for the same reason we had to use three in Figure 9.3; 2 is not enough to give us a good cover. So 3 is  really the minimum there. And three on the front should now be obvious. 
You can now begin from Figure 9.6. and visualize 3,4 and 5 as three  intersecting circles (like a ven diagram). Also imagine 0, 1, 2 as overlapping annulus. Try to expand them toward the center to cover the projective plane while being mindful of the identification along the boundary. You should be able to see that in fact, all the triangles in the figure are "full" meaning you do get triple intersections (10 of them) but no quadruples. 
We are ready to determine the cohomology:  There are 6 vertices and hence the dimension of direct first direct product is 6. The kernel of the first map is 1-dimensional, you have basically 15 equations where the differences have to be zero and just 6 variables for the forms; they all have to be equal. So $H^0=R$. 
The image of the first map which is the kernel of the second map is 6-1=5 dimensional and so the the image of the second map is 15-5=10 (15 because if you count all the edges of Figure 9.6, you get 15). This means $H^1=0$ (10-10=0). There are no higher order intersections, and so everything else is zero as well.   
