Hatcher Exercise 2.2.9
We compute homology for a few two-dimensional CW-complexes:
(a) Let with the north and south poles identified. We’ve previously seen that this space is homotopy equivalent to , and so the homology should be:
But, let’s do it using cellular homology. To understand the CW-complex structure, first consider the CW-complex structure for (without identification) consisting of two 0-cells (the poles), a 1-cell between them, and a 2-cell attached by mapping one half-interval of the boundary along from to , and the consecutive half-interval along with opposite orientation – that is, from to (imagine the 2-cell in a Pacman shape – the attaching map is zipping the mouth shut by sending both halves of the mouth to the single line ).
Then, we can get a CW-complex on by modifying the this structure – just start with one 0-cell and have the attaching map of the 1-cell map both endpoints to – keep the attaching map the same as described above. Then, the cellular chain complex is:
Clearly is zero since there is only one zero-cell. is also zero because it maps the 2-cell along twice in opposite orientation. So, the boundary maps are zero and the homology groups are just the groups in the chain complex, which are the same as what we stated at the beginning.
(b) Let . This is two concentric tori meeting at their outer and inner edge, respectively. We can get a CW-complex structure as follows:
Where all vertices represent the single vertex , and the 2-cells are attached according to the word around their boundary. Then, the cellular chain complex is:
Again, is zero because there is only one 0-cell. Examining how maps the 2-cells onto each 1-cell, notice that every 1-cell appears around the boundary of both and exactly twice in opposite orientation (or no times). So, according to the cellular boundary formula sends both and to zero, so . Then, the homology is:
(c) Let be the space obtained from by deleting the interiors of two distinct subdiscs in the interior and then identifying all three resulting boundary circles together in an orientation preserving way. We can put a CW-complex structure on by making ‘cuts’ up to the two removed holes, so that the space can be described by a polygon with edges identified according to . Then, we have a CW-complex structure with one 0-cell, three 1-cells, and a 2-cell attached according to the boundary word. Then, the cellular chain complex is:
As with previous problems, is zero because there is only one 0-cell. By the cellular boundary formula, ( and appear exactly in opposite-orientation pairs). Then, , and . Then, we have . So,
(d) Let be the space obtained from by identifying the points differing by on one copy of , and the points differing by on the other copy of .
When we look at this identification on one circle, its clear that we get a wedge of (respectively ) circles. If we view as a square with boundary identified, this can be interpreted as dividing up each edge into unique new edges. Then, we can put a CW-complex structure on like this (for and ):
Where all the points are identified to a single vertex, and as such the internal edges are identified in the obvious way (to the ones they are parallel with). This gives a CW-complex structure with one 0-cell, 1-cells, and 2-cells, each attached in a miniature torus patch according to . The cellular chain complex is:
Again, is zero. As it turns out, is zero as well since every edge appears exactly in an opposite-orientation pair in the boundary word(s) for each face. Then, we have:
Indeed, notice that this coincides with the homology of the torus when . Furthermore, the space from part (b) is precisely this construction when and , and the homology coincides there as well.