We compute homology for a few two-dimensional CW-complexes:

(a) Let $X = S^2$ with the north and south poles identified. We've previously seen that this space is homotopy equivalent to $S^2 \vee S^1$, and so the homology should be:

$$H_0(X) = \mathbb{Z} \qquad H_1(X) = \mathbb{Z} \quad H_2(X) = \mathbb{Z}$$

But, let's do it using cellular homology. To understand the CW-complex structure, first consider the CW-complex structure for $S^2$ (without identification) consisting of two 0-cells $p,q$ (the poles), a 1-cell $e$ between them, and a 2-cell attached by mapping one half-interval of the boundary along $e$ from $p$ to $q$, and the consecutive half-interval along $e$ with opposite orientation -- that is, from $q$ to $p$ (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 $e$).

Then, we can get a CW-complex on $X$ by modifying the this structure -- just start with one 0-cell $p$ and have the attaching map of the 1-cell map both endpoints to $p$ -- keep the attaching map the same as described above. Then, the cellular chain complex is:

$$0 \rightarrow \mathbb{Z} \xrightarrow{d_2} \mathbb{Z} \xrightarrow{d_1} \mathbb{Z} \rightarrow 0$$

Clearly $d_1$ is zero since there is only one zero-cell.

$$d_2$$ is also zero because it maps the 2-cell along $$e$$ 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 $$X = S^1 \times (S^1 \vee S^1)$$. This is two concentric tori meeting at their outer and inner edge, respectively. We can get a CW-complex structure as follows: <div class="math-figure"><img src="/assets/math_solutions/hatcher/e2-2-9_1.svg" width="300"/></div> Where all vertices represent the single vertex $$p$$, and the 2-cells are attached according to the word around their boundary. Then, the cellular chain complex is:$$

0 \rightarrow \mathbb{Z}^2 \xrightarrow{d_2} \mathbb{Z}^3 \xrightarrow{d_1} \mathbb{Z} \rightarrow 0

$$Again, $$d_1$$ is zero because there is only one 0-cell. Examining how $$d_2$$ maps the 2-cells onto each 1-cell, notice that every 1-cell appears around the boundary of both $$F$$ and $$G$$ exactly twice in opposite orientation (or no times). So, according to the cellular boundary formula $$d_2$$ sends both $$F$$ and $$G$$ to zero, so $$d_2 = 0$$. Then, the homology is:$$

H_0(X) = \mathbb{Z} \qquad H_1(X) = \mathbb{Z}^3 \qquad H_2(X) = \mathbb{Z}^2

$$--- **(c)** Let $$X$$ be the space obtained from $$D^2$$ by deleting the interiors of two distinct subdiscs in the interior and then identifying all three resulting boundary circles together in an orientation preserving way. We can put a CW-complex structure on $$X$$ by making 'cuts' up to the two removed holes, so that the space can be described by a polygon with edges identified according to $$aba^{-1} b^{-1} c a^{-1} c^{-1}$$. Then, we have a CW-complex structure with one 0-cell, three 1-cells, and a 2-cell $$F$$ attached according to the boundary word. Then, the cellular chain complex is:$$

0 \rightarrow \la F \ra \xrightarrow{d_2} \la a,b,c \ra \xrightarrow{d_1} \la p \ra \rightarrow 0

$$As with previous problems, $$d_1$$ is zero because there is only one 0-cell. By the cellular boundary formula, $$d_2(F) = -a$$ ($$b$$ and $$c$$ appear exactly in opposite-orientation pairs). Then, $$\im d_2 = \la -a \ra$$, and $$\ker d_2 = 0$$. Then, we have $$H_1(X) = \la a,b,c \ra / \la a \ra \cong \mathbb{Z}^2$$. So,$$

H_0(X) = \mathbb{Z} \qquad H_1(X) = \mathbb{Z}^2 \qquad H_2(X) = 0

$$**(d)** Let $$X$$ be the space obtained from $$S^1 \times S^1$$ by identifying the points differing by $$2\pi/m$$ on one copy of $$S^1$$, and the points differing by $$2 \pi / n$$ on the other copy of $$S^1$$. When we look at this identification on one circle, its clear that we get a wedge of $$n$$ (respectively $$m$$) circles. If we view $$S^1 \times S^1$$ as a square with boundary identified, this can be interpreted as dividing up each edge into $$n$$ unique new edges. Then, we can put a CW-complex structure on $$X$$ like this (for $$n=3$$ and $$m=2$$): <div class="math-figure"><img src="/assets/math_solutions/hatcher/e2-2-9_2.svg" width="450"/></div> Where all the points are identified to a single vertex, and as such the internal edges are identified in the obvious way (to the ones they are parallel with). This gives a CW-complex structure with one 0-cell, $$n+m$$ 1-cells, and $$nm$$ 2-cells, each attached in a miniature torus patch according to $$a_i b_j a_i^{-1} b_j^{-1}$$. The cellular chain complex is:$$

0 \rightarrow \la F_{ij} \; : \; i \in [n], j \in [m] \ra \xrightarrow{d_2} \la a_i, b_j \; : \; i \in [n], j \in [m] \ra \xrightarrow{d_1} \la p \ra \rightarrow 0

$$Again, $$d_1$$ is zero. As it turns out, $$d_2$$ is zero as well since every edge appears exactly in an opposite-orientation pair in the boundary word(s) for each face. Then, we have:$$

H_0(X) = \mathbb{Z} \qquad H_1(X) = \mathbb{Z}^{n+m} \qquad H_2(X) = \mathbb{Z}^{nm}

$$

Indeed, notice that this coincides with the homology of the torus when $n=1,m=1$. Furthermore, the space from part (b) is precisely this construction when $n=2$ and $m=1$, and the homology coincides there as well.