The following examples illustrate why computations done with cellular homology are often more efficient than those calculated by using singular homology alone.
The n-sphere The
n-dimensional sphere Sn admits a CW structure with two cells, one 0-cell and one
n-cell. Here the
n-cell is attached by the constant mapping from S^{n-1} to the 0-cell. Since the generators of the cellular chain groups {H_{k}}(S^n_{k},S^{n}_{k - 1}) can be identified with the
k-cells of
Sn, we have that {H_{k}}(S^n_{k},S^{n}_{k - 1})=\Z for k = 0, n, and is otherwise trivial. Hence for n>1, the resulting chain complex is :\dotsb\overset{\partial_{n+2}}{\longrightarrow\,}0 \overset{\partial_{n+1}}{\longrightarrow\,}\Z \overset{\partial_n}{\longrightarrow\,}0 \overset{\partial_{n-1}}{\longrightarrow\,} \dotsb \overset{\partial_2}{\longrightarrow\,} 0 \overset{\partial_1}{\longrightarrow\,} \Z {\longrightarrow\,} 0, but then as all the boundary maps are either to or from trivial groups, they must all be zero, meaning that the cellular homology groups are equal to :H_k(S^n) = \begin{cases} \mathbb Z & k=0, n \\ \{0\} & \text{otherwise.} \end{cases} When n=1, it is possible to verify that the boundary map \partial_1 is zero, meaning the above formula holds for all positive n.
Genus g surface Cellular homology can also be used to calculate the homology of the
genus g surface \Sigma_g. The
fundamental polygon of \Sigma_g is a 4n-gon which gives \Sigma_g a CW-structure with one 2-cell, 2n 1-cells, and one 0-cell. The 2-cell is attached along the boundary of the 4n-gon, which contains every 1-cell twice, once forwards and once backwards. This means the attaching map is zero, since the forwards and backwards directions of each 1-cell cancel out. Similarly, the attaching map for each 1-cell is also zero, since it is the constant mapping from S^0 to the 0-cell. Therefore, the resulting chain complex is : \cdots \to 0 \xrightarrow{\partial_3} \mathbb{Z} \xrightarrow{\partial_2} \mathbb{Z}^{2g} \xrightarrow{\partial_1} \mathbb{Z} \to 0, where all the boundary maps are zero. Therefore, this means the cellular homology of the genus g surface is given by : H_k(\Sigma_g) = \begin{cases} \mathbb{Z} & k = 0,2 \\ \mathbb{Z}^{2g} & k = 1 \\ \{0\} & \text{otherwise.} \end{cases} Similarly, one can construct the genus g surface with a crosscap attached (a non-orientable genus g surface) as a CW complex with one 0-cell, g 1-cells \{a_1, \dotsm, a_g\} , and one 2-cell which is attached along the word a_1^1\dotsm a_g^2 . Therefore, the resulting chain complex is: \cdots \to 0 \xrightarrow{\partial_3} \mathbb{Z} \xrightarrow{\partial_2} \mathbb{Z}^{g} \xrightarrow{\partial_1} \mathbb{Z} \to 0, where the boundary maps are \partial_3=\partial_1 =0 and \partial_2(1)=2a_1+2a_2+\dotsm + 2a_g = 2(a_1+\dotsm+a_g). Its homology groups are H_k(N_g) = \begin{cases} \mathbb{Z} & k = 0 \\ \mathbb{Z}^{g-1} \oplus \Z_2 & k = 1 \\ \{0\} & \text{otherwise.} \end{cases}
Torus The n-torus (S^1)^n can be constructed as the CW complex with one 0-cell, n 1-cells, ..., and one n-cell. The chain complex is 0\to \Z^{\binom{n}{n}} \to \Z^{\binom{n}{n-1}} \to \cdots \to \Z^{\binom{n}{1}} \to \Z^{\binom{n}{0}} \to 0 and all the boundary maps are zero. This can be understood by explicitly constructing the cases for n = 0, 1, 2, 3, then see the pattern. Thus, H_k((S^1)^n) \simeq \Z^{\binom{n}{k}} .
Complex projective space If X has no adjacent-dimensional cells, (so if it has n-cells, it has no (n-1)-cells and (n+1)-cells), then H_n^{CW}(X) is the
free abelian group generated by its n-cells, for each n. The complex projective space \mathbb CP^n is obtained by gluing together a 0-cell, a 2-cell, ..., and a (2n)-cell, thus H_k(\mathbb CP^n) = \Z for k = 0, 2, ..., 2n, and zero otherwise.
Real projective space The
real projective space \mathbb{R} P^n admits a CW-structure with one k-cell e_k for all k \in \{0, 1, \dots, n\}. The attaching map for these k-cells is given by the 2-fold covering map \varphi_k \colon S^{k - 1} \to \mathbb{R} P^{k - 1}. (Observe that the k-skeleton \mathbb{R} P^n_k \cong \mathbb{R} P^k for all k \in \{0, 1, \dots, n\}.) Note that in this case, H_k(\mathbb{R} P^n_k, \mathbb{R} P^n_{k - 1}) \cong \mathbb{Z} for all k \in \{0, 1, \dots, n\}. To compute the boundary map : \partial_k \colon H_k(\mathbb{R} P^n_k, \mathbb{R} P^n_{k - 1}) \to H_{k - 1}(\mathbb{R} P^n_{k - 1}, \mathbb{R} P^n_{k - 2}), we must find the degree of the map : \chi_k \colon S^{k - 1} \overset{\varphi_k}{\longrightarrow} \mathbb{R} P^{k - 1} \overset{q_k}{\longrightarrow} \mathbb{R} P^{k - 1}/\mathbb{R} P^{k - 2} \cong S^{k - 1}. Now, note that \varphi_k^{-1}(\mathbb{R} P^{k - 2}) = S^{k - 2} \subseteq S^{k - 1}, and for each point x \in \mathbb{R} P^{k - 1} \setminus \mathbb{R} P^{k - 2}, we have that \varphi^{-1}(\{x\}) consists of two points, one in each connected component (open hemisphere) of S^{k - 1}\setminus S^{k - 2}. Thus, in order to find the degree of the map \chi_k, it is sufficient to find the local degrees of \chi_k on each of these open hemispheres. For ease of notation, we let B_k and \tilde B_k denote the connected components of S^{k - 1}\setminus S^{k - 2}. Then \chi_k|_{B_k} and \chi_k|_{\tilde B_k} are homeomorphisms, and \chi_k|_{\tilde B_k} = \chi_k|_{B_k} \circ A, where A is the antipodal map. Now, the degree of the antipodal map on S^{k - 1} is (-1)^k. Hence,
without loss of generality, we have that the local degree of \chi_k on B_k is 1 and the local degree of \chi_k on \tilde B_k is (-1)^k. Adding the local degrees, we have that : \deg(\chi_k) = 1 + (-1)^k = \begin{cases} 2 & \text{if } k \text{ is even,} \\ 0 & \text{if } k \text{ is odd.} \end{cases} The boundary map \partial_k is then given by \deg(\chi_k). We thus have that the CW-structure on \mathbb{R} P^n gives rise to the following chain complex: : 0 \longrightarrow \mathbb{Z} \overset{\partial_n}{\longrightarrow} \cdots \overset{2}{\longrightarrow} \mathbb{Z} \overset{0}{\longrightarrow} \mathbb{Z} \overset{2}{\longrightarrow} \mathbb{Z} \overset{0}{\longrightarrow} \mathbb{Z} \longrightarrow 0, where \partial_n = 2 if n is even and \partial_n = 0 if n is odd. Hence, the cellular homology groups for \mathbb{R} P^n are the following: : H_k(\mathbb{R} P^n) = \begin{cases} \mathbb{Z} & \text{if } k = 0 \text{ and } k=n \text{ odd}, \\ \mathbb{Z}/2\mathbb{Z} & \text{if } 0 == Functoriality ==