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Write $S^n$ as the union of slightly enlarged northern and southern hemispheres $U$ and $V$. Both are contractible, while $U\cap V$ deformation retracts onto $S^{n-1}$. The reduced Mayer–Vietoris theorem therefore givesStarting from ${\tilde{H}}_0(S^0;\mathbb{Z})\cong \mathbb{Z}$ proves the homology of a sphere:This uses no cellular homology. A reflection of $S^n$ reverses its orientation and has degree of a continuous mapping $-1$, so it induces the identity on $H_0$, multiplication by $-1$ on $H_n$, and the unique map between zero groups in every other degree.

For a CW complex with skeleta $X^k$, its cellular chain complex isThe differential is the connecting map to $H_{k-1}(X^{k-1})$ followed by passage to $H_{k-1}(X^{k-1},X^{k-2})$. Equivalently, the coefficient of a $(k-1)$-cell in the boundary of a $k$-cell is the degree obtained from its attaching map after collapsing the complement of that lower cell. This is the cellular boundary formula.

The quotient $D^k/(x\sim -x\text{ on }{S}^{k-1})$ builds ${\mathbb{RP}}^k$ from ${\mathbb{RP}}^{k-1}$ by one $k$-cell, so ${\mathbb{RP}}^n$ has one cell in each dimension $0,\dots ,n$. The two lifts of the attaching map contribute with relative sign $(-1{)}^k$, and the cellular homology of real projective space has differentialConsequentlyWith ${\mathbb{F}}_2=\mathbb{Z}/2$ coefficients every differential vanishes, and hencewith zero homology outside that range.

The standard CW complex structure on Real projective space ${\mathbb{RP}}^n$ has one cell in each dimension $0,1,\dots ,n$. Its cellular chain complex has $C_q\cong \mathbb{Z}$ and differentialConsequentlyAfter tensoring the cellular complex with ${\mathbb{F}}_2=\mathbb{Z}/2$, every differential vanishes, so

The quotient is obtained from its image $Z\cong S^1$ by attaching the interior of $D^2$. The boundary map winds $n$ times around $Z$, so this is the cyclic Moore space in dimension one $M(\mathbb{Z}/n,1)$. Its cellular chain complex isTherefore

For a CW complex $X$, the cellular chain complex isthe free abelian group generated by the oriented $n$-cells. Its differential is the compositeNaturality of the long exact sequences of the triples $X^{n-2}\subset X^{n-1}\subset X^n$ makes two successive connecting maps compose to zero, so $d_{n-1}{d}_n=0$.

The cellular boundary formula says that the coefficient of an $(n-1)$-cell $e_{\alpha }^{n-1}$ in the boundary of $e_{\beta }^n$ is the degree of

Use the cellular chain complex or simplicial chain complex over $\mathbb{Q}$, rather than the infinite singular chain complex. Put $r_i=\mathrm{rank}{\partial }_i$, with $r_0=r_{d+1}=0$. The rank-nullity theorem gives $\dim \ker {\partial }_i=\dim C_i-r_i$; quotienting the chain cycles by the chain boundaries then gives $b_i=\dim C_i-r_i-r_{i+1}$. Each boundary operator's rank occurs in adjacent degrees with opposite signs, soFiniteness makes this cancellation an ordinary finite sum. This is the algebra behind the cell-count formula for Euler characteristic.

An open $q$-cell of a CW complex is a copy of an open $q$-dimensional ball. Its characteristic map is a map from a closed ball whose interior parametrizes that cell and whose boundary lands in the preceding skeleton. The characteristic map need not be injective on the boundary. Each oriented cell gives one generator of the cellular chain complex.

The abelianization of a knot group is infinite cyclic. Nielsen changes of a two-generator free basis reduce its exponent vector to $(1,0)$. In the cyclic-cover cellular chain complex, the first boundary is then $(t-1,0)$, whose kernel is a free rank-one Laurent module. Quotienting by the lifted relator boundaries gives a cyclic module, regardless of the number of relators. A noncyclic Alexander module therefore obstructs every two-generator knot-group presentation.

For a connected space, this is the connected covering corresponding to the kernel of the homomorphism from its fundamental group to its abelianization. Its deck transformation group is $H_1(Y;\mathbb{Z})$. Lifted cells make its cellular chain complex a complex over the group ring $\mathbb{Z}[H_1(Y)]$.