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For a CW complex, the qth cellular chain group is the relative homology group of successive skeleta, a free abelian group with one generator for each q-cell. The attaching maps determine the cellular boundary, and the resulting cellular chain complex computes singular homology.

Identify opposite sides of a square by translations to obtain a torus. All four corners become one vertex; the horizontal sides give one topological cell $a$ and the vertical sides one topological cell $b$. The open square supplies one two-dimensional topological cell. Thus $(c_0,c_1,c_2)=(1,2,1)$ and $\chi =0$. The cellular boundary of each one-dimensional topological cell is zero because its endpoints agree. Traversing the face boundary gives the group commutator $ab{a}^{-1}{b}^{-1}$, whose signed counts of both $a$ and $b$ vanish. Hence its cellular boundary is also zero, and $(b_0,b_1,b_2)=(1,2,1)$.^[ref]

The boundary of a tetrahedron is a two-sphere, with four vertices, six edges and four triangular topological cells: $\chi =4-6+4=2$. Its rational Betti numbers are $(1,0,1)$ by homology of a sphere. The torus's two independent one-dimensional homology classes cancel its degree-zero and degree-two contributions. The torus and annulus both have Euler characteristic zero, yet different homology groups; one integer cannot recover all of homology.

A CW complex is built by attaching q-dimensional discs to the $(q-1)$-skeleton along maps of their boundary spheres, with the closure-finite and weak-topology conditions. Write $X^q$ for its q-skeleton and $X^{-1}=\varnothing$. The cellular chain group isThe cellular boundary is the composite of the connecting map and the relative quotient map:Consecutive maps in the relevant exact sequences compose to zero, so these maps make a cellular chain complex. Equivalently, a boundary coefficient is the degree of the attaching sphere followed by collapse onto the sphere of a selected lower cell. Its cellular homology is $\ker d_q/\mathrm{im}{d}_{q+1}$. The relative groups of successive skeleta vanish off their cell dimension; their exact sequences identify this homology with singular homology, the cellular homology theorem.

For $k\ge 1$, the sphere $S^k$ has a zero-cell and a k-cell with zero boundary. The Complex projective space ${\mathbb{CP}}^n$ has one cell in each dimension $0,2,\dots ,2n$ and no odd cells, so all cellular boundaries vanish. ConsequentlyAt $k=0$, the sphere consists of two points and $H_0(S^0)={\mathbb{Z}}^2$; at $n=0$, projective space is a point.

Put $a=2n-2$ and $Z=S^a\times S^3$. Its product CW complex has cells in dimensions $0,3,a,a+3=2n+1$. For $n>3$, these positive dimensions are distinct and never consecutive, so all cellular boundaries vanish. ThusIts $2n$-skeleton is $S^a\vee S^3$. Collapsing it gives the quotientIt induces an isomorphism on top homology, with degree $+1$ after choosing the quotient orientation. Let $p:S^{2n+1}\to {\mathbb{CP}}^n$ be the Hopf fibration and set $\varphi =pq$, the Hopf projection of a product smash quotient.

On reduced homology, the collapse kills the two lower-dimensional classes, and the remaining top class maps to zero because projective space has no odd-dimensional homology. Therefore ${\varphi }_{\ast }=0$ in every reduced degree. On homotopy groups, ${\pi }_j(Z)\cong {\pi }_j(S^a)\oplus {\pi }_j(S^3)$ for $j\ge 2$, with summands induced by the two factor inclusions. Both inclusions are collapsed by $q$, so $q_{\ast }$ and hence ${\varphi }_{\ast }$ vanish on both summands. The domain is simply connected and connected, which also handles the lower groups.

It remains to prove that $\varphi$ is not null-homotopic. The Hopf fibration has fibre $S^1$ and has the homotopy lifting property for the CW domain $Z$: a homotopy downstairs and a lift of its initial map have a lift with that specified initial value. If $\varphi$ were homotopic to a constant, lift this homotopy starting at $q$. The ending map would land in the single fibre $S^1$. Every map $Z\to S^1$ is null-homotopic, since the simply connected domain permits its lift to the contractible universal cover $\mathbb{R}\to S^1$. Thus $q$ would be null-homotopic in $S^{2n+1}$, contradicting its nonzero top homology map. Hence

Use the standard CW complex structure on the real projective plane, with one cell $e_r$ for each $r=0,1,2$. The attaching map of the two-cell winds twice around the one-cell, so the cellular chain complex has $d{e}_2=2e_1$ and $d{e}_1=0$.

The product cells $e_{ij}=e_i\times e_j$, $0\le i,j\le 2$, give nine cells, in dimensions $0,1,2,3,4$ with counts $1,2,3,2,1$. The cellular chains of a product of finite CW complexes have boundary $d(e_i\times e_j)=d{e}_i\times e_j+(-1{)}^i{e}_i\times d{e}_j$. In the ordered basesthe nonzero matrices areIn particular $d_2{d}_3=d_3{d}_4=0$. The degree-one homology is ${\mathbb{Z}}^2/2{\mathbb{Z}}^2$. The degree-two cellular cycles are $\mathbb{Z}{e}_{11}$, and the cellular boundaries there are $2\mathbb{Z}{e}_{11}$. The degree-three cycles are $\mathbb{Z}(e_{21}+e_{12})$, whose double is the boundary of $e_{22}$. Finally $d_4$ is injective. This computes the integral homology of two real projective planes:The degree-three torsion is the Tor functor contribution in the Künneth theorem; it would be missed by retaining only tensor products of the two factors' homology groups.