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For a CW complex $X$, let $X^k$ be its $k$-skeleton. The cellular chain complex isThe isomorphism uses excision and $H_k(D^k,S^{k-1})\cong \mathbb{Z}$, after choosing an orientation for each cell. In degree zero, take $X^{-1}=\varnothing$ and the free group on the vertices. The differential is the compositeThe long exact sequences of skeleton pairs imply $d_{k-1}{d}_k=0$. The cellular boundary formula describes its coefficient at a $(k-1)$-cell as the mapping degree of the attaching sphere after all other $(k-1)$-cells are collapsed. In degree one this is the signed difference of the two endpoints. The cellular homology theorem identifies the homology of this complex with singular homology: relative homology of successive skeletons is concentrated in their cell dimension, and the exact skeleton sequences leave precisely these kernels modulo images. This is the construction, rather than merely a count of cells.

The filtration ${\mathbb{CP}}^0\subset {\mathbb{CP}}^1\subset {\mathbb{CP}}^2$ gives one cell in each of dimensions zero, two and four, since ${\mathbb{CP}}^k\setminus {\mathbb{CP}}^{k-1}\cong {\mathbb{C}}^k$. There are no odd-dimensional cells, so every cellular differential vanishes. ThusThe complex projective line has one zero-cell and one two-cell. Its product with itself has one zero-cell, two two-cells and one four-cell. Its cellular chain complex again has zero differentials, givingA homeomorphism induces isomorphisms on homology groups. The second groups have different ranks, and therefore

Now orient the Complex projective plane by its complex coordinates. The inclusion of its two-skeleton ${\mathbb{CP}}^1$ identifies the fundamental class of a projective line with a generator $h$ of $H_2({\mathbb{CP}}^2;\mathbb{Z})\cong \mathbb{Z}$: the cellular generator has no incoming three-cell boundary and no outgoing boundary.

Take the projective lines $L_0=\{z_2=0\}$ and $L_1=\{z_1=0\}$. A path of unitary coordinate transformations takes one to the other, so their oriented homology classes are both $h$. They meet in the single point $[1:0:0]$. In its affine chart, let $w=z_1/z_0$ and $v=z_2/z_0$. Their tangent spaces are the complex $w$-axis and complex $v$-axis, so the intersection is a transverse intersection. The ordered real basesconcatenate to the complex orientation of the ambient four-manifold. Thus the local sign is $+1$, andThis computes the intersection form using distinct representatives of the same class, avoiding an ill-defined attempt to count a line intersecting itself as a set. Bilinearity now givesEquivalently, the map $H_2\to \mathrm{Hom}(H_2,\mathbb{Z})$ sending $a$ to $\lambda (a,-)$ is an isomorphism. This is exactly a unimodular intersection pairing, and proves projective lines generate a unimodular intersection form. Reversing the ambient orientation would change the matrix to $(-1)$ and would still be unimodular.

Let $V_4=⟨u,v\mid u^2=v^2=[u,v]=1⟩$. Define a surjectionIts kernel determines a connected regular four-sheeted cover of a wedge of circles and a real projective plane, say ${\tilde{X}}_n\to X_n$.

Use the CW structure on $X_n$ with one vertex, one edge for every $a_i$ and for $b$, and one two-cell attached along $b^2$. In the regular cover, the degree-two cellular chain group isThe cellular boundary formula gives the lifted attaching word $b^2$ the boundaryin the $b$-edge summand of $C_1$. ThereforeThere are no three-cells, so

If $X_n$ were homotopy equivalent to a compact connected surface $S$, the subgroup corresponding to this cover would produce a connected four-sheeted cover $\tilde{S}$ homotopy equivalent to ${\tilde{X}}_n$. A finite cover of a compact surface is again a compact connected surface. But the integral top homology of a compact connected surface has rank at most one, contradicting the displayed rank two. Hence $X_n$ is not homotopy equivalent to a compact connected surface for any $n\ge 1$.

Choose the intersection point $p=[({\theta }_0,{\theta }_1)]$ as the zero-cell. Let $a$ be the horizontal one-cell $S^1\times \{{\theta }_1\}$ and $b$ the vertical one-cell $\{{\theta }_0\}\times S^1$. The usual CW complex structure on the torus has one two-cell $c$ attached by the word $ab{a}^{-1}{b}^{-1}$, and the extra disk gives a two-cell $d$ attached along $b$ with degree one. Collapsing $a$ leaves one zero-cell $p$, one one-cell $b$, and the two two-cells $c,d$. The cellular boundary formula givesTherefore

The Excision theorem says that if $Z\subseteq A\subseteq X$ and $\overline{Z}\subseteq \mathrm{int}A$, then inclusion inducesIt lets us compute local homology in arbitrarily small neighborhoods.

Let $C\subseteq X$ be the image of $\{{\theta }_0\}\times S^1$. At a point of $X\setminus C$, a neighborhood is a disk, whose link is a circle. At a point of $C\setminus \{p\}$, three half-disks meet along their diameters, and the link is a Theta graph, with first homology ${\mathbb{Z}}^2$. At $p$, the two folds created by collapsing the horizontal circle give a dumbbell graph as link: two circles joined by an interval. Its first homology is again ${\mathbb{Z}}^2$. The local homology from a link therefore givesBecause local homology is invariant under a homeomorphism, every self-homeomorphism of $X$ preserves the rank-two locus $C$.

It must also preserve $p$. Indeed, $p$ is the unique point of $C$ whose sufficiently small local link is a dumbbell graph; every other point of $C$ has a theta-graph link. The connecting edge of a dumbbell graph is a bridge in a graph, whereas no edge of a theta graph is a bridge, so these local topological types are distinct.

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 ${\mathbb{CP}}^n$ has one cell in every even dimension from $0$ to $2n$, and ${\mathbb{CP}}^k$ is its $2k$-skeleton. Collapsing that subcomplex leaves one zero-cell and one cell in each dimension $2i$ for $k+1\le i\le n$. All cellular boundaries vanish, so

Let $x\in H^2({\mathbb{CP}}^n;\mathbb{Z})$ be the usual generator. For $i>k$, choose $u_i$ whose pullback under the quotient map is $x^i$. Naturality of the cup product and the cohomology ring of complex projective space giveTogether with the unit, this determines the ring; equivalently, its reduced part is the ideal $(x^{k+1})/(x^{n+1})$ with the inherited multiplication. This is the cohomology ring of a collapsed projective subspace.

If $n=k+1$, the quotient has just a zero-cell and a $2n$-cell, so it is $S^{2n}$ and is a compact manifold. Conversely, suppose $n\ge k+2$ and the quotient is homotopy equivalent to a compact manifold. Its top cohomology is $\mathbb{Z}$, so that manifold must be closed, orientable, and $2n$-dimensional. But $H^{2n-2}\cong \mathbb{Z}$ while $H^2=0$, contradicting Poincare duality. Therefore

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