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Let $X$ be a finite CW complex, with $c_i$ topological cells of dimension $i$ and none above $d$. Its Euler characteristic is the displayed alternating sum. For a finite simplicial complex, $c_i$ counts its $i$-dimensional simplexes, each counted once rather than once for each containing simplex. In dimension two this becomes $V-E+F$. A topological cell's interior is counted; identified parts of its boundary are counted in their own lower-dimensional topological cells.^[ref]

The cellular homology theorem identifies cellular homology with singular homology, so the count equals the alternating rational Betti numbers. These are the free ranks of integral homology; torsion subgroups do not contribute to them. Over any field, the alternating homology dimensions still give the same Euler characteristic, although individual Betti numbers can depend on the field.

A CW filtration builds a CW complex by its skeletawhere each quotient $X^n/X^{n-1}$ is a wedge of $n$-spheres, one for each $n$-cell.

The Euler characteristic compresses a finite CW complex's homology into one integer: count its topological cells with alternating signs, or count its Betti numbers with the same signs. A filled triangle has Euler characteristic one; its cycle graph boundary has Euler characteristic zero. The missing face changes the answer even though the vertices and edges are identical.

For a finite CW complex $X$, including the geometric realization of a finite simplicial complex,Here $b_i={\dim }_{\mathbb{Q}}{H}_i(X;\mathbb{Q})$ is the rational Betti number. More generally this ordinary homology definition makes sense whenever every rational homology group is finite-dimensional and only finitely many are nonzero. An arbitrary infinite CW complex need not satisfy these assumptions; its formal infinite alternating cell count is not a definition of this invariant.

The usefulness is that a finite description can reveal something about a space without computing every homology group. By the cell-count formula for Euler characteristic, count pieces rather than solve all the boundary operators. If two spaces give different answers, topological invariance of Euler characteristic rules out a homeomorphism, and even a homotopy equivalence. Equal answers are less decisive: the Euler characteristic of an annulus and the Euler characteristic of a torus are both zero, while their homology groups differ. This is a compact diagnostic, with a clearly defined limit on what it can tell us.

The boundary of a cube has eight vertices, twelve edges and six square faces. Treating each open face as a two-dimensional topological cell givesThis agrees with its being a two-sphere: stretching the hollow polyhedral surface changes its geometry but not its topology. The interior is excluded from this space.

Now include the solid interior. The cube has the same lower-dimensional topological cells, together with one open three-dimensional topological cell. The cell-count formula for Euler characteristic becomesThe filled cube is contractible, so its homology agrees with that of a point. The term $-1$ is essential: $V-E+F$ by itself is the full count only for a two-dimensional CW complex.

Adding a diagonal within one square face replaces that face by two triangular faces and introduces one edge. Thus the Euler characteristic changes by $-1+1=0$, as in subdivision invariance of Euler characteristic. Including an interior is a different operation: it adds points to the space, rather than giving the same space a finer description. This distinction matters whenever a geometric model could mean either a surface or the solid it encloses.

A filled triangle is a disk with three vertices, three edges and one two-dimensional topological cell, so $\chi =3-3+1=1$. Its simplicial homology has $(b_0,b_1,b_2)=(1,0,0)$: the filled triangle is contractible, and the three-edge chain cycle is the chain boundary of its face. Keeping only the cycle graph on those three vertices gives $\chi =3-3=0$ and $(b_0,b_1)=(1,1)$, since that chain cycle no longer bounds a face. The surrounding plane's exterior is not part of either simplicial complex; do not add it to $F$.

Build a closed annulus from two nested $n$-sided polygons, $n\ge 3$, with $n$ radial edges joining corresponding vertices. There are $2n$ vertices, $3n$ edges and $n$ quadrilateral topological cells, giving $\chi =2n-3n+n=0$. The central missing region and the exterior are not faces of the annulus. A deformation retraction onto a circle gives $(b_0,b_1,b_2)=(1,1,0)$, agreeing with the cell count. This is a useful reduction: the annulus is two-dimensional, yet its homology is that of a one-dimensional cycle graph.

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.

For a concrete example of cellular homology of real projective space, the real projective plane ${\mathbb{RP}}^2$ has one topological cell in each of dimensions zero, one and two. Its one-dimensional topological cell is a loop, and the face attaches around that loop twice. Choose orientations so that the cellular chain complex over the integers iswith nonzero chain groups in degrees two, one and zero. The twice-around attachment is why the first boundary operator shown is multiplication by two.^[ref]

Over the rational numbers, multiplication by two is invertible, so $H_2$ and $H_1$ vanish, while $H_0\cong \mathbb{Q}$. Over the finite field ${\mathbb{F}}_2$, two equals zero. Both boundary operators then vanish, giving one-dimensional homology groups in all three degrees. Their respective Betti numbers areNevertheless,The direct cell count is also $1-1+1=1$.

This calculation makes the strength and loss of information visible. The Euler characteristic survives a change of the field used for coefficients, while individual homology groups can change. Over the integers, the same chain complex has $H_1\cong \mathbb{Z}/2\mathbb{Z}$ and $H_2=0$: a torsion subgroup that rational Betti numbers do not record. The rank cancellation in the Euler-Poincare formula explains why the alternating total still agrees with the cell count.

For finite-dimensional Complex projective space ${\mathbb{CP}}^r$, use the filtration by its linear projective subspaces. The difference ${\mathbb{CP}}^j\setminus {\mathbb{CP}}^{j-1}$ is ${\mathbb{C}}^j$, so there is one topological cell in every even dimension $0,2,\dots ,2r$, and none in odd dimensions. Every cellular boundary vanishes for degree reasons. The integral cohomology is therefore $\mathbb{Z}$ in those even degrees and zero elsewhere.

For the cohomology ring, let $x\in H^2({\mathbb{CP}}^r;\mathbb{Z})$ be the Poincare dual of a projective hyperplane, using the complex orientation. Generic hyperplanes meet transversely, and a collection of $j$ of them intersects in ${\mathbb{CP}}^{r-j}$. The geometric interpretation of the cup product identifies $x^j$ with the dual of that intersection. Restricting to a linear ${\mathbb{CP}}^j$, the $j$ hyperplanes meet in one point, with positive sign because the transverse normal directions are complex lines. HenceThus $x^j$ is a primitive generator in degree $2j$, not merely a nonzero multiple. There are no classes beyond degree $2r$, soThe point case $r=0$ has ring $\mathbb{Z}$. Taking the union of the finite projective spaces gives $H^{\ast }({\mathbb{CP}}^{\infty };\mathbb{Z})=\mathbb{Z}[x]$, since each fixed cohomological degree stabilizes. This proves the cohomology ring of complex projective space computation.

For the rank bound for a nonsingular complex bilinear map, write $a={\dim }_{\mathbb{C}}U$, $b={\dim }_{\mathbb{C}}V$, $R=\mathrm{im}\varphi$, and $s={\dim }_{\mathbb{C}}R$. Assume $a,b\ge 1$. Injectivity with either nonzero factor fixed implies $\varphi (u\otimes v)\ne 0$ whenever $u,v\ne 0$. Thereforeis well defined and continuous. Scaling either vector only scales the output, and in local projective charts the formula is given by bilinear coordinates with no simultaneous zero. Let $h$ be the target's degree-two generator and $x,y$ the pullbacks of the two factor generators. The Künneth theorem givesFixing $[v]$ or $[u]$ makes $F$ a projective linear embedding induced by an injective linear map. Such an embedding pulls back the hyperplane class to the factor's hyperplane class. Since the two restrictions determine $H^2$ of the product, $F^{\ast }h=x+y$, with the generator of a point factor interpreted as zero.

Put $d=a+b-2$. In the source cohomology ring, every term in $(x+y{)}^d$ vanishes except the one with exponents $a-1,b-1$, andThe binomial coefficient is a positive integer and the final monomial generates the top integral cohomology. If $s\le d$, the class $h^d$ in ${\mathbb{CP}}^{s-1}$ would be zero. This contradiction provesWhen $a=b=1$, the nonzero pure-tensor image already gives $s\ge 1$. A positivity convention is genuinely necessary: if $U=0$, $V={\mathbb{C}}^2$, and $W=0$, all restrictions to the zero tensor subspaces are injective, but the literal bound would read $0\ge 1$. The displayed theorem is therefore the nonzero-factor version. If a nonzero factor is infinite-dimensional, either factor restriction already bounds the image dimension below by that factor dimension; the sum of two positive cardinal dimensions, with at least one infinite, is their maximum, giving the corresponding cardinal bound.

Label the six vertices $A,B,C,D,E,F$ clockwise, starting at the upper-left vertex. The arrows identify $AB$ with $ED$, $BC$ with $FE$, and $FA$ with $DC$, preserving the indicated directions. Consequently the quotient space has exactly two vertices,Orient its three topological cells by $a:A\to B$, $b:B\to C$, and $c:F\to A$. The clockwise boundary of the hexagonal topological cell traversesEvery topological cell of dimension one therefore occurs once with each sign. The cellular chain complex, with integral coefficients, isHere the rows correspond to $P,Q$. The image of ${\partial }_1$ is the primitive subgroup generated by $Q-P$, and its kernel is freely generated by $a+b$ and $a+c$. Thus the homology groups areThe two vertex classes are important: identifying all six vertices would give an incorrect cellular chain complex. The result agrees with the homology of a torus.

The $q$-skeleton of a CW complex is the subcomplex consisting of its topological cells of dimension at most $q$. A cellular map preserves these skeleta. The cellular approximation theorem lets continuous maps respect this filtration up to homotopy.

Collapse the indicated standard skeleton to a point. Besides its basepoint, the quotient has one cell in dimensions $k+1$ through $m$. Its integral cellular boundary is two in even dimensions and zero in odd dimensions. Steenrod operations on its relative cohomology are inherited from real projective space.