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A Betti number is the dimension of a real singular homology group; on a smooth manifold it is also the dimension of the corresponding de Rham cohomology group. For a finite-dimensional cohomology group on a closed Riemannian manifold, it equals the dimension of its space of harmonic representatives. Betti numbers can increase under a finite cover, so equality is a separate hypothesis, not a general covering-space theorem.

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.

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.

For a CW complex, the homology of its cellular chain complex naturally equals its singular homology. The relative groups of successive skeleton pairs, free on the cells in the corresponding dimension and zero in other dimensions, yield this identification through their connecting maps.

The Hawaiian earring is the union of a countable family of circles tangent at one point, with their radii tending to zero, endowed with its planar subspace topology. Equivalently it is the one-point compactification of a countable disjoint union of real lines. Every neighbourhood of the common point contains all but finitely many whole circles. This compact topological space is not locally contractible there and has different singular homology from the infinite CW complex wedge of circles.

Use integral singular homology. The singular chain group $C_n(B)$ is the free abelian group on continuous maps ${\Delta }^n\to B$, and its boundary operator isFor $i<j$, deleting vertex $j$ and then vertex $i$ has sign $(-1{)}^{i+j}$, while deleting $i$ and then the shifted vertex $j-1$ has sign $(-1{)}^{i+j-1}$. Both yield the same codimension-two face. Pairing these terms proves ${\partial }^2=0$. The inclusion $A\subset B$ identifies $C_n(A)$ with the subgroup spanned by those simplices having image in $A$, so it is injective and preserved by $\partial$.

The relative chain complex is $C_n(B,A)=C_n(B)/C_n(A)$ with induced boundary $\partial \bar{b}=\overline{\partial b}$. Thus there is a degreewise short exact sequence of chain complexesInclusion and quotient take homological cycles to homological cycles and boundaries to boundaries, hence define the group homomorphisms $i_{\ast }$ and $q_{\ast }$ on homology. We construct and prove exactness of the resulting sequence directly, rather than invoke an unproved algebraic lemma:ending in $H_0(A)\to H_0(B)\to H_0(B,A)\to 0$.

For the relative homology connecting homomorphism, lift a relative cycle $\bar{b}\in C_n(B,A)$ to $b\in C_n(B)$. Since $\partial \bar{b}=0$, the chain $\partial b$ belongs to $C_{n-1}(A)$; it is a homological cycle because ${\partial }^2=0$. DefineChanging the lift by $a\in C_n(A)$ changes the boundary by $\partial a$, hence does not change its homology class. More generally, if a different relative cycle represents the same relative homology class, lifts satisfy $b^{\prime }-b=\partial c+a$ for some $c\in C_{n+1}(B)$ and $a\in C_n(A)$. Then $\partial b^{\prime }-\partial b=\partial a$. This proves independence both of the lift and of the homology representative. Lifting a sum by the sum of lifts proves additivity, so $\delta$ is a well-defined group homomorphism.

At $H_n(A)$, an element in the image of $\delta$ is represented by $\partial b$ for some $b\in C_{n+1}(B)$, so its image in $H_n(B)$ is zero. Conversely, if a homological cycle in $A$ $a$ has $i_{\ast }[a]=0$, then $a=\partial b$ for a $B$-chain $b\in C_{n+1}(B)$. Its quotient $\bar{b}$ is a relative cycle and $\delta [\bar{b}]=[a]$. ThusFor completeness, the other terms are exact by the same elementary chain argument. At $H_n(B)$, clearly $q_{\ast }{i}_{\ast }=0$. If a homological cycle in $B$ $b$ has zero relative class, then $\bar{b}=\partial \bar{c}$ for a quotient chain $\bar{c}$. Lifting $c$ gives $a=b-\partial c\in C_n(A)$, and $\partial a=0$. Hence $[b]=i_{\ast }[a]$, proving $\ker q_{\ast }=\mathrm{im}{i}_{\ast }$. At $H_n(B,A)$, absolute homological cycles have zero connecting image. Conversely, if $\delta [\bar{b}]=0$, then $\partial b=\partial a$ for some $a\in C_n(A)$; the chain $b-a$ is an absolute homological cycle and maps to $[\bar{b}]$. Therefore $\ker \delta =\mathrm{im}{q}_{\ast }$. In degree zero every quotient chain lifts to an absolute homological cycle, because the differential to degree $-1$ is zero, so $q_{\ast }:H_0(B)\to H_0(B,A)$ is surjective. These proofs establish the whole long exact sequence in relative homology and all the algebraic facts used in its construction.

A chain complex consists of abelian groups $C_q$ and homomorphisms ${\partial }_q:C_q\to C_{q-1}$ with ${\partial }_{q-1}{\partial }_q=0$. Its groups of chain cycles and chain boundaries are $Z_q=\ker {\partial }_q$ and $B_q=\mathrm{im}{\partial }_{q+1}$; the identity ensures $B_q\subseteq Z_q$. Its homology groups areFor a nonnegative complex, take $C_q=0$ for $q<0$.

The singular chain group $C_q(X)$ is the free abelian group on continuous maps $\sigma :{\Delta }^q\to X$, called singular simplices. With ${\delta }_i$ the inclusion of the face opposite vertex $i$, defineDeleting any two vertices in the two possible orders gives the same face with opposite signs, proving ${\partial }^2=0$. This defines the singular chain complex and its singular homology. A continuous map $f$ induces the chain map $f_{\#}(\sigma )=f\circ \sigma$; it commutes with the boundary because composition commutes with restriction to faces.

To prove homotopy invariance of homology, let $F:X\times I\to Y$ have endpoints $f,g$. Triangulate ${\Delta }^q\times I$ by the $q+1$ simplices with ordered verticesLet $p_i:{\Delta }^{q+1}\to {\Delta }^q\times I$ be their affine parameterizations. The singular prism operator isOn taking boundaries, shared internal faces of consecutive prism simplices cancel. The top and bottom faces give $g_{\#}\sigma -f_{\#}\sigma$, and the remaining faces are $-P_{q-1}(\partial \sigma )$ with their induced signs. ThusFor a chain cycle $z$, $g_{\#}z-f_{\#}z=\partial Pz$ is a chain boundary, so $f_{\ast }=g_{\ast }$ on homology. If $f$ is a homotopy equivalence, its homotopy inverse therefore induces an inverse to $f_{\ast }$, proving that singular homology depends only on homotopy type.

Use the standard CW complex for an orientable genus-three surface: one vertex, six edges $a_1,b_1,a_2,b_2,a_3,b_3$, and a two-cell whose attaching word is ${\prod }_{j=1}^3[a_j,b_j]$. The cellular homology theorem identifies this cellular calculation with singular homology. The boundary of every edge is zero, and each edge's total exponent in the attaching word is zero, so the two-cell boundary is zero. HenceChoose $C_1$ to be the nonseparating curve represented by $a_1$, and choose $C_2$ as a small circle bounding a disc in a surface patch. Then $[C_1]$ is a nonzero primitive homology class, whereas the inclusion of $C_2$ induces zero on $H_1$. If a homotopy equivalence carried $C_1$ onto $C_2$, its restriction would factor through $C_2$ and kill $[C_1]$, contradicting injectivity on first homology. This is the homology obstruction to carrying surface curves onto one another.

There is nevertheless such a continuous map. Map the edge $a_1$ homeomorphically onto $S^1$ and all other edges to its base point. The two-cell attaching word then has winding number zero in $S^1$, hence is null-homotopic and the map extends to $r:{\Sigma }_3\to S^1$. Compose $r$ with a homeomorphism $S^1\to C_2$ and the inclusion of $C_2$ in the surface. This gives $g(C_1)=C_2$. Thus an arbitrary map can carry the first curve onto the second, but a homotopy equivalence cannot.

For an unambiguous drawing, realize ${\Sigma }_3$ as the double of a manifold formed from a disc with three holes. Its two copies are glued along all four boundary circles. The red curve below follows an arc from the outer boundary to the first hole on the front copy and returns along the matching arc on the back copy; it is a nonseparating curve. The blue circle is entirely in a disc patch on the front copy and is contractible.

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

The Excision theorem for singular homology states that if $Z\subseteq A\subseteq X$ and $\overline{Z}\subseteq {\mathrm{int}}_{X}A$, inclusion induces isomorphisms $H_q(X\setminus Z,A\setminus Z;\mathbb{Z})\cong H_q(X,A;\mathbb{Z})$ for all $q$. For an open cover $X=U\cup V$, its small-chain form says that the subcomplex of singular chains whose individual simplices lie in $U$ or $V$ computes $H_{\ast }(X)$. Concretely, repeated barycentric subdivision puts each finite chain in that subcomplex, and subdivision is chain-homotopic to the identity. There is then a short exact sequence of chain complexesIts long exact sequence in homology, with excision identifying the last complex's homology with $H_{\ast }(X)$, is the Mayer–Vietoris sequence:

The Real projective space ${\mathbb{RP}}^1$ is a circle. The projective plane has one cell in each dimension $0,1,2$, with cellular homology boundary maps $d_2=2$ and $d_1=0$. Thus, writing $A_q=H_q({\mathbb{RP}}^2;\mathbb{Z})$, we have $A_0=\mathbb{Z}$, $A_1=\mathbb{Z}/2$ and $A_q=0$ for $q\ge 2$. Cover the circle by two open arcs with a two-component intersection and multiply this cover by the projective plane. In the Mayer–Vietoris sequence the overlap-to-cover map isIts kernel and cokernel are each $A_q$. Consequently $0\to A_q\to H_q(X)\to A_{q-1}\to 0$ for $q\ge 1$. In degree one the quotient is free, so this sequence splits; in degree two it identifies the group directly with $A_1$. The integral homology of a circle times a real projective plane is therefore

Without a compactness restriction, the answer to the final existence question is yes: there is a noncompact oriented four-manifold. Explicitly, takeThe involution is free and reverses orientation on both $S^2$ and $\mathbb{R}$, hence preserves their product orientation. Its quotient is consequently an oriented three-manifold, the orientable total space of the orientation line bundle of the projective plane. The contraction $[v,t]\mapsto [v,(1-s)t]$ is a deformation retraction onto ${\mathbb{RP}}^2$, so $Y\simeq X$.

If a closed oriented manifold is intended, the answer instead is no. A closed connected oriented $d$-manifold has $H_d\cong \mathbb{Z}$ and no homology above $d$. Here the highest nonzero positive-dimensional homology is the torsion group $H_2=\mathbb{Z}/2$, so no dimension can supply that required top fundamental class. The distinction is necessary because the printed wording does not require the oriented manifold to be closed or three-dimensional.

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.

For the sphere homology calculation use singular homology, homotopy invariance of homology, and the reduced Mayer–Vietoris sequence, not cellular homology. For a point, its chain group for singular homology is $\mathbb{Z}$ in every nonnegative degree, and its boundary operator is multiplication by ${\sum }_{i=0}^q(-1{)}^i$, hence alternately zero and one in positive degrees. Its homology is therefore $\mathbb{Z}$ in degree zero and zero above. Singular chains of a disjoint union split into the summand chain complexes. Start with $S^0$, the disjoint union of two points: $H_0(S^0;\mathbb{Z})={\mathbb{Z}}^2$, all positive groups vanish, and ${\tilde{H}}_0(S^0;\mathbb{Z})=\mathbb{Z}$. For $n\ge 1$, remove the north and south poles to obtain an open cover $U,V$ of $S^n$. Each member is contractible, while $U\cap V$ has a deformation retraction to $S^{n-1}$, including the disconnected intersection when $n=1$. The reduced exact sequence therefore suppliesAlso $S^n$ is connected, so its reduced degree-zero group vanishes. Induction givesThe separate $S^0$ calculation avoids conflating its two degree-zero generators.

Choose an orientation of $S^n$, with $n\ge 1$, and let $[S^n]$ be its fundamental class. The mapping degree is the integer determined by $f_{\ast }[S^n]=\mathrm{deg}(f)[S^n]$. To define the local degree at $x_0$, set $y_0=f(x_0)$ and require $x_0$ to be isolated in $f^{-1}(y_0)$. Choose an open neighborhood $B$ containing no other point of that fiber. Then $f$ defines a map of pairsThe degree-$n$ relative homology groups on both sides are $\mathbb{Z}$, by a manifold chart and excision. The local orientation of a manifold chooses their generators. The induced map multiplies these generators by an integer, denoted ${\mathrm{deg}}_{x_0}f$. Excision and naturality show independence of the choice of $B$. Differentiability is unnecessary; if $f$ is smooth and $D{f}_{x_0}$ is invertible, the inverse function theorem gives ${\mathrm{deg}}_{x_0}f=\mathrm{sign}\det D{f}_{x_0}$ relative to the orientations. Without isolation in the fiber this pointwise definition need not apply.

For the degree as a sum of local degrees, assume $y\in S^n$ has finite fiber $F=\{x_1,\dots ,x_s\}$. Disjoint coordinate neighborhoods and excision identifyThe absolute fundamental class maps to the tuple of local orientation generators. Under the relative map induced by $f$, the $j$th generator goes to ${\mathrm{deg}}_{x_j}f$ times the generator at $y$, and the resulting map from the direct sum adds these contributions. On the other hand, naturality says that first applying the absolute map $f_{\ast }$ and then passing to the local group at $y$ gives $\mathrm{deg}(f)$ times that generator. ThusIf the fiber is empty the sum is zero: the map factors through the contractible punctured sphere and has degree zero. If every point of a fiber is isolated, compactness makes that closed fiber finite, so the theorem applies. In particular, a regular value of a smooth map gives the signed count of its inverse images. For $n=0$, define a mapping degree on the reduced generator $[p_{+}]-[p_{-}]$: the identity has mapping degree one, the swap of the two points minus one, and either constant map zero. With local orientation signs positive at $p_{+}$ and negative at $p_{-}$, the same sum formula holds. The connected-sphere convention above is the standard positive-dimensional one.

Use singular homology with integer coefficients. The Künneth theorem for triangulable spaces gives a split short exact sequence with tensor terms ${⨁}_{i+j=k}{H}_i(X)\otimes H_j(Y)$ and torsion terms ${⨁}_{i+j=k-1}{\mathrm{Tor}}_1(H_i(X),H_j(Y))$. When one factor's homology groups are free abelian, the torsion terms vanish and the tensor map is an isomorphism.

The circle has $H_0(S^1)=H_1(S^1)=\mathbb{Z}$ and all other groups zero. Inductively, the homology groups of a torus are free, andPascal's identity givesGeometrically the generators correspond to choosing $k$ of the $n$ circle factors and taking their product, consistent with the cellular model having one $k$-cell for each such subset and cancelling opposite boundary faces.

The first integral singular homology of the Hawaiian earring has a direct summand isomorphic to $\mathbb{Q}$. This consequence of its structural homology theorem suffices, through the universal coefficient theorem for cohomology and nonzero Ext of the rationals with integer coefficients, to show that its integral $H^2$ is nonzero. The precise structural theorem is Theorem 3.1 of www.sas.rochester.edu/mth/sites/doug-ravenel/otherpapers/eda-kawamura2.pdf.

The chain complex whose degree-$n$ group is freely generated over $R$ by the singular simplices $\sigma :{\Delta }^n\to X$. Its boundary is the alternating sum of restrictions to codimension-one faces. The homology of this complex is singular homology.

Triangulate ${\Delta }^q\times I$ by the simplices with vertices $(v_0,0),\dots ,(v_i,0),(v_i,1),\dots ,(v_q,1)$ and coefficients $(-1{)}^i$. Composing with a homotopy between f and g gives a degree-one map P on singular chains. Shared faces cancel; the remaining faces give the displayed chain homotopy identity. Hence homotopic maps induce the same map on singular homology.