Homology (mathematics)

Homology assigns abelian groups $H_i(X)$ to a space, measuring cycles modulo boundaries in each dimension.

For a locally compact space, replace finite singular chains by chains for which every compact subset meets the images of only finitely many simplices with nonzero coefficients. The boundary operator preserves this local finiteness, giving the Borel-Moore chain complex and its homology. On a compact space it reduces to ordinary singular homology. A noncompact manifold has a locally finite fundamental class, with its orientation local system when necessary. The chain groups and homology are zero in negative degrees.

A generalized homology theory obeys the homotopy, exactness, suspension and wedge axioms, without imposing the ordinary dimension axiom. A represented homology theory is obtained by taking stable homotopy groups after a smash product of spectra.

A topological spectrum $E$ represents the reduced generalized homology theory ${\tilde{E}}_k(X)={\pi }_k(E\wedge {\Sigma }^{\infty }X)$ on based spaces, where the smash product of spectra is derived. It is covariant in $X$.

An element of a cycle module is null-homologous when it represents zero in homology, equivalently when it lies in the image of the boundary operator. A closed submanifold bounding a submanifold one dimension higher is null-homologous.

A positive-degree class whose reduced homology diagonal vanishes. In a torsion-free homology Hopf algebra, powers of an even primitive class have binomial diagonals. Dualizing that diagonal produces the multiplication of a divided power algebra.

A nonempty space is acyclic with the chosen coefficients if its reduced homology vanishes in every degree. Equivalently it has the homology of a point. It need not be contractible.

Suppose $X={\bigcup }_a{X}_a$ is directed by inclusion and every compact subset of $X$ lies in some $X_a$. Every finite singular chain then lies in one $X_a$, so $C_{\ast }(X)={\underrightarrow{\lim }}_a{C}_{\ast }(X_a)$. Exactness of filtered colimits gives

Every open subset $U\subseteq {\mathbb{R}}^N$ is the directed union of finite unions of closed rational cubes contained in $U$. Every compact subset of $U$ lies in one such finite polyhedron, whose homology is finitely generated. The homology of a directed union therefore expresses each $H_i(U;\mathbb{Z})$ as a direct limit over a countable family of countable groups, so it is countable.

Reduced homology modifies degree zero so that a one-point space has zero homology in every degree. For a nonempty space, ${\tilde{H}}_k(X)=H_k(X)$ when $k>0$, while $H_0(X)\cong {\tilde{H}}_0(X)\oplus \mathbb{Z}$.

A homology group is one of the abelian groups produced by a homology theory. Its elements are homology classes, represented in singular homology by cycles modulo boundaries.

The zeroth homology group is the free abelian group on the path components of $X$. In particular, if $X$ has finitely many path components, then

A homology class is an equivalence class of chain cycles modulo chain boundaries in a chain complex.

A continuous map $f:X\to Y$ induces homomorphisms $f_{\ast }:H_n(X)\to H_n(Y)$ satisfying $(g\circ f{)}_{\ast }=g_{\ast }\circ f_{\ast }$ and $({\mathrm{id}}_X{)}_{\ast }={\mathrm{id}}_{H_n(X)}$.

Homotopic maps induce the same homomorphism on every homology group. Consequently, homotopy equivalent spaces have isomorphic homology groups.

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.

The first homology group $H_1(X)$ is the group of one-dimensional cycles modulo boundaries. For a path-connected space it is the abelianization of the fundamental group.

Singular homology forms a chain complex from formal sums of continuous maps from standard simplices into a topological space. Its homology groups are invariant under homotopy.

For an open cover $\mathcal{U}$ of a space $X$, let $C_{\ast }^{\mathcal{U}}(X)$ be the subgroup generated by singular simplices whose images lie in a single member of the cover. It is a chain complex, and its inclusion into the full singular chain complex induces an isomorphism on homology.

For a finite chain, enough iterations of barycentric subdivision make every simplex small: pull the cover back to each compact standard simplex and use its Lebesgue number. The subdivision operator commutes with the boundary and is related to the identity by the standard subdivision chain homotopy, which keeps each produced simplex inside its original image. Therefore any cycle becomes homologous to a small cycle, proving surjectivity.

If a small cycle $z$ bounds a full chain $b$, choose $N$ so that ${\mathrm{Sd}}^{N}b$ is small. Write ${\mathrm{Sd}}^N-1=\partial Q+Q\partial$. Since the homotopy preserves simplex images, $Qz$ is also small. Then $z=\partial ({\mathrm{Sd}}^{N}b-Qz)$ is a small boundary, proving injectivity. This is the chain-level locality input in the proof of the Mayer–Vietoris sequence.

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.

For an open cover, take the subcomplex generated by singular simplices whose images lie in single cover members. Repeated barycentric subdivision makes each finite singular chain small, and the standard subdivision chain homotopy shows that inclusion induces homology isomorphisms. For a two-set cover, the short exact sequence of these small chains gives the Mayer–Vietoris sequence; excising the complement of one open set from the pair with the other set gives its equivalent relative-homology formulation.

A local coefficient system assigns an Abelian group or module to each point, with compatible transport isomorphisms along paths depending only on homotopy relative endpoints. Equivalently it is a functor from the fundamental groupoid. On a connected manifold it is described by a representation of the fundamental group after fixing one fiber.

Its fiber at $x$ is $H_n(M,M\setminus \{x\};\mathbb{Z})\cong \mathbb{Z}$. Transport changes its generator by the sign of local orientation along the path. For a connected nonorientable manifold, its degree-zero cohomology is zero, because no nonzero integer is invariant under sign reversal, whereas its degree-zero homology is $\mathbb{Z}/2$, the coinvariants under that reversal.

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.

For a space with finite-dimensional homology over a specified field and only finitely many nonzero Betti numbers, this polynomial records their graded dimensions. Its value at one is the total Betti number, and its value at minus one is the Euler characteristic. The Künneth theorem gives $P_{M\times N}(t)=P_M(t)P_N(t)$ over a field. For integer homology ranks, use the rational Betti numbers.

A singular $n$-chain is a finite formal integer linear combination of singular simplices. Its support is contained in a compact subset of the ambient space.

The singular chain group is the free abelian group on all continuous maps ${\Delta }^q\to X$, the singular simplices. In particular it is free even when the set of simplices is infinite. This freeness makes applying $\mathrm{Hom}(C_q(X),-)$ to a short exact coefficient sequence exact.

A singular $n$-simplex in a space $X$ is a continuous map $\sigma :{\Delta }^n\to X$ from the standard $n$-simplex. Its alternating restrictions to the faces form its boundary.

For an oriented compact manifold of dimension $4k$, the intersection form is the bilinear pairing on middle-dimensional homology obtained by representing classes by transverse cycles and counting their signed intersection points. On an oriented four-manifold it is equivalently the cup-product pairing on $H^2$ evaluated on the fundamental class.

The intersection form of a 4-manifold is the four-dimensional case of this middle-dimensional intersection pairing.

For a simply connected closed four-manifold with second Betti number one, Poincare duality and the universal coefficient theorem for cohomology make $H_2$ torsion-free of rank one. The integral intersection form is unimodular, so its one-by-one matrix is $[1]$ or $[-1]$. Hence the square of a generator $u\in H^2$ generates $H^4$, giving the displayed cohomology ring. This is a ring isomorphism; it need not preserve a preassigned orientation class.

For a closed oriented smooth four-manifold, $b_2^{+}$ is the dimension of a maximal positive-definite subspace of its real intersection form. A symplectic form forces $b_2^{+}\ge 1$, because its cohomology class has positive square. Reversing orientation exchanges the positive and negative indices.

Here $H$ is the hyperbolic form with matrix $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. The degree-two cohomology consists of the two factor orientation classes and $4gh$ products of degree-one classes. The factor classes pair once and have zero squares. In symplectic bases $a_i,b_i$ and $c_j,d_j$, $(a_i{c}_j)(b_k{d}_l)=-{\delta }_{ik}{\delta }_{jl}{\omega }_1{\omega }_2$ and $(a_i{d}_j)(b_k{c}_l)={\delta }_{ik}{\delta }_{jl}{\omega }_1{\omega }_2$. Each pair $(i,j)$ contributes two hyperbolic forms; the fiber classes contribute one more. The sign in the first formula comes from interchanging two odd-degree factors.

An integral intersection form on a finite-rank free abelian group $L$ is unimodular if its adjoint $x\mapsto \lambda (x,-)$ is an isomorphism to the integral dual. In an integral basis, this is equivalent to the pairing matrix having determinant $1$ or $-1$, not merely nonzero determinant. For example, projective lines generate a unimodular intersection form on the Complex projective plane, with matrix $(1)$ in its complex orientation.

When compact oriented four-manifolds are glued along an entire closed boundary three-manifold, the signature of the result is the sum of their intersection form signatures. It applies to $X_0{\cup }_{\partial X_0}(-X_1)$ with the difference of the two signatures. Gluing along only part of a boundary can require a correction; that is a different situation.

An intersection pairing counts signed intersections of transverse representatives of complementary homology classes, using one relative representative when the manifold has boundary. With ${\mathbb{F}}_2$ coefficients it is defined without orientation choices. Poincare-Lefschetz duality makes this a perfect pairing for compact manifolds.

The intersection matrix records an intersection pairing on chosen classes. Its nonsingularity proves linear independence: a relation between the classes, paired with each class in turn, gives a vector in the matrix's kernel.

In the product $(S^{2d}{)}^{2\ell }$, for $d,\ell \ge 1$, vary the factors indexed by an $\ell$-element subset $I$ and fix points in the other factors. Orient the resulting submanifold $S_I$ in increasing factor order. These homology classes form an integral middle-dimensional basis by the Künneth theorem. Complementary subsets give a transverse intersection at one point, with positive sign since all factor dimensions are even. For noncomplementary subsets some factor is fixed in both representatives; moving that point makes them disjoint. In particular every basis class has zero self-intersection number. Complementary pairs give hyperbolic blocks in the intersection form.

For a compact three-manifold $W$, let $L=\ker (H_1(\partial W;{\mathbb{F}}_2)\to H_1(W;{\mathbb{F}}_2))$. The boundary intersection pairing satisfies $L^{\perp }=L$: the long exact sequence in relative homology identifies $L$ with boundaries of relative surfaces, and the boundary-interior adjoint identity identifies its orthogonal complement with the same kernel. Hence $2\dim L=\dim H_1(\partial W;{\mathbb{F}}_2)$. This also implies that a closed surface with odd first mod-two Betti number, such as ${\mathbb{RP}}^2$, cannot be the entire boundary of a compact three-manifold.

For a map $f:M\to N$ of connected closed oriented four-manifolds, naturality of the cup product givesWhen $\mathrm{deg}f\ne 0$, nondegeneracy makes $f^{\ast }:H^2(N;\mathbb{Q})\to H^2(M;\mathbb{Q})$ injective. Consequently the target intersection form, multiplied by the mapping degree, must embed into the source form. For equal second Betti numbers, a matrix $P$ for pullback satisfies $P^{\mathsf{T}}{Q}_{M}P=(\mathrm{deg}f)Q_N$; taking determinants often rules out nonzero degrees. In particular, a two-dimensional indefinite source form cannot realize a nonzero multiple of a two-dimensional definite target form.

On a compact oriented smooth manifold $N$ of dimension four without boundary, set $Q([\alpha ],[\beta ])={\int }_N\alpha \wedge \beta$ for closed real two-forms. The Stokes theorem makes the expression independent of representatives, and the even degrees make it symmetric. It is the real de Rham cohomology version of the intersection form. The Hodge decomposition theorem and Hodge star operator express it as $Q(h,k)=(h,\ast k{)}_{L^2}$ on harmonic representatives.

On a closed oriented Riemannian manifold of dimension four, the Hodge star operator preserves harmonic two-forms and splits them into orthogonal spaces ${\mathcal{H}}^{+}$ and ${\mathcal{H}}^{-}$ with eigenvalues $+1$ and $-1$. The real de Rham intersection form in dimension four equals the positive $L^2$ inner product on ${\mathcal{H}}^{+}$ and its negative on ${\mathcal{H}}^{-}$. It is nondegenerate, since $Q(h,\ast h)=\Vert h{\Vert }_2^2$. Its signature pair is $(\dim {\mathcal{H}}^{+},\dim {\mathcal{H}}^{-})$.

Integral homology is homology with coefficients in $\mathbb{Z}$. It retains both the free and torsion parts of each homology group.

For $n>0$,Writing $S^n$ as two contractible hemispheres whose intersection deformation retracts to $S^{n-1}$ gives the calculation inductively through the Mayer–Vietoris theorem.

For the standard coordinate inclusion of spheres, write points as $(u,v)\in {\mathbb{R}}^{m+1}\times {\mathbb{R}}^{n-m}$, with $v\ne 0$ in the complement. The map $(u,v)\mapsto (u,v/\Vert v\Vert )$ is a homeomorphism to $B_{\mathrm{open}}^{m+1}\times S^{n-m-1}$, with inverse $(u,w)\mapsto (u,\sqrt{1-\Vert u{\Vert }^2}w)$. Contracting the ball factor proves the displayed homotopy equivalence. In codimension one there are two contractible components, so integral $H_0$ is ${\mathbb{Z}}^2$; in higher codimension the only nonzero homology groups are $H_0=\mathbb{Z}$ and $H_{n-m-1}=\mathbb{Z}$.

For a map $f:S^m\to S^m$, the degree is the integer by which $f_{\ast }$ multiplies a chosen generator of $H_m(S^m;\mathbb{Z})$. It is invariant under homotopy.

For a map $f:M\to N$ between closed connected $n$-manifolds, the mod-two degree is defined by $f_{\ast }[M{]}_2={\mathrm{deg}}_2(f)[N{]}_2$. Every manifold has a mod-two orientation class, so no orientability assumption is needed. It is invariant under homotopy. A nonsurjective map has mod-two degree zero: mapping the fundamental classes into the local relative homology at a missed target point makes the source contribution zero but leaves the target local generator nonzero.

Suppose a map $S^m\to M$ of closed connected oriented manifolds has nonzero mapping degree. A product decomposition with $0<k<m$ would have oriented closed factors: an orientation-reversing loop in either factor, with the other coordinate fixed, would reverse the product orientation. Pull back their top orientation classes to obtain $\alpha \in H^k(M;\mathbb{Z})$ and $\beta \in H^{m-k}(M;\mathbb{Z})$ with $\alpha ⌣\beta$ a generator of $H^m(M;\mathbb{Z})$. Both pullbacks to the sphere vanish in their intermediate degrees, contradicting the nonzero-degree pullback of their product.

Define mapping degree for $S^0$ using its rank-one reduced homology group ${\tilde{H}}_0(S^0;\mathbb{Z})$. Its generator is the difference of its two points. The four maps of that two-point set consist of the identity, the interchange, and two constant maps; their degree multipliers are $1,-1,0,0$. Thus sphere maps of arbitrary integer degree require positive sphere dimension.

For a map of oriented disk pairs, the multiplier on top relative homology is its relative mapping degree. Naturality of the connecting homomorphism identifies it with the mapping degree of the boundary map.

For a smooth map of disk pairs whose graph of a function is transverse to the zero slice, orient the graph by its domain. With the tangent space of the zero slice ordered before that of the graph, the local smooth intersection number at a zero is $\mathrm{sgn}\det DF$. Summing these signs gives the relative mapping degree and hence the boundary mapping degree. Reversing the order of these two $m$-dimensional tangent spaces changes the sign by $(-1{)}^m$.

Collapsing the boundary identifies the top relative homology of a disk pair with the top reduced homology of its quotient sphere. Together with the boundary connecting homomorphism, this shows that the induced quotient map and the boundary map have the same mapping degree, using the corresponding orientations.

Let $f:M\to N$ be a continuous map between oriented $d$-dimensional manifolds, with $d\ge 1$. Suppose $x$ is an isolated point of the fiber over $y=f(x)$. Choose an open neighborhood $U$ with $U\cap f^{-1}(y)=\{x\}$. The map of pairsinduces a map between rank-one degree-$d$ local homology groups. Relative to the local orientation of a manifold, this map is multiplication by an integer, the local degree. Excision makes it independent of the neighborhood. A smooth map between manifolds with invertible derivative at $x$ has local degree $\mathrm{sign}\det D{f}_x$, by the inverse function theorem and the chosen orientations.

In dimension two, $z\mapsto z^k$ for positive $k$ has local degree $k$ at zero, and $z\mapsto {\overline{z}}^{|k|}$ has degree $k$ for negative $k$. The proper map $z\mapsto |z{|}^2\in \mathbb{C}$ has isolated zero fiber and local degree zero, since its image misses every nearby nonreal target. These proper maps extend to the two-sphere. Taking their product with the identity of ${\mathbb{R}}^{n-2}$ realizes the same local degrees in every dimension $n\ge 2$. In dimension one an isolated local degree is restricted to $0,1,-1$.

For a continuous map $f:M\to N$ of closed connected oriented $d$-dimensional manifolds, suppose $f^{-1}(y)=\{x_1,\dots ,x_s\}$ is finite. ThenBy excision, the degree-$d$ relative homology group $H_d(M,M\setminus f^{-1}(y);\mathbb{Z})$ is the direct sum of the local groups at $x_j$. The fundamental class maps to the tuple of local orientation generators. Mapping this tuple to $H_d(N,N\setminus \{y\};\mathbb{Z})$ adds their local degrees. Naturality identifies the result with the image of $(\mathrm{deg}f)[N]$. If the fiber is empty, the same argument gives degree zero.

If $f:S^m\to S^m$, then its suspension $\Sigma f:S^{m+1}\to S^{m+1}$ has the same degree. This follows from naturality of the suspension isomorphism in reduced homology.

Every integer $m$ is the mapping degree of a continuous map $S^n\to S^n$ in every dimension $n\ge 1$. Start with the circle map $z\mapsto z^m$, whose argument winds $m$ times, and use degree under suspension repeatedly. Negative $m$ reverses the winding, and $m=0$ is constant before suspending. The degrees of maps of the zero-sphere are a genuine exceptional case.

For a map $f:M\to N$ between closed connected oriented $d$-manifolds, its degree is the integer determined by $f_{\ast }[M]=(\mathrm{deg}f)[N]$ in top-dimensional homology.

Let $M,N$ be closed connected oriented manifolds of the same dimension, and suppose $a^r$ generates the top integral cohomology of $N$. If every class of degree $|a|$ on $M$ has zero rth cup power, every continuous map $f:M\to N$ has mapping degree zero. Indeed, $f^{\ast }(a^r)=(f^{\ast }a{)}^r=0$, and evaluation on the fundamental class gives $\mathrm{deg}f=0$. For $S^2\times S^4\to {\mathbb{CP}}^3$, the degree-two class is a multiple of the sphere class $u$, whose square vanishes.

For $k>0$, an epimorphism ${\pi }_1({\Sigma }_2)\to \mathbb{Z}/k$ gives a connected $k$-sheeted cover. Its Euler characteristic is $-2k$, hence its genus is $k+1$, and the lifted orientation gives degree $k$. For negative $k$ compose the $|k|$-sheeted cover with an orientation-reversing homeomorphism of the target. For $k=0$ take a constant map from the torus.

Choose an oriented embedded closed disc in a closed oriented positive-dimensional manifold and collapse its complement to a point. A target point away from the collapsed value has one preimage with local degree $+1$, proving global degree one by the degree as a sum of local degrees.

On the unit two-sphere this formula maps onto the torus: every torus point has arguments corresponding to $x,y\in [-1/2,1/2]$, and $x^2+y^2\le 1/2$ leaves a real $z$ on the sphere. Scaling both arguments continuously to zero is a null-homotopy, so the degree is zero. Surjectivity alone does not force nonzero degree.

For maps of oriented connected closed manifolds of the same dimension, apply functoriality of homology to $f_{\ast }[M]=(\mathrm{deg}f)[N]$ and $g_{\ast }[N]=(\mathrm{deg}g)[L]$. The composite coefficient is their product. This recovers the multiplication of signed sheet counts for compositions of finite covering maps.

For maps between oriented connected closed manifolds of the same dimension, a homotopy induces equal maps on top-dimensional homology, hence equal coefficients on the fundamental class. For smooth maps and a smooth homotopy, the degree by integration of a pullback volume form gives another proof: the Generalized Stokes theorem on the homotopy cylinder makes the difference of endpoint pullback integrals zero because the target top form is closed. This is an obstruction to extending a degree-one sphere identity over a ball.

Although topological degree classifies maps $S^n\to S^n$ up to homotopy, it is not a complete invariant for other manifolds. On the torus, the identity and the integer shear $\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$ both have degree one, but induce distinct maps on $H_1(T^2;\mathbb{Z})$ and cannot be homotopic.

For a smooth map $f:M\to N$ between oriented connected closed manifolds of the same dimension, a normalized volume form on the target gives $\mathrm{deg}f={\int }_M{f}^{\ast }\omega$. The formula agrees with the degree as a sum of local degrees. A change of normalized top-form adds an exact form by top-dimensional de Rham cohomology, whose pullback has zero integral by Stokes theorem.

For a smooth map between oriented unit spheres, integrate the pullback of the target area form. Localizing a normalized top-degree form near a regular value, and using Stokes theorem to discard exact-form differences, identifies this integral with the degree as a sum of local degrees.

A degree of a map between oriented manifolds equal to one makes the pullback on degree-one cohomology injective by the Poincare duality pairing. Thus a degree-one map of closed oriented surfaces requires $2h\le 2g$. Conversely, collapse the extra handles in ${\Sigma }_h\#{\Sigma }_{g-h}$ onto a point. The retained oriented disc shows that this map has degree one.

If $S^n\to N$ has prime degree $p$ and $N$ is a closed connected oriented $n$-manifold, every integral homology group of $N$ in degrees strictly between zero and $n$ is a finite $p$-primary abelian group. Consequently one power of $p$ annihilates all these groups. The proof combines Poincare duality over each field ${\mathbb{F}}_q$, $q\ne p$, with the universal coefficient theorem for homology and finite generation.

For $n\ge 2$, a self-map of $S^n$ factoring through Real projective space ${\mathbb{RP}}^n$ has degree zero when $n$ is even, and can have exactly the even degrees when $n$ is odd. In the odd case the first map lifts through the double covering and the covering has degree two. The dimension-one exception allows every integer, since ${\mathbb{RP}}^1\cong S^1$.

The antipodal map $x\mapsto -x$ on $S^m$ has degree $(-1{)}^{m+1}$.

If a continuous map $f:S^d\to S^d$ has no fixed point, the formula $H_t(x)=((1-t)f(x)-tx)/\Vert (1-t)f(x)-tx\Vert$ is a homotopy from $f$ to the antipodal map. The denominator could vanish only at $t=1/2$ and $f(x)=x$, which is excluded. The homotopy invariance of mapping degree therefore gives $\mathrm{deg}f=(-1{)}^{d+1}$.

An odd map between spheres is a continuous map $f:S^n\to S^m$ satisfying $f(-x)=-f(x)$. Thus it is equivariant for the two antipodal maps. It descends to a continuous map of Real projective spaces.

Suppose a continuous map $g:S^n\times S^n\to S^n$ changes sign on negating either input. Its quotient $\bar{g}:{\mathbb{RP}}^n\times {\mathbb{RP}}^n\to {\mathbb{RP}}^n$ pulls back the real tautological line bundle to the tensor product of vector bundles $p_1^{\ast }{\gamma }_n\otimes p_2^{\ast }{\gamma }_n$. The fiber map sends $(sx)\otimes (ty)$ to $stg(x,y)$. The First Stiefel–Whitney class of a tensor product of real line bundles consequently gives ${\bar{g}}^{\ast }a=a_1+a_2$. The Künneth theorem and the mod-two cohomology ring of real projective space implyEvery interior binomial coefficient in row $n+1$ must be even. By binomial coefficients with even interior terms, $n=2^k-1$ for some $k\ge 0$. This is a necessary condition, not an assertion of existence in every such dimension.

For an odd map between spheres $f:S^n\to S^m$ and its quotient $\bar{f}$, there is an isomorphism ${\gamma }_n\cong {\bar{f}}^{\ast }{\gamma }_m$ of real tautological line bundles. With $x$ a unit vector, send $tx$ to $tf(x)$; replacing $x$ by $-x$ and $t$ by $-t$ gives the same vector. This proves both linearity on fibers and well-definedness. Therefore ${\bar{f}}^{\ast }{w}_1({\gamma }_m)=w_1({\gamma }_n)$. The mod-two cohomology ring of real projective space then implies $n\le m$, since a vanishing $(m+1)$st power must pull back to a vanishing power.

Let a finite group $G$ act smoothly on a smooth manifold, and suppose that a $G$-invariant differential form $\omega$ is an exact differential form, say $\omega =d\eta$. Averaging gives the invariant primitiveIf the action is free, invariant forms descend uniquely through the resulting covering map.

For the double covering $\pi :S^{2n}\to {\mathbb{RP}}^{2n}$ and the antipodal map $a$, every pulled-back top form satisfies $a^{\ast }{\pi }^{\ast }\omega ={\pi }^{\ast }\omega$. Since $\mathrm{deg}a=-1$, its integral over the sphere is its own negative and hence vanishes. It is therefore exact on the sphere. Averaging a primitive under $a$ and descending it proves that $\omega$ is exact on real projective space.

A Euclidean homeomorphism is a proper map and extends to its one-point compactification $S^n$. Its degree is the degree of a continuous mapping of this extension, and belongs to $\{1,-1\}$. A linear isomorphism has degree equal to the sign of its determinant.

The factor swap ${\mathbb{R}}^m\times {\mathbb{R}}^n\to {\mathbb{R}}^n\times {\mathbb{R}}^m$ has degree $(-1{)}^{mn}$. Moving an oriented basis of the first factor past one of the second requires $mn$ transpositions.

The relative chain complex is $C_{\ast }(X,A)=C_{\ast }(X)/C_{\ast }(A)$, and its homology is the relative homology $H_{\ast }(X,A)$. A pair $A\subseteq U\subseteq X$ produces a long exact sequence for the triple.

Let $A$ be a nonempty union of disjoint embedded circles in a closed oriented topological surface ${\Sigma }_g$, and let $c$ be the number of components after cutting along them. Excision and Poincare-Lefschetz duality identify the relative homology with the complementary-degree cohomology of the cut surface. Its components have boundary, so the relative groups are free, with the displayed ranks; $H_0$ and higher groups vanish. The rank of the circle classes in $H_1({\Sigma }_g)$ is $|{\pi }_0(A)|+1-c$. For a single circle these groups detect whether its class is zero, but for genus at least two cannot distinguish a disk boundary from an essential separating circle.

A relative homology class is represented by a singular chain whose boundary lies in $A$. Two representatives give the same class when their difference is a boundary operator image plus a chain in $A$. Thus a path with endpoints in $A$ can represent a class even when its endpoints are different.

For a subspace $A\subseteq X$, the relative chain complex is the quotient group in each degree,with boundary operator $[c]\mapsto [\partial c]$. This is well defined because $C_{\ast }(A)$ is a chain subcomplex of $C_{\ast }(X)$.

A relative cycle for $A\subset B$ is a chain $b\in C_n(B)$ whose boundary lies in $A$. Equivalently its coset is a homological cycle in $C_{\ast }(B,A)$. Unlike an absolute cycle, its boundary need not be zero. It represents a relative homology class.

If $L$ is a simplicial subcomplex of $K$, thenIts homology is the simplicial relative homology $H_k(K,L)$.