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The Cartan product formula says that total Steenrod squares and total Steenrod reduced powers preserve the cup product:It is named after Henri Cartan and is distinct from Cartan's magic formula for the Lie derivative of a differential form, named after Élie Cartan.

For a CW complex $X$ and coefficient abelian group $A$, the cohomology of its cellular cochain complex naturally equals its singular cohomology:The cellular cochain group is $\mathrm{Hom}(C_q^{\mathrm{cell}}(X;\mathbb{Z}),A)$, and its coboundary map is precomposition with the cellular boundary.

If $X$ has only even-dimensional cells, every cellular coboundary map is zero. If, in addition, there are finitely many cells in each dimension, then its integral cohomology groups are free abelian groups, with one generator for each cell in that degree. For infinitely many cells in one degree, the cochain group is a direct product of groups of copies of $\mathbb{Z}$, rather than the direct sum used for cellular chains; the finite-cell freeness conclusion must not be applied without that qualification. Even with all additive groups known, the cup product still requires a separate computation.

The Chern-Weil homomorphism turns a polynomial in curvature into a topological cohomology class. For a principal bundle with Lie group $G$, a degree-$k$ polynomial $P$ on its Lie algebra that is invariant under the adjoint action determineswhere $F$ is the curvature form of any principal connection. The Bianchi identity makes $P(F)$ a closed differential form. Changing the principal connection changes it by an exact differential form, so the de Rham cohomology class depends on the principal bundle, not that choice. Products of polynomials become cup products.^[ref]

For example, the trace-square polynomial produces $\mathrm{tr}(F\wedge F)$, underlying the first Pontryagin class. This is the bridge between the locally measured curvature and a global topological invariant.

For a closed smooth embedding $i:N\hookrightarrow M$ of codimension $r$ whose normal bundle is oriented over a coefficient ring $R$, a tubular neighborhood, excision and the Thom isomorphism theorem identifyThe relative-to-absolute map defines $i_{!}:H^j(N;R)\to H^{j+r}(M;R)$. For closed oriented $M,N$, it is characterized bywith compatible orientations. It obeys $i_{!}(b⌣i^{\ast }c)=i_{!}b⌣c$. If $j:M\setminus N\hookrightarrow M$ and $\delta :H^q(M\setminus N;R)\to H^{q+1-r}(N;R)$ is the connecting map followed by the inverse Thom isomorphism, then, over ${\mathbb{F}}_2$,These identities follow from naturality of the relative cup product and show how the exact sequence determines multiplication in complements. Over ${\mathbb{F}}_2$ every real normal bundle has the required orientation.

A cohomology theory assigns abelian groups $H^q(X;R)$ to a topological space $X$. For singular cohomology with coefficients in a commutative ring $R$, form the cochain complex $C^q(X;R)$ with coboundary $\delta$ and takeA cocycle represents a cohomology class; adding a coboundary leaves that class unchanged.

A cohomology class evaluates on a homology class by applying its cochain to a representing cycle. Replacing either representative changes the result by zero, explaining the relation to homology. A continuous map $f:X\to Y$ induces cochain pullback and hence $f^{\ast }:H^q(Y;R)\to H^q(X;R)$: the direction reverses.

The cup product combines degree-$p$ and degree-$q$ classes into degree $p+q$, giving a graded cohomology ring. The circle has one integral generator in degrees zero and one. A two-dimensional torus has two independent degree-one classes whose cup product generates degree two. That multiplication distinguishes it from $S^1\vee S^1\vee S^2$, which has the same additive cohomology groups but zero products between its positive-degree generators.

For classes on $X$ and $Y$, apply the induced maps on cohomology of the coordinate projections and take their cup product on $X\times Y$. The result has degree $|a|+|b|$. On a tensor product of graded rings, the sign rule $(a\otimes b)(c\otimes d)=(-1{)}^{|b||c|}ac\otimes bd$ makes this a ring homomorphism.

A degree-$q$ cohomology group is the quotient of degree-$q$ cocycles by coboundaries in a cochain complex. Topological cohomology groups arise from singular or cellular cochains. Their direct sum carries the cohomology ring structure when equipped with the cup product.

A cohomology operation is a natural transformation between cohomology functors. Thus for every continuous map $f$, its pullback commutes with the operation: $\theta (f^{\ast }a)=f^{\ast }\theta (a)$. This makes operations additional invariants of homotopy equivalence, beyond the additive groups and their cup products.

The cohomology ring is the graded ring whose multiplication is the cup product. It records intersection information that the graded cohomology groups alone do not detect.

Pulling the external product of two normally complex submanifolds back along the diagonal gives their transverse intersection. Its normal complex structure is the direct sum of the restricted normal complex structures. Multiplication of Thom classes for a direct sum identifies this geometric representative with the cup product in complex cobordism. Empty intersection represents zero.

// Target: geometry-and-topology.bigb

The largest number of positive-degree cohomology classes whose cup product is nonzero, with a specified coefficient ring $R$; it can be infinite. It is zero if there is no nonzero positive-degree class. The vanishing cup product from an open cover argument implies that Lusternik–Schnirelmann category is at least cup length plus one. For Complex projective space ${\mathbb{CP}}^r$, its integral cup length is $r$.

For even $n\ge 2$, attach an $(n+1)$-cell to $S^n\times S^n$ along the diagonal. The cellular chain complex has boundary $(1,1)$ for the new cell. The degree-$n$ cohomology generator restricts to $a-b$, and the top generator restricts to $ab$. Naturality of the cup product therefore gives $u^2=-2v$, since $(a-b{)}^2=-2ab$. Thus the resulting cohomology ring is $\mathbb{Z}[u,v]/(u^2+2v,uv,v^2)$, with $|u|=n$ and $|v|=2n$. This distinguishes the attachment from a wedge of spheres, whose positive-degree products vanish.

For an oriented rank-$r$ vector bundle, let $U\in H^r(D(E),S(E);\mathbb{Z})$ be its Thom class. Forgetting relative supports sends $U$ to ${\pi }^{\ast }e(E)$, where $e(E)$ is the Euler class. Compatibility of relative and mixed cup products givesFor odd $r$, graded commutativity of the cup product makes $2U^2=0$. The Thom isomorphism theorem is injective, so $2e(E)=0$. This explains why an odd-rank Euler class can have only two-torsion.

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.

Reduction $\rho$ of the integral Bockstein homomorphism associated to $0\to \mathbb{Z}\stackrel{2}{\to }\mathbb{Z}\to {\mathbb{F}}_2\to 0$ equals the cup product square on degree-one classes. One can see this on an ordered two-simplex: lift the values of a mod-two one-cocycle to zero or one. The half-coboundary is one precisely when both consecutive edge values are one, giving the cup-square cocycle.

A space is formal over $\mathbb{Q}$ if its rational cochain commutative differential graded algebra is joined to $(H^{\ast }(X;\mathbb{Q}),0)$ by quasi-isomorphisms. Equivalently in the usual simply connected Sullivan-model setting, its minimal model has a quasi-isomorphism to its cohomology algebra with zero differential. This says its rational homotopy type can be recovered from the cohomology algebra; zero cup products alone do not imply formality of a topological space without an additional argument.

A generalized cohomology theory assigns graded groups contravariantly to suitable topological pairs, with homotopy invariance, excision and a natural long exact sequence for each pair. It omits the dimension axiom of ordinary cohomology, so its coefficient groups at a point can occur in degrees other than zero. A multiplicative theory also has compatible external and relative cup products.

A continuous map $f:X\to Y$ induces a pullback $f^{\ast }:H^q(Y;R)\to H^q(X;R)$. It preserves sums and cup products, and $(g\circ f{)}^{\ast }=f^{\ast }\circ g^{\ast }$.

Integral cohomology is cohomology with integer coefficients, $H^{\ast }(X;\mathbb{Z})$. Its cup product makes it a graded cohomology ring, retaining torsion and integral information that can disappear after changing coefficients to a field.

For a closed manifold $M$ with an orientation and dimension $4k$, its integral intersection form is the symmetric bilinear form onobtained by pairing complementary middle-dimensional homology classes. Under Poincare duality, it is the cup product pairingon degree-$2k$ integral cohomology modulo its torsion subgroup. The even degree makes the pairing symmetric; integral duality makes it a unimodular intersection pairing. When representatives are transverse oriented submanifolds, the same value is the signed count of their intersection points.

The intersection form of a 4-manifold is the case $k=1$. On the Complex projective plane its matrix is $(1)$, whereas on $S^2\times S^2$ the factor spheres give $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$.

For a compact manifold with boundary, the perfect intersection pairing pairs absolute and relative homology, $H_{2k}(M)$ with $H_{2k}(M,\partial M)$; the induced absolute self-pairing can be degenerate. For instance the core sphere in $S^2\times D^2$ generates $H_2$ but has self-intersection zero.