Cup product ($⌣$)

The cup product is a natural bilinear operation $H^p(X;R)\times H^q(X;R)\to H^{p+q}(X;R)$. It is graded-commutative: $a⌣b=(-1{)}^{pq}b⌣a$.

For a cohomology class $a$, its rth cup power is its r-fold product in the cohomology ring. Naturality gives $f^{\ast }(a^r)=(f^{\ast }a{)}^r$. A nilpotent class has some positive cup power zero; the nilpotence order is preserved by ring isomorphisms and can obstruct a homotopy equivalence or a nonzero mapping degree.

The cross product takes classes on two factors to a class on their product. Its multiplication rule has the graded tensor product sign: $(a\times b)(c\times d)=(-1{)}^{|b||c|}(ac)\times (bd)$. The Künneth theorem makes these products an additive basis when the integral cohomology of the finite cell complexes is torsion-free.

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 open subsets $U,V$ of a topological space $Y$, the relative cup product givesCompatibility with the maps to absolute cohomology identifies its image with the ordinary cup product. More generally the construction requires an excisive pair of subsets. The open-set case can be proved using subdivision to compute on chains subordinate to the open cover.

If a pointed CW complex is covered by $l$ contractible subcomplexes containing the basepoint, every reduced class lifts to a relative class for each appropriate subcomplex. Their relative cup product lies in $h^{\ast }(X,{\bigcup }_j{A}_j)=h^{\ast }(X,X)=0$, so every product of $l$ reduced classes vanishes. The conclusion also holds if $X$ is path connected. Without either condition a two-point space covered by its two points is a counterexample in ordinary degree-zero cohomology.

Let $U_1,\dots ,U_k$ be an open cover of $Y$, and suppose ${\alpha }_i\in H^{d_i}(Y;R)$ restricts to zero on $U_i$. The pair long exact sequence lifts each ${\alpha }_i$ to $H^{d_i}(Y,U_i;R)$. Their relative cup product lies in $H^{\sum d_i}(Y,{\bigcup }_i{U}_i;R)=H^{\sum d_i}(Y,Y;R)=0$, so ${\alpha }_1⌣\cdots ⌣{\alpha }_k=0$. This proves characteristic-class products vanish from local trivializations without cancelling possible zero divisors in the coefficient ring.

Homogeneous cohomology classes satisfyIn particular, an odd-degree class has square killed by two, and its square vanishes whenever two is invertible in the coefficient ring.

For projections $p_X:X\times Y\to X$ and $p_Y:X\times Y\to Y$, the exterior product isThe same definition gives $H^p(X,A;R)\otimes H^q(Y;R)\to H^{p+q}(X\times Y,A\times Y;R)$.

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.

The property that every degree-one element has square zero is invariant under a graded ring isomorphism. Over ${\mathbb{F}}_2$ it distinguishes the cohomology ring of a torus from the mod-two cohomology ring of an inversion mapping torus of dimension at least two.

For positive $n$, let $x,y$ be the pullbacks of the orientation classes from the two factors. Then $x^2=y^2=0$, $xy$ generates degree $2n$, and $yx=(-1{)}^{n}xy$. When $n$ is even, every graded ring automorphism restricts on $H^n$ to a signed permutation matrix; when $n$ is odd the ring admits every matrix in $G{L}_2(\mathbb{Z})$.

For positive unequal dimensions $p,q$, let $u,v$ be the pullbacks of the integral orientation classes of the factors. The Künneth theorem gives the free abelian basis $1,u,v,uv$, with degrees $0,p,q,p+q$. The products satisfy $u^2=v^2=0$ and $vu=(-1{)}^{pq}uv$, by naturality and graded commutativity of the cup product. In particular, for $p=2,q=4$, the cohomology ring is $\mathbb{Z}[u,v]/(u^2,v^2)$, with degrees two and four.

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.

Two finite CW complexes can have isomorphic integral cohomology rings but different cohomology rings modulo $p$. For example, attaching a three-cell of degree $p$ to the projective line in ${\mathbb{CP}}^2$ preserves the nonzero square of the degree-two class modulo $p$, whereas the corresponding Moore space wedged with $S^4$ has zero square. The spaces therefore cannot be homotopy equivalent.

Choose degree-one classes ${\alpha }_i,{\beta }_i$ dual to a symplectic basis of one-cycles and let $\omega$ generate $H^2({\Sigma }_g;\mathbb{Z})$. The only nonzero products of positive-degree basis elements are