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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.

The exactness axiom of a generalized cohomology theory gives a natural long exact sequence for each CW pair:For $(X,X)$, the restriction $h^n(X)\to h^n(X)$ is the identity. Its preceding relative-to-absolute map has zero image. The previous connecting map is zero too, since it follows the surjective identity in degree $n-1$. Exactness at the relative group therefore makes its map both injective and zero, givingin particular $h^n(x_0,x_0)=0$. No dimension axiom is being used; a generalized theory can have nonzero $h^n(x_0)$ in many degrees.

The wedge-product notation for pairs here means the smash product of CW pairs,Its quotient is the based smash product $(X/A)\wedge (Y/B)$. A multiplicative theory has an external productIf the two spaces are the same, pullback by the diagonal gives the relative cup product in $h^{k+m}(X,A\cup B)$. This is different from the wedge sum of two based spaces.

First read the covering subcomplexes as pointed subcomplexes, so $x_0\in A_j$. Contractibility makes the inclusion $\{x_0\}\hookrightarrow A_j$ a homotopy equivalence, and thus $h^r(A_j,x_0)=0$ in every degree, by exactness and homotopy invariance of cohomology. The relative exact sequence for the triple $\{x_0\}\subseteq A_j\subseteq X$ givesConsequently each homogeneous $w_j$ lifts to $v_j\in h^{r_j}(X,A_j)$. Repeated relative multiplication givesNaturality maps this product to the product of the $w_j$ in $h^{\ast }(X,x_0)$. ThereforeFor nonhomogeneous classes the same conclusion follows by distributivity. This is nilpotence from a contractible subcomplex cover.

There is an alternative sufficient convention: if $X$ is path connected, the $A_j$ need not contain $x_0$. The absolute image $\bar{w}$ of a reduced class restricts to zero at $x_0$, hence at any other point since the point inclusions are homotopic along a path. On a nonempty contractible $A_j$ it therefore restricts to zero, so exactness lifts ${\bar{w}}_j$ to $h^{\ast }(X,A_j)$. Their relative product vanishes as above. The map $h^{\ast }(X,x_0)\to h^{\ast }(X)$ is injective: restriction to the basepoint is split surjective by the map $X\to \mathrm{pt}$, making the previous connecting map zero. Thus vanishing of the absolute product gives vanishing of the reduced product too.

Without either the pointed-subcomplex or connectedness convention, the printed conclusion is false. Take $X=\{x_0,x_1\}$ covered by its two singleton subcomplexes and $h^{\ast }=H^{\ast }(-;\mathbb{Z})$. Both subcomplexes are contractible, but $H^0(X,x_0)$ contains the function $w$ with values $0,1$. Pointwise multiplication gives $w^2=w\ne 0$. This states precisely the needed interpretation rather than assuming a contractible component lies in the basepoint component.

Take ${\mathbb{CP}}^3$. It is a closed smooth manifold and is simply connected: its CW complex structure has no one-cells, and attaching cells of dimension at least two to a point cannot create a fundamental group. Its cohomology ring is $\mathbb{Z}[x]/(x^4)$ with $|x|=2$, so $x^3\ne 0$.

The theorem used is the Lusternik–Schnirelmann critical point theorem: every smooth real-valued function on a closed connected smooth manifold has at least its unnormalized Lusternik–Schnirelmann category many distinct critical points, without a nondegeneracy assumption. Here category is the minimum number of open subsets covering the manifold whose inclusions are null-homotopic. The cup length bound gives category at least four. To justify that bound, a cover by three categorical sets would make each positive-degree class vanish on its assigned set. The long exact sequence of each pair lifts three copies of $x$ to relative classes, and their relative cup product would lie in $H^6(M,M;\mathbb{Z})=0$, contradicting $x^3\ne 0$. Thus every smooth function has at least four critical points, including functions with degenerate critical points. The critical-point theorem in this formulation is described in Conley pairs in geometry—Lusternik–Schnirelmann theory and more.

Use integral cohomology and let $k\ge 1$. On the projective bundle $Y=\mathbb{P}(E)$ define the complex tautological line bundlePut $t=e(\mathcal{L})\in H^2(Y;\mathbb{Z})$, using the canonical complex orientation. On each fibre, $t$ is the Euler class of the tautological line over ${\mathbb{CP}}^{k-1}$, so $1,t,\dots ,t^{k-1}$ restrict to an integral basis of its cohomology.

Here is the finite-cover Leray-Hirsch theorem proof in this case. For each open set $U\subseteq X$, defineIf $E$ is trivial on $U$, its projective bundle is $U\times {\mathbb{CP}}^{k-1}$ and $\mathcal{L}$ is pulled back from the tautological line on the second factor. The Künneth theorem makes ${\Phi }_U$ an isomorphism, since the fibre has finite free integral cohomology. The same holds for every open subset of $U$.

Compactness of $X$ provides a finite trivializing cover $U_1,\dots ,U_N$. Induct on its size. If the result holds on $V=U_1\cup \cdots \cup U_{N-1}$, it holds on $U_N$ and on $V\cap U_N$, both lying in a trivializing chart. Form the diagram of Mayer–Vietoris sequences for the base, with the finitely many degree shifts on the left, and the total space on the right. Naturality of pullback and multiplication by the global even-degree classes $t^i$ makes the diagram commute. The Five lemma gives the isomorphism on $V\cup U_N$. ThusThis is the claimed free module statement, with its graded degree shifts made explicit.

The module basis expresses $t^k$ uniquely using the lower powers, with homogeneous coefficients. Define the Chern classes $c_j(E)\in H^{2j}(X;\mathbb{Z})$ by the unique relationThe pullbacks are suppressed when $f$ is regarded as a polynomial over $H^{\ast }(X)$. Evaluation at $t$ gives a surjective map $H^{\ast }(X)[T]\to H^{\ast }(Y)$. Since $f(T)$ is monic, monic polynomial division over a ring writes any polynomial as $q(T)f(T)+r(T)$ with degree of $r$ below $k$. If its evaluation is zero, module independence forces every coefficient of $r$ to vanish. Hence the kernel is exactly the ideal generated by $f$, provingThe even-degree coefficients are central in the graded commutative algebra, so this division and ideal statement also apply when the base has odd-degree cohomology. Uniqueness of the coefficients proves their naturality under pullback, by pulling back the relation and using the same module basis. With the hyperplane convention $u=-t$, the relation has the usual all-positive Chern coefficients; the alternating signs here correspond to the tautological line itself.

Now suppose $E={⨁}_i{L}_i$. The sections of a projective bundle $s_i:X\to Y$ choose the line $(L_i{)}_x\subset E_x$. They satisfy $s_i^{\ast }\mathcal{L}\cong L_i$, so $s_i^{\ast }t=e_i$ and pulling back the relation gives $f(e_i)=0$.

To obtain the full factorization over the possibly torsion-containing base ring, also use the associated open chartsThey contain the images of the sections $s_i$ and cover $Y$. Projection identifies $\mathcal{L}{|}_{V_i}$ with ${\pi }^{\ast }{L}_i{|}_{V_i}$, so $t-{\pi }^{\ast }{e}_i$ restricts to zero on $V_i$. The long exact sequence of the pair lifts this class to $H^2(Y,V_i)$. The relative cup product of the $k$ lifted classes lies inIts absolute image is ${\prod }_i(t-{\pi }^{\ast }{e}_i)$, so that product vanishes. The polynomial ${\prod }_i(T-e_i)$ is monic of degree $k$ and lies in the kernel of evaluation. Subtracting the monic generator $f(T)$ leaves degree below $k$, and module independence again makes the difference zero. ThereforeThe open-cover argument proves the factorization without a non-zero-divisor assumption on the differences $e_i-e_j$.

For CW pairs $(X,A)$ and $(Y,B)$, write their product pair asUse the compactly generated product, equivalently the weak CW complex topology on products of cells. For arbitrary infinite CW complexes this can be finer than the ordinary product topology. The ordinary product topology already agrees when one factor is locally compact, or when both factors have countably many cells; in particular there is no distinction for finite CW complexes.^[ref]

The quotient of this pair is $(X/A)\wedge (Y/B)$. Setwise, the collapsed subset is exactly the set of pairs with at least one coordinate in its collapsed subspace; the compatible quotient constructions give the homeomorphism in the compactly generated space convention. In a multiplicative generalized cohomology theory, external products of relative classes of degrees $k,m$ lie in degree $k+m$ for this pair. Pullback along a diagonal map gives the relative cup product.

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.