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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.