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The integral Künneth theorem gives a natural short exact sequence containing tensor and Tor terms. Here all cohomology groups are free abelian, so the Tor terms vanish and the external cup product is a ring isomorphism. Thus

Solved by gpt-5.6-sol high.

A rank-$d$ vector bundle $\pi :E\to B$ is $R$-oriented when its fiber groups $H^d(E_b,E_b\setminus \{0\};R)$ admit a locally coherent choice of generator. Equivalently, it has a Thom class $u_E\in H^d(D(E),S(E);R)$ restricting to that generator on every fiber. The Thom isomorphism theorem states thatThe Euler class of a vector bundle is $e(E)=s^{\ast }{u}_E\in H^d(B;R)$, where $s$ is the zero section. The Gysin sequence of a sphere bundle is

Over ${\mathbb{F}}_2=\mathbb{Z}/2$, the required rings areas in the cohomology ring of complex projective space and the mod-two cohomology ring of real projective space. By the Künneth theorem, the base has ring ${\mathbb{F}}_2[x,y]/(x^{n+1},y^{m+1})$.

Let $L$ be the underlying real plane bundle of the complex tautological bundle on ${\mathbb{CP}}^n$, and let $\lambda$ be the real tautological line bundle on ${\mathbb{RP}}^m$. Under the splitting principle, write the formal Stiefel-Whitney roots of $L$ as $a,b$, so $a+b=w_1(L)=0$ and $ab=w_2(L)=x$. Tensoring with $\lambda$ adds $y=w_1(\lambda )$ to each root. Thus the mod-two Euler class of $L{\otimes }_{\mathbb{R}}\lambda$ is

For $n=m=2$, put $B={\mathbb{CP}}^2\times {\mathbb{RP}}^2$. Multiplication by $e=x+y^2$ onhas ranks $1,1,2,1,1,0,0$ from degrees $0$ through $6$. More explicitly, it is injective through degree three; in degree four its kernel is generated by $x^2+x{y}^2$, and all of degrees five and six lie in its kernel. The Gysin sequence therefore yieldsfor the unit sphere bundle $X=S(L\otimes \lambda )$.

Solved by gpt-5.6-sol high.