Write $\alpha=pa+qb$. In the <Gysin sequence of a sphere bundle> for the oriented circle bundle $S(E_\alpha)\to S^2\times S^2$, multiplication by $\alpha$ is
$$
\mathbb Z\longrightarrow\mathbb Z^2,qquad1\longmapsto(p,q),
$$
in degrees $0$ to $2$, and
$$
\mathbb Z^2\longrightarrow\mathbb Z,qquad(u,v)\longmapsto qu+pv,
$$
in degrees $2$ to $4$. If $\alpha\neq0$ and $d=\gcd(|p|,|q|)$, taking the relevant kernels and cokernels gives
$$
\boxed{H^i(S(E_\alpha);\mathbb Z)\cong
\begin{cases}
\mathbb Z,&i=0,3,5,\\
\mathbb Z\oplus\mathbb Z/d,&i=2,\\
\mathbb Z/d,&i=4,\\
0,&\text{otherwise}.
\end{cases}}
$$
When $\alpha=0$, the bundle is trivial and the groups in degrees $0$ through $5$ have ranks $1,1,2,2,1,1$, respectively. These are precisely the groups recorded in <integral cohomology of a circle bundle over a product of two spheres>.

The additive cohomology for nonzero $\alpha$ depends only on $d$. On the other hand, part (a) shows that the homeomorphism group acts on $(p,q)$ only by signed permutations. For example, $(1,0)$ and $(1,1)$ both have $d=1$, so their sphere bundles have isomorphic additive cohomology, but no signed permutation carries one Euler class to the other. The cohomology therefore does not determine the homeomorphism-group orbit of $\alpha$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 114 / 4 / c / Solution

Let $a,b$ be the standard basis of $H^2(S^2\times S^2;\mathbb Z)$ and let $E\to S^2\times S^2$ be an oriented circle bundle with Euler class $pa+qb\neq0$. If $d=\gcd(|p|,|q|)$, its <Gysin sequence of a sphere bundle>[Gysin sequence] gives
$$
H^i(E;\mathbb Z)\cong
\begin{cases}
\mathbb Z,&i=0,3,5,\\
\mathbb Z\oplus\mathbb Z/d,&i=2,\\
\mathbb Z/d,&i=4,\\
0,&\text{otherwise}.
\end{cases}
$$
For zero Euler class the bundle is trivial and its cohomology is that of $(S^2\times S^2)\times S^1$.

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 Incoming links: Integral cohomology of a circle bundle over a product of two spheres