= Solution
Call this <simultaneous antipodal quotient of two spheres> $M$. The simultaneous <antipodal map> acts freely on $S^2\times S^2$, so $M$ is a connected <closed manifold> of dimension four. Each antipodal factor has <mapping degree> $-1$; their product preserves orientation. Hence $M$ is orientable.
The product $S^2\times S^2$ is <simply connected>, so this double <covering space> is the <universal cover>. Consequently
$$
\pi_1(M)=\mathbb Z/2,\qquad H_1(M;\mathbb Z)=\mathbb Z/2,
$$
using the <abelianization> of the <fundamental group>. The <Euler characteristic under a finite covering> gives $\chi(M)=4/2=2$. Its rational first <Betti number> is zero, and <Poincare duality> gives $b_3=b_1=0$ and $b_0=b_4=1$. Thus $b_2=0$ as well.
The <universal coefficient theorem for cohomology> now gives $H^1(M;\mathbb Z)=0$ and
$$
H^2(M;\mathbb Z)\cong\operatorname{Ext}^1_{\mathbb Z}(H_1(M;\mathbb Z),\mathbb Z)\cong\mathbb Z/2.
$$
Here $\operatorname{Hom}(H_2(M;\mathbb Z),\mathbb Z)=0$ because $b_2=0$. Integral <Poincare duality> further gives $H^3(M;\mathbb Z)\cong H_1(M;\mathbb Z)=\mathbb Z/2$, and $H^4(M;\mathbb Z)=\mathbb Z$.
Let $a,b,c$ generate these groups in degrees $2,3,4$ respectively. The only potentially nonzero positive-degree <cup product> is $a^2$. But $2a^2=(2a)a=0$, while $H^4(M;\mathbb Z)$ has no nonzero <torsion subgroup>, so $a^2=0$. Every other positive product vanishes by dimension. Thus
$$
\boxed{H^*(M;\mathbb Z)=\mathbb Z\{1,c\}\oplus(\mathbb Z/2)\{a,b\},\quad |a|=2,\ |b|=3,\ |c|=4,}
$$
\b[with every product of positive-degree elements equal to zero.] The degree-three torsion is essential: it would be lost by computing only rational <cohomology>.
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