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Choose orientations on the two sphere factors whose product is the specified orientation of $X$. Give the four factors of $Y=(S^2{)}^4$ the corresponding orientations in the order $1,2,3,4$. The Künneth theorem has no Tor functor terms here and givesLet $u_i\in H^2(Y;\mathbb{Z})$ be the pullback of the positively normalized cohomology class from factor $i$. The cohomology ring iswith $⟨u_1{u}_2{u}_3{u}_4,[Y]⟩=1$. Its generators commute by graded commutativity of the cup product.

Fix a point in each sphere. For each pair $i<j$, let $S_{ij}$ be the four-dimensional submanifold in which factors $i,j$ vary and the remaining factors equal their fixed points. Orient $S_{ij}$ by the ordered factors $i,j$. The six homology classesgive a coordinate sphere intersection basis by the homology cross product description in the Künneth theorem. Moving a fixed point in one of the other two factors produces a disjoint homologous representative of each $S_{ij}$. Hence every basis element has zero self-intersection number.

More precisely, the intersection pairing isFor complementary pairs the representatives intersect transversely at one point. The sign is positive because swapping two-dimensional factors contributes $(-1{)}^{2\cdot 2}=1$. For pairs with a common index, a factor fixed in both representatives can be displaced to make them disjoint. In the paired order $(12,34),(13,24),(14,23)$, the intersection form is three copies of $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$.

For the diagonal map $\delta :X\to Y$, write $a,b\in H^2(X;\mathbb{Z})$ for the two sphere generators. ThenThe coefficient of $[S_{ij}]$ in $[\Delta ]$ is $⟨u_i{u}_j,[\Delta ]⟩$. Pullback and the preceding cup product relations show that these coefficients are one for $12,14,23,34$ and zero for $13,24$. ThereforeThere are exactly two complementary pairings among these four terms, each counted twice in the square. ThusThis computes both the explicit homology class and its intersection pairing, including all orientation signs.