 Solution

Write $A=S^2\times S^2$ and let $i:A\to X$ be its inclusion after the attachment. The cohomology ring of a product of two spheres iswhere $a,b$ come from the first and second factors and $ab$ is the chosen orientation class. The diagonal pulls both $a$ and $b$ back to the same generator of $H^2(S^2;\mathbb{Z})$.

There is one new three-cell. In the cellular chain complex, its boundary has coordinates $(1,1)$ in the two-dimensional cells. Thus the relevant differential isand the original four-cell still has zero boundary. This gives $H_2(X;\mathbb{Z})=\mathbb{Z}$, $H_3(X;\mathbb{Z})=0$, $H_4(X;\mathbb{Z})=\mathbb{Z}$, and no other positive homology groups. Equivalently, the relative cohomology sequence of $(X,A)$ gives an injective restriction in degree two with image $\mathbb{Z}(a-b)$, and an isomorphism in degree four.

Choose $u\in H^2(X;\mathbb{Z})$ and $v\in H^4(X;\mathbb{Z})$ by $i^{\ast }u=a-b$, $i^{\ast }v=ab$. Naturality of the cup product givesSince restriction is injective in degree four, $u^2=-2v$. All further positive-degree cup products vanish by dimension. ThereforeChanging the sign of $v$ would instead give $u^2=2v$; the displayed sign uses the product orientation $ab$. The factor $2$ is the characteristic feature of a cup square after a diagonal sphere attachment.