I am trying to compute the cohomology of the product of two spheres with two points removed.
The first idea that came to my mind was to use the Mayer-Vietoris sequence with a decomposition of $S^2 \times S^2$ but it seems that it does not work when one looks at the intersection.
Edit: This is the attempt for my computing. Let $X$ be the space that I am interested in, let $Y = (U_1 \cup V_1) \times (U_2 \cup V_2)$ where $U_i$, $V_i$ are contractible neighborhoods in $S^2$ of $\pi_i(p_j)$ respectively. Then $X \cup Y = S^2 \times S^2$ and $X \cap Y = Y\setminus \{p_1, p_2\}$. Basically. I am stuck in the cohomology of such intersection.
Is there other technique that I could use?
Appreciate any help.
Recall that $S^2\times S^2$ can be obtained from $S^2\vee S^2$ by attaching a $4$-cell, that is, as the pushout of a diagram
$\require{AMScd}$ \begin{CD} D^4 @<i<< S^3 @>>> S^2\vee S^2\\ \end{CD}
Now you have a map of two diagrams of the form $\bullet \leftarrow \bullet \rightarrow \bullet$ as follows:
$\require{AMScd}$ $$\small{\begin{CD} D^4\setminus\{p_1,p_2\} @<i<< S^3 @>>> S^2\vee S^2\\ @V\simeq VV @VidVV @VidVV\\ S^3\vee S^3 @<j<< S^3 @>>>S^2\vee S^2 \end{CD}}$$
The upper inclusion is the inclusion as the boundary (we delete points in the interior), and the bottom map is the inclusion of one of the wedge summands. Since $i$ and $j$ are cofibrations and all vertical maps are homotopy equivalences, you get an induced homotopy equivalence of pushouts. Do you see why the pushout of the bottom row is $S^2\vee S^2\vee S^3$?