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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.

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    Could you make the suggested failure explicit? Where do you get stuck? – 2017-01-13
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    @PedroTamaroff I edited it including my attempt. – 2017-01-13
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    Hint: This space is homotopy equivalent to $S^2\vee S^2\vee S^3$ – 2017-01-14
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    @iwriteonbananas with that hint the computations turns out quite straightforward: however I don't see that clear such homotopy; would you mind extending a little bit on how can I prove it. Thanks – 2017-01-16

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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 @>> 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:

$\require{AMScd}$ $$\small{\begin{CD} D^4\setminus\{p_1,p_2\} @>> S^2\vee S^2\\ @V\simeq VV @VidVV @VidVV\\ S^3\vee S^3 @>>S^2\vee S^2 \end{CD}}$$

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 $S^2\vee S^2\vee S^3$?