2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Could you make the suggested failure explicit? Where do you get stuck? $\endgroup$ – Pedro Tamaroff Jan 13 '17 at 21:43
  • $\begingroup$ @PedroTamaroff I edited it including my attempt. $\endgroup$ – C. Zhihao Jan 13 '17 at 21:52
  • $\begingroup$ Hint: This space is homotopy equivalent to $S^2\vee S^2\vee S^3$ $\endgroup$ – iwriteonbananas Jan 14 '17 at 8:32
  • $\begingroup$ @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 $\endgroup$ – C. Zhihao Jan 16 '17 at 17:39
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.