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I want to know a 4-skelecton of quotient space $(\prod_{i=1}^n S^2/\bigvee_{i=1}^n S^2 )^4 = \bigvee_{1\leq i < j\leq n} S^4$

If $n=2$, then I can realize. But in $ n=3$ case I have no idea.

Give me a hint. Thank you in advance.

[$n$=2]

$S^2\times S^2/S^2 = S(S^1\vee S^2)=S(S^3)$

In this process, we can naturally see $S^2\times S^2/S^2\vee S^2 = S^4$

HK Lee
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There is a problem with your calculation in the case $n=2$. You should be quotienting by the wedge $S^2\vee S^2$ rather than by $S^2$, and then you get $S^4$. The reason is that $S^2\vee S^2$ is the 2-skeleton of a natural cell structure on $S^2\times S^2$ when each 2-sphere is viewed as composed of a 0-cell and a 2-cell.

In general, $\prod_{i=1}^n S^2$ will have ${n \choose 2}$ 4-dimensional cells corresponding to Cartesian products of pairs of 2-spheres, and the subset $\bigvee_{i=1}^n S^2$ is precisely the 2-skeleton. If you quotient the 4-skeleton by the 2-skeleton you will get a wedge of 4-spheres $\bigvee_{1\leq i< j\leq n } S^4$.

However, the way you defined it, your space on the left hand side is not just the 4-skeleton and will have additional cells of dimensions 6, 8, etc. I don't think your claim is true without taking the 4-skeleton on the left-hand-side.

Mikhail Katz
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  • Thank you for $n=2$ and correct statement of problem. I editted. – HK Lee Aug 05 '13 at 09:07
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    Incidentally, I assume this material is preparatory to the introduction of Whitehead products. An example of this is the attaching map of the 4-cell to the wedge of a pair of 2-spheres. – Mikhail Katz Aug 05 '13 at 09:33
  • First, your answer is good for me. Thank you. And your background for this problem fits to me, since I am studying the above product. – HK Lee Aug 05 '13 at 09:40