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The statement. If $X,Y$ are $CW$ complexes, then $\Sigma (X\times Y)$ is homotopically equivalent to $\Sigma X\vee \Sigma Y\vee \Sigma (X\wedge Y)$.
Here $\Sigma$ means the reduced suspension.
The proof is along the lines mentioned in this Math.SE answer.

Questions. I don't understand the following in the proof:

  1. Why do they take reduced join, i.e., why won't the join work?
  2. What is a reduced join geometrically?
  3. How is a reduced join different from the actual join?
  4. Why does contracting $CX$ and $CY$ result in $\Sigma(X\times Y)$?
  5. Why does contracting $X*y_0$ and $x_0*Y$ result in $\Sigma X \vee \Sigma (X\wedge Y)\vee \Sigma Y$?

Can anyone please explain me the proof? I appreciate your help, thanks.

Daniele Tampieri
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Koro
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    There are too many questions in your post. You need to narrow things down, either by thinking a bit more about it or starting with one of the questions and later tackling the others. On another note, 4. is already asked here for example. Did you try reading that? There is a very good answer there that should help. – Pedro Sep 07 '23 at 13:26
  • @Pedro: No, I hadn't seen that post. Thanks for sharing. I'll go through it. – Koro Sep 07 '23 at 13:42

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