Here is what I want to prove:
$$X \star Y \cong \Sigma (X \wedge Y) $$
Could anyone give me a hint about how to prove this, please?
EDIT:
I corrected the title of the question to reduced join not join only because the first version of the question statement was wrong.
Now I need a proof for the corrected version.
EDIT:
The proof that $$X * Y \simeq \Sigma (X \wedge Y) \quad \quad (1)$$ is also acceptable if someone can help me in proving so.
EDIT:
I found this question here Showing that $X * Y$ is homotopically equivalent to $\sum (X \wedge Y)$ . but the answer that is given does not fit for me as I did not studied the homotopy extension property yet.
EDIT:
Also, I got a hint in the proof of $(1)$ to use Prop. $4I.1$ on pg.467 of AT which states that:
If $X$ and $Y$ are CW complexes, then $$\sum (X \times Y) \simeq \sum X \vee \sum Y \vee \sum (X \wedge Y).$$
But still, I do not know how to use it. Any help will be appreciated.