Homology group of the join

Prove that $$\tilde{H_n}(X*Y) \cong H_{n-1}(X \times Y/X \vee Y)$$

Firstly, I set $$A=X \times Y \times [0,1) / (x,y_0,0) \sim (x,y_1,0)$$ $$B=X \times Y \times (0,1] / (x_0,y,1) \sim (x_1,y,1)$$ Where $A$ deformation retracts to $X$ and $B$ deformation retracts to $Y$.

Then I use MV-sequence to get a LES.

But I'm not sure how to prove this SES splits. $$0 \to H_{n+1}(X * Y) \to H_n(A \cap B) \to H_n(A) \oplus H_n(B) \to 0$$

My guess about the map $H_n(A) \oplus H_n(B) \to H_n(A \cap B)$ is: $f:(a,b) \to (p_x a, -p_y b)$ for $a \in C_n(A)$ and $b \in C_n(B)$.

• Alternatively you can use the fact that the join is homotopy equivalent to the suspension of the smash product. – Ayman Hourieh Jul 18 '14 at 9:15
• @AymanHourieh: What you say is essentially a stronger version of the statement, so I guess OP should prove than $X * Y \simeq \Sigma(X \wedge Y)$. – Najib Idrissi Jul 18 '14 at 9:28
• @NajibIdrissi I agree. It depends on what the OP already knows. This fact is good to have in one's toolbox anyway. – Ayman Hourieh Jul 18 '14 at 9:49