Exact sequence for join and product

Statement
Let $$X$$ and $$Y$$ be topological spaces. Denote by $$X * Y$$ the join and by $$X \times Y$$ the product. The long exact sequence of reduced homology obtained using Mayer-Vietoris then splits into short exact sequences of the form:

$$0 \to \tilde{H}_{q+1}(X * Y) \to \tilde{H}_q(X \times Y) \to \tilde{H}_q(X) \oplus \tilde{H}_q(Y) \to 0$$

Moreover, this short exact sequence splits so we get:

$$\tilde{H}_q(X \times Y) \cong \tilde{H}_{q+1}(X * Y) \oplus \tilde{H}_q(X) \oplus \tilde{H}_q(Y)$$

Note that the above is not true for unreduced homology at $$q = 0$$.

Applications
In the case when either $$X$$ or $$Y$$ is a sphere, the homology of $$X \times Y$$ can be computed in terms of the homology of $$X$$. This is because taking a join with a sphere, is equivalent to an iterated suspension, and the homology of an iterated suspension is simply obtained by displacing the homology of the original space.

In symbols:

$$\tilde{H}_q(X \times S^m) = \tilde{H}_{q-m}(X) \oplus \tilde{H}_q(X) + \tilde{H}_q(S^m)$$

The utility of this exact sequence in computing the homology of the product breaks down when we do not understand the homologies of the join. In this case, we use a more advanced tool such as the Kunneth formula.

Related results

 * Kunneth formula
 * Eilenberg-Zilber theorem

Proof
We can view $$X * Y$$ as the double mapping cylinder for the coordinate projections from $$X \times Y$$ to $$X$$ and to $$Y$$ and then apply the exact sequence for double mapping cylinder. This is a long exact sequence. To show that it separates into several short exact sequence, and that each one splits, it suffices to construct a section of the map from $$\tilde{H}_q(X) \oplus \tilde{H}_q(Y)$$ to $$\tilde{H}_q(X \times Y)$$. For $$q \ge 1$$, this section can be constructed geometrically, and for $$q = 0$$ it can be constructed explicitly in terms of the description of reduced homology.