Exact sequence for join and product

Template:Exact sequence for construction

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:

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

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

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 $q \geq 1$ , this section can be constructed geometrically, and for $q = 0$ it can be constructed explicitly in terms of the description of reduced homology.