Kunneth formula for homology

Statement

Suppose $X$ and $Y$ are topological spaces. We then have the following relation for the homology groups of $X$, $Y$, and the product space $X \times Y$.

For any $n \ge 0$ and any module $M$ over a principal ideal domain $R$ for coefficients, we have:

$H_n(X \times Y;M) \cong \left(\sum_{i + j = n} H_i(X;M) \otimes H_j(Y;M)\right) \oplus \left(\sum_{p + q = n-1} \operatorname{Tor}(H_p(X;M),H_q(Y;M))\right)$

Here, $\operatorname{Tor}$ is torsion of modules over the ring $R$.

Related facts

• Kunneth formula for cohomology
• Universal coefficients theorem for homology
• Universal coefficients theorem for cohomology
• Dual universal coefficients theorem (computes cohomology in terms of homology)

Particular cases

Case of free modules

If all the homology groups $H_i(X;M), 0 \le i \le n - 1$ are free (or more generally torsion-free) modules over $R$, and/or all the homology groups $H_j(X;M), 0 \le j \le n - 1$, are free (or more generally torsion-free) modules over $R$, then all the torsion part vanishes and we get:

$H_n(X \times Y;M) \cong \sum_{i + j = n} H_i(X;M) \otimes H_j(Y;M)$

In particular, if all $H_i(X;M)$ and $H_j(Y;M)$ are free modules over $R$ and $b_i(X;M)$ and $b_j(Y;M)$ denote the respective free ranks, and all these are finite, we obtain that:

$b_n(X \times Y;M) = \sum_{i + j = n} b_i(X;M)b_j(Y;M)$

Note that if $R$ is a field, then the above holds.

Impact for ranks even in case of torsion

When $R$ is a principal ideal domain and all the homologies are finitely generated modules over $R$, we can consider the rank as the rank of the torsion-free part of the homology modules. If $b_i(X;M)$ denotes the free rank of the torsion-free part of $H_i(X;M)$, we get:

$b_n(X \times Y;M) = \sum_{i + j = n} b_i(X;M)b_j(Y;M)$

Note that this applies even if the homology modules have torsion.

In the special case that $R = M = \mathbb{Z}$, the numbers $b_i$ are called Betti numbers, and we get:

$b_n(X \times Y) = \sum_{i+j=n}b_i(X)b_j(Y)$

In particular, this yields that Poincare polynomial of product is product of Poincare polynomials.

Facts used

The Kunneth formula combines the Kunneth theorem and the Eilenberg-Zilber theorem.

Proof

Fill this in later