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)

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.