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 $n\geq 0$ 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 $H_{i}(X;M),0\leq i\leq n-1$ are free (or more generally torsion-free) modules over $R$, and/or all the homology groups $H_{j}(X;M),0\leq j\leq n-1$, 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