Kunneth formula for homology

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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

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

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