TITLE: Calculate Bockstein homomorphism
QUESTION [10 upvotes]: I would like to ask how to compute the Bockstein homomorphism $\beta: H^k(M^d,\mathbb{Z}_n) \to H^{k+1}(M^d,\mathbb{Z}_n)$? Can one compute the Bockstein homomorphism from the universal coefficient theorem
$$0 \to H^d(X, \mathbb{Z}) \otimes M \to H^d(X, M) \to \operatorname{Tor}(H^{d+1}(X, \mathbb{Z}), M) \to 0$$
and/or other needed information (such as the cohomology rings $H^*(M^d, \mathbb{Z}_n)$, $H^*(M^d, \mathbb{Z})$, and homology groups $H_*(M^d, \mathbb{Z})$)?

REPLY [2 votes]: The Bockstein homomorphism (between homologies with coefficients in $\mathbb{Z}_n$) has some relation to the universal coefficient theorem. Roughly speaking, it is$$\text{H}_k(X, \mathbb{Z}_n) \to \text{Tor}(\text{H}_{k - 1}(X), \mathbb{Z}_n) = \text{Tor}(\text{H}_{k - 1}(X)) \otimes \mathbb{Z}_n \to \text{H}_{k - 1}(X) \otimes \mathbb{Z}_n \to \text{H}_{k -1 }(X, \mathbb{Z}_n)$$(I say "roughly" because the equality in this composition is not canonical). But it seems to be more natural to relate the Bockstein homomorphism to the coefficient homology sequence induced by the short exact sequence$$0 \to \mathbb{Z} \to \mathbb{Z} \to \mathbb{Z}_n \to 0.$$Certainly, the cohomological Bockstein homomorphism is related to the cohomological multiplication by the usual product rule.