For $r\geq 1$, we have the following short exact sequence
$$0\rightarrow \mathbb{Z}/2^r\mathbb{Z} \xrightarrow{\cdot 2^r}\mathbb{Z}/2^{2r}\mathbb{Z} \xrightarrow{\bmod2^r}\mathbb{Z}/2^r\mathbb{Z}\rightarrow 0$$
which induces a long exact sequence in cohomology:
$$\rightarrow H^k(-;\mathbb{Z}/2^r\mathbb{Z}) \rightarrow H^k(-;\mathbb{Z}/2^{2r}\mathbb{Z})\rightarrow H^k(-;\mathbb{Z}/2^r\mathbb{Z})\xrightarrow{\beta_r}H^{k+1}(-;\mathbb{Z}/2^r\mathbb{Z})\rightarrow$$
Here $\beta_r$ is the Bockstein homomorphism. I'm curious how this behaves with respect to the Kunneth theorem. That is, suppose we have two space $X,Y$ such that
$$H^\cdot(X\times Y; \mathbb{Z}/2^r\mathbb{Z}) \cong H^\cdot (X;\mathbb{Z}/2^r\mathbb{Z})\otimes H^\cdot (Y;\mathbb{Z}/2^r\mathbb{Z})$$
Since Bocksteins are derivations, for any $a\otimes b \in H^\cdot (X\times Y;\mathbb{Z}/2^r\mathbb{Z})$
$$\beta_r(a\otimes b) = \beta_r(a)\otimes b + a\otimes \beta_r(b) $$
Do we have $\beta_r(a)$ given by the Bockstein for the long exact sequence
$$\rightarrow H^k(X;\mathbb{Z}/2^r\mathbb{Z})\rightarrow H^k(X;\mathbb{Z}/2^{2r}\mathbb{Z})\rightarrow H^k(X;\mathbb{Z}/2^r\mathbb{Z})\xrightarrow{\beta_r}H^{k+1}(X;\mathbb{Z}/2^r\mathbb{Z})\rightarrow$$
Best Answer
Yes, this is true. (Let's assume $Y$ is nonempty for a silly reason.) To check that this is true, it suffices to check it when $b=1$, verifying that $\beta_r(a \otimes 1) = \beta_r(a) \otimes 1$.
Let $p: X \times Y \to X$ be the projection map. The induced map on cohomology is given by $p^\ast(a) = a \otimes 1$ under the identification you've given. Therefore, the identity we want to prove is that $\beta_r p^\ast = p^\ast \beta_r$. However, this is implied by naturality of the Bockstein.