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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:

$$\rightarrow H^k(-;\mathbb{Z}/2^r\mathbb{Z}) \rightarrow H^k(-;\mathbb{Z}/2^{2^r}\mathbb{Z})\rightarrow H^k(-;\mathbb{Z}/2^r\mathbb{Z})\xrightarrow{\beta_r}H^{k+1}(-;\mathbb{Z}/2^r\mathbb{Z})\rightarrow$$

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

$$\rightarrow H^k(X;\mathbb{Z}/2^r\mathbb{Z})\rightarrow H^k(X;\mathbb{Z}/2^{2r}\mathbb{Z})\rightarrow H^k(X;\mathbb{Z}/2^r\mathbb{Z})\xrightarrow{\beta_r}H^{k+1}(X;\mathbb{Z}/2^r\mathbb{Z})\rightarrow$$

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

$$\rightarrow H^k(X;\mathbb{Z}/2^r\mathbb{Z})\rightarrow H^k(X;\mathbb{Z}/2^{2r}\mathbb{Z})\rightarrow H^k(X;\mathbb{Z}/2^r\mathbb{Z})\xrightarrow{\beta_r}H^{k+1}(X;\mathbb{Z}/2^r\mathbb{Z})\rightarrow$$

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:

$$\rightarrow H^k(-;\mathbb{Z}/2^r\mathbb{Z}) \rightarrow H^k(-;\mathbb{Z}/2^{2^r}\mathbb{Z})\rightarrow H^k(-;\mathbb{Z}/2^r\mathbb{Z})\xrightarrow{\beta_r}H^{k+1}(-;\mathbb{Z}/2^r\mathbb{Z})\rightarrow$$

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 $\beta_r(a) =\beta_r’(a)$, for example, where $\beta’_r$ is the Bockstein for the long exact sequence

$$\rightarrow H^k(X;\mathbb{Z}/2^r\mathbb{Z})\rightarrow H^k(X;\mathbb{Z}/2^{2r}\mathbb{Z})\rightarrow H^k(X;\mathbb{Z}/2^r\mathbb{Z})\xrightarrow{\beta’_r}H^{k+1}(X;\mathbb{Z}/2^r\mathbb{Z})\rightarrow$$

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:

$$\rightarrow H^k(-;\mathbb{Z}/2^r\mathbb{Z}) \rightarrow H^k(-;\mathbb{Z}/2^{2^r}\mathbb{Z})\rightarrow H^k(-;\mathbb{Z}/2^r\mathbb{Z})\xrightarrow{\beta_r}H^{k+1}(-;\mathbb{Z}/2^r\mathbb{Z})\rightarrow$$

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

$$\rightarrow H^k(X;\mathbb{Z}/2^r\mathbb{Z})\rightarrow H^k(X;\mathbb{Z}/2^{2r}\mathbb{Z})\rightarrow H^k(X;\mathbb{Z}/2^r\mathbb{Z})\xrightarrow{\beta_r}H^{k+1}(X;\mathbb{Z}/2^r\mathbb{Z})\rightarrow$$

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:

$$\rightarrow H^k(-;\mathbb{Z}/2^r\mathbb{Z}) \rightarrow H^k(-;\mathbb{Z}/2^{2^r}\mathbb{Z})\rightarrow H^k(-;\mathbb{Z}/2^r\mathbb{Z})\xrightarrow{\beta_r}H^{k+1}(-;\mathbb{Z}/2^r\mathbb{Z})\rightarrow$$

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 $a\otimes b \in H^\cdot (X\times Y;\mathbb{Z}/2^r\mathbb{Z})$

$$\beta_r(a\otimes b) = \beta_r(a)\otimes b + a\otimes \beta_r(b) $$

Is the value of $\beta_r(a)$, for example, given by the Bockstein for the long exact sequence

$$\rightarrow H^k(X;\mathbb{Z}/2^r\mathbb{Z})\rightarrow H^k(X;\mathbb{Z}/2^{2r}\mathbb{Z})\rightarrow H^k(X;\mathbb{Z}/2^r\mathbb{Z})\xrightarrow{\beta_r}H^{k+1}(X;\mathbb{Z}/2^r\mathbb{Z})\rightarrow$$

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:

$$\rightarrow H^k(-;\mathbb{Z}/2^r\mathbb{Z}) \rightarrow H^k(-;\mathbb{Z}/2^{2^r}\mathbb{Z})\rightarrow H^k(-;\mathbb{Z}/2^r\mathbb{Z})\xrightarrow{\beta_r}H^{k+1}(-;\mathbb{Z}/2^r\mathbb{Z})\rightarrow$$

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 $\beta_r(a) =\beta_r’(a)$, for example, where $\beta’_r$ is the Bockstein for the long exact sequence

$$\rightarrow H^k(X;\mathbb{Z}/2^r\mathbb{Z})\rightarrow H^k(X;\mathbb{Z}/2^{2r}\mathbb{Z})\rightarrow H^k(X;\mathbb{Z}/2^r\mathbb{Z})\xrightarrow{\beta’_r}H^{k+1}(X;\mathbb{Z}/2^r\mathbb{Z})\rightarrow$$