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The formula for the Bockstein $\beta:H_n(X;\mathbb{Z}/p\mathbb{Z})\to H_{n-1}(X;\mathbb{Z}/p\mathbb{Z})$ is $$\beta[c\otimes 1]=[\frac{1}{p}\partial c\otimes 1]$$ (McCleary page 456)

How about for the higher Bockstein $$\beta_r: H_n(X;\mathbb{Z}/p^r\mathbb{Z})\to H_{n-1}(X;\mathbb{Z}/p^r\mathbb{Z})$$?

I read that it is related to the connecting homomorphism, but am quite confused about the formula. Is it (just a guess based on my limited and possibly wrong understanding):

$$\beta_r[c\otimes 1]=[\frac{1}{p^r}\partial c\otimes 1]$$

I read some texts but strangely none of them seem to write the explicit formula. Thanks for any help.

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    $\begingroup$ I think you want to divide by $p$ and not $p^r$, since the Bockstein is related to the coefficient sequence $\mathbb{Z}/p^r\mathbb{Z}\to \mathbb{Z}/p^{r+1}\mathbb{Z}\to \mathbb{Z}/p^r\mathbb{Z}$, where the first map is multiplication by $p$. In fact, it is the boundary map in the long exact sequence associated to a short exact sequence of chain complexes, and is defined using the snake lemma. $\endgroup$
    – Mark Grant
    Commented Nov 17, 2017 at 14:53
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    $\begingroup$ @MarkGrant Thanks. But I thought it is the coefficient sequence $\mathbb{Z}/p^r\to\mathbb{Z}/p^{2r}\to\mathbb{Z}/p^r$ instead? (pg 460 McCleary's book) $\endgroup$
    – yoyostein
    Commented Nov 17, 2017 at 15:11
  • $\begingroup$ You're right, my previous comment was stupid. Still, you can easily work out the explicit formula from the definition of the boundary map. $\endgroup$
    – Mark Grant
    Commented Nov 17, 2017 at 16:11
  • $\begingroup$ I tried working out the explicit formula (using the connecting homomorphism), and it seems that it is indeed $\beta_r[c\otimes 1]=[\frac{1}{p^r}\partial c\otimes 1]$. If anyone finds a mistake, please alert me, thanks. $\endgroup$
    – yoyostein
    Commented Nov 18, 2017 at 7:09

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