Note on p-divisibility in Bockstein Spectral Sequence

If $u\in H_n(X)$ generates a copy of $\mathbb{Z}$ , then $u\notin\ker p^r$ .

Write $u=a+b$ , where $a\in pH_n(X)$ , $b\in\ker p^r$ . Note that since $b=u-a\in\ker p^r$ , hence $b=u-a$ does not generate a copy of $\mathbb{Z}$ . The only way that is possible is when $b=u-a=0$ , i.e $u=a$ .

Author: mathtuition88

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Author: mathtuition88

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