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.

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