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I was looking through my notes for a homotopy theory course and found the following mysterious statement (K is of course the Eilenberg-Maclane space):

$$H^{n+1}(K(\mathbb Z_p,n);\mathbb Z_p) \cong \mathbb Z_p.$$

(This would be obvious if n+1 were replaced with n. This is supposed to imply that the natural transformations $H^n(X; \mathbb Z_p)\to H^{n+1}(X; \mathbb Z_p)$ are all multiples of the Bockstein homomorphism).

I'm at a loss trying to understand why. Spectral sequences haven't been covered yet, so there should be some simple reason. Also, is there a way to see the Bockstein in all this?

Thank you!

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it is correct that those elements are multiples of the bockstein. –  Sean Tilson Aug 8 '10 at 18:22
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1 Answer

up vote 12 down vote accepted

Universal coefficient theorem + $H_{n+1}(K(Z_p,n);Z)=0$. An elementary way to see the latter is that the single $n+1$ cell added to $S^n$ to kill $p$ times the generator is not a cellular cycle.

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Thank you! This seems to sum it up pretty well. –  Ilya Grigoriev Feb 8 '10 at 4:54
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