Let $\mathbb{Z}/p^{\infty}$ denote the Prufer group. By $p$-completion properties, it follows that $(B\mathbb Z/p^{\infty})^{\wedge}_p\simeq K(\mathbb{Z}^{\wedge}_p,2)\simeq(BS^1)^{\wedge}_p$. But, why does the inclusion $\mathbb{Z}/p^{\infty}\hookrightarrow S^1$ induce such a homotopy equivalence?.
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5
Think of the Prüfer group as $\mathbb{Q}/\mathbb{Z}$ and of $S^1$ as $\mathbb{R}/\mathbb{Z}$. Here $\mathbb{Q}$ is discrete, $\mathbb{R}$ has its usual topology and is contracible. The map $\mathbb{Q} \to\mathbb{R} $ becomes an equivalence after $p$-completion.
answered Sep 2, 2020 at 22:50
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$\begingroup$ Thank you!, do you mean $B\mathbb{Q}^{\wedge}_p\rightarrow B\mathbb{R}^{\wedge}_p$ is a homotopy equivalence? $\endgroup$ Commented Sep 3, 2020 at 1:28
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5$\begingroup$ Yes, I think if you want to turn this idea into an actual proof you look at the induced map of fiber sequences $B\mathbb{Q}\to B\mathbb{Q/Z} \to K(Z,2)$ and $B\mathbb{R}\to B\mathbb{S^1}\to K(Z,2)$, and observe that the map on the fibers becomes an equivalence after $p$-completion. $\endgroup$ Commented Sep 3, 2020 at 7:30
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$\begingroup$ but $B\mathbb{R}$ is contractible, and $B\mathbb{Q}$ is $p$-good. So far, their $p$-completions cannot be homotopy equivalent. $\endgroup$ Commented Sep 3, 2020 at 22:38
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1$\begingroup$ I'm not sure I correctly remember what $p$-good means, but note that the mod $p$ homology of $B\mathbb{Q}$ is of course zero. $\endgroup$ Commented Sep 4, 2020 at 5:47
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1$\begingroup$ Fine!, I was wrong, sorry. Thank you again. $\endgroup$ Commented Sep 4, 2020 at 18:02