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?.



$(B\mathbb Z/p^{\infty})^{\wedge}_p\rightarrow (BS^1)^{\wedge}_p$ induced by inclusion is an equivalence
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.


