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

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} …
Achim Krause's user avatar
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