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Is there any topological space whose fundamental group contains $\mathbb{Q}$ or $\mathbb{R}$? In case of (singular) homology or cohomology, we can change its coefficients to any abelian groups (with suitable modification, such as universal coefficient theorem). But for a fundamental group, I've never seen any space that has fundamental group $\mathbb{Q}$ or $\mathbb{R}$. I think it is impossible, but how to prove it? Thanks in advance.

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    $\begingroup$ Note that, as a group, $(\mathbb{R}, +)$ is just the direct sum of an uncountable number of copies of $(\mathbb{Q},+)$. $\endgroup$ Jul 22, 2016 at 0:22

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Any group (including $\mathbb{Q}$ and $\mathbb{R}$) can be realized as the fundamental group of some space - in fact, for every group $G$, there is a two-dimensional CW complex with fundamental group $G$. See Proposition 1.28 of Hatcher here.

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    $\begingroup$ (+1) For the record, an explicit not-too-hard-to-see construction of a space with fundamental group $\Bbb Q$ is the following homotopy colimit: take $\Bbb N$-many copies of $S^1 \times [0, 1]$, glue the front of the $k$-th copy to the end of the $(k+1)$-th copy by a degree $k$ map, for all $k$. So, e.g., running along the top of the $2$nd copy of the cylinder twice is homotopic to running along the front of the first cylinder once, i.e., that element represents $1/2$. $\endgroup$ Jul 24, 2016 at 22:32
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Endow $Q$ with the discrete topology $Q_d$ and consider the classifying space corresponding to $Q_d$. Same construction with $R$.

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  • $\begingroup$ If the OP knew what a classifying space was they wouldn't have asked this question. $\endgroup$ Jul 23, 2016 at 3:10
  • $\begingroup$ so i hope, he will learn it,by the way if op new the aswer of his question he would not have asked it $\endgroup$ Jul 23, 2016 at 3:12

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