# Fundamental group contains $\mathbb{R}$ or $\mathbb{Q}$

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.

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

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.
• (+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 $1/2$. Jul 24, 2016 at 22:32
Endow $Q$ with the discrete topology $Q_d$ and consider the classifying space corresponding to $Q_d$. Same construction with $R$.