The fundamental group of $\mathbb{R}/\mathbb{Q}$ What is the fundamental group of $\mathbb{R}/\mathbb{Q}$? 
Here $\mathbb{R}$ is equipped with general topology and the quotient space is in the meaning of topology instead of additive subgroup.

It is not difficult too see $\mathbb{R}/\mathbb{Q}$ is path connected, and $\mathbb{R}/\mathbb{Q}$ is not trivial. Because $\mathbb{Q}$ is dense in $\mathbb{R}$, $U$ in $\mathbb{R}/\mathbb{Q}$ is open if and only if the preimage $\pi^{-1}(U)$ is a open subset of $\mathbb{R}$ containing $\mathbb{Q}$, where $\pi$ denote the quotient map.
 A: The fundamental group is trivial, since the space is contractible. Denote $X:= \mathbb{R} / \mathbb{Q}$. 
Let $\xi:= \pi(\mathbb{Q})$. Then note that $\{\xi\}$ is dense since for every $x \in X$, the pre-image of every open neighborhood $U$ of $x$ contains $\mathbb{Q}$ i.e. $\xi = \pi(\mathbb{Q}) \in U$
Let $\phi: X \rightarrow \{\xi\}$, and $\iota: \{\xi\} \hookrightarrow X$ the inclusion, then we have $\phi \circ \iota = \text{id}_{\{\xi\}}$ and we want to show $\iota \circ \phi \sim \text{id}_{X}$ via the following homotopy:
$$ H: [0,1] \times X \rightarrow X ~~, \begin{cases} (t,x) \mapsto x, & t \in [0,\frac{1}{2}] \\
(t,x) \mapsto \xi, & t \in (\frac{1}{2},1] \end{cases}
$$
We have $H(0, -) = \text{id}_X$ and $H(1,-) = \iota \circ \phi$. To check continuity, let $(t_i, x_i)_{i \in I}$ be a net in $[0,1] \times X$, s.t. $t_i \rightarrow t$ and $x_i \rightarrow x$, then we want to show that $H(t_i, x_i) \rightarrow H(t,x)$:


*

*if $t \in [0, \frac{1}{2})$, then for large enough $i_0$ we have $H(t_i,x_i) = x_i$ for every $i \geq i_0$ and thus $H(t_i,x_i)  \rightarrow x = H(t, x)$ 

*if $t \in (\frac{1}{2},1]$, then for large enough $i_0$ we have $H(t_i,x_i) = \xi$ for every $i \geq i_0$ and thus $H(t_i, x_i) \rightarrow \xi = H(t, x)$

*if $t = \frac{1}{2}$, then we want to show that $H(t_i,x_i) \rightarrow H(\frac{1}{2}, x) = x$. So let $U \subseteq X$ be some neighborhood of $x$ and choose $i_0 \in I$ s.t. $\forall i \geq i_0: x_i \in U$. Such a $i_0$ exists since $x_i \rightarrow x$. Then we have that for every $i \geq i_0$ either $t_i \in [0, \frac{1}{2}]$ and thus $H(t_i, x_i) = x_i \in U$ or $t_i \in (\frac{1}{2},1]$ and thus $H(t_i, x_i) = \xi$, which is in $U$ since $\{\xi\}$ was dense in $X$. 
