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.

  • $\begingroup$ "Here $\mathbb R$ is equipped with general topology and the quotient space is in the meaning of topology instead of additive subgroup." – but with the equivalence classes given by the cosets of the additive subgroup? $\endgroup$
    – joriki
    Mar 16, 2020 at 10:45
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    $\begingroup$ @joriki No, for irrational number $x$, its equivalent class is one-element set $\overline{x}=\{x \}$. $\endgroup$
    – Landau
    Mar 16, 2020 at 10:52

1 Answer 1


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 $X$.

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    $\begingroup$ I think the crucial of this proof is that $\mathbb{R}/\mathbb{Q}$ is just $T_0$, and the poor separation leads to a violation of my intuition. In fact $\{\xi\} $ is automatically included in each open set of $\mathbb{R}/\mathbb{Q}$. $\endgroup$
    – Landau
    Mar 16, 2020 at 13:36
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    $\begingroup$ The space looks as follows (or at least I like to think of it this way): there is one point $\xi$ which lies in the middle of uncountable many points, each corresponding to a $r \in \mathbb{R} \setminus \mathbb{Q}$ where the "distance between" $\xi$ and each of the $r$ is infinitessimal. $\endgroup$ Mar 16, 2020 at 13:57
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    $\begingroup$ Indeed, the crucial point was that $\mathbb{Q}$ lies dense in $\mathbb{R}$ and this $\{ \xi\}$ lies dense in $X$ i.e. $\xi$ lies infinitessimally close to any other point. $\endgroup$ Mar 16, 2020 at 13:58
  • $\begingroup$ Maybe to help intuition one can think of the simplest example of this phenomenon: the Sierpinski space. This is a space with two points, where one point is open and the other is not. The open point is "infinitessimaly close" to the non-open point (more precisely it dense in the whole space) so the homotopy which is the identity for half the time and constant for the other half is continuous. Unless I'm mistaken, the space $\mathbb{R}/\mathbb{Q}$ is homeomorphic to an uncountable wedge product of Sierpinski spaces. $\endgroup$
    – William
    Mar 16, 2020 at 16:29
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    $\begingroup$ By this prove, it is easy to see that for arbitrary topology space $X$, if $Y$ is dense in $X$, then $X/Y$ is contractiable. $\endgroup$
    – Landau
    Mar 17, 2020 at 5:02

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