Showing that fundamental group of the rational is trivial.

Question

I am trying rigorously to prove that, for any $x_0\in \mathbb{Q}$, $\pi_1(\mathbb{Q},x_0)=\left\{[e_{x_0}]\right\}$ with $e_{x_0}$ loop in $x_0$. I just have the following intuition.

Let $[f]\in \pi_1(\mathbb{Q},x_0)$ then $f:[0,1]\to \mathbb{Q}$ continuous and $f(0)=f(1)=x_0$.
Since $\mathbb{Q}$ is discrete and $f$ continuous this forces the function $f$ to look like the constant function $e_{x_0}:[0,1]\to\mathbb{Q}$, $e_{x_0}(t)=x_0$ for all $t\in [0,1]$.

Is it possible to explicitly find a homotopy between $f$ and $e_{x_0}$? Or should it be concluded from some fact of the style $\mathbb{Q}$ is not path connected etc.

A small correction: $\mathbb Q$ is totally disconnected, but not discrete. — Chris Culter, Sep 25 '20 at 04:56

score 1 · Accepted Answer · answered Sep 25 '20 at 03:59

1

Every continous nonconstant image of $[0,1]$ is of size continuum. So the fund group consists of one (constant) path.

answered Sep 25 '20 at 03:59

markvs

19,653

thanks. If $f:[0,1]\to\mathbb{Q}$ continuous then $f$ is constant, we are done. – eraldcoil Sep 25 '20 at 05:11

Showing that fundamental group of the rational is trivial.

1 Answers1