2
$\begingroup$

Let $\mathbb{Q}$ be the set of rational numbers endowed with a topology (let's say subspace topology). What could we say about its fundamental $\pi_{1}$ and homology groups?

Presumably, $\pi_{0}(\mathbb{Q}) = \mathbb{Q}$ since rationals are totally disconnected. I would also think that $\pi_{1}(\mathbb{Q},x_{0}) = \mathbb{Q}$ for some $x_{0} \in \mathbb{Q}$ since no two points of rationals can be connected by a loop. What about the homology groups?

Could anyone confirm? Or does anyone have an idea on how to proceed? Or maybe some relevant references?

EDIT: We need of course a basepoint for a fundamental group of space that is not path-connected and applying Hurewicz was a very poor choice. Thanks for correcting.

  • 0
    Thanks, guys. Now that I got the point, I'm embarrassed to see how I could have possibly miscalculated these groups in such a false way.2011-07-08

2 Answers 2

5

First note, that the only continuous maps $S^n \to \mathbb Q$ and $D^{n-1} \to \mathbb Q$ of the $n$-dimensional sphere and the $(n-1)$-dimensional Disk to the rationals are constant ($n\geq 1$). Therefore $\pi_k(\mathbb Q, *) = H_k(\mathbb Q, \mathbb Z) = H^k(\mathbb Q, \mathbb Z) = 0$ for all $k \geq 1$ and every basepoint $* \in \mathbb Q$. The statement on (co-)homology can easily be seen by considering singular (co-)homology.

  • 0
    Thanks. This see$m$s to be very clear $n$ow. Lesson learned: never ever forget basepoints. But I fran$k$ly did not expect $\$m$$a$th$b$b{Q}$ to behave so si$m$ply.2011-07-08
4

I can find very few correct sentences in your question!

First, $\pi_0$ counts the number of path-connected components, therefore, in this case, Hurewicz theorem is not applicable, since $\mathbb{Q}$ is not path-connected. Also, as you said, rationals is totally disconnected, so how did you write the expression of $\pi_1$ without choosing a base point?