# Homotopy/Homology groups of rationals

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.

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Hint: describe the continuous functions $f:X\to \mathbb{Q}$ if $X$ is a connected topological space. –  Amitesh Datta Jul 8 '11 at 13:35
I would suggest solving this problem without the use of any technical machinary (such as the Hurewicz theorem); i.e., use only the basic definitions. This will provide good (and hopefully easy) practice with the homology and homotopy groups. –  Amitesh Datta Jul 8 '11 at 13:38
Remember that $\pi_1$ needs a basepoint. –  Grumpy Parsnip Jul 8 '11 at 13:38
@Liudas: if there is only one homotopy class of closed loops passing through $x_0$, then how did you get $\pi_1(\mathbb{Q}, x_0) = \mathbb{Q}$? –  Alexei Averchenko Jul 8 '11 at 14:42
@Liudas: it is not true that all homology groups of your space vanish. $H_0$, for example vanishes very, very rarely! –  Mariano Suárez-Alvarez Jul 8 '11 at 15:17

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.
Thanks. This seems to be very clear now. Lesson learned: never ever forget basepoints. But I frankly did not expect $\mathbb{Q}$ to behave so simply. –  johnny Jul 8 '11 at 20:19
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?