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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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    On the other hand, I'm pretty sure we cannot apply Hurewicz on a space that is totally disconnected.2011-07-08
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    Hint: describe the continuous functions $f:X\to \mathbb{Q}$ if $X$ is a connected topological space.2011-07-08
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    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.2011-07-08
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    Remember that $\pi_1$ needs a basepoint.2011-07-08
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    So your computation of $\pi_1$ (and also $H_1$) are not correct.2011-07-08
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    Indeed, any $f: [0,1] \rightarrow \mathbb{Q}$ has to be constant. If we fix any $x_{0} \in \mathbb{Q}$ as a basepoint, then it should be true that $\pi_{1}(X,x_{0}) = \mathbb{Q}$. Then, by considering simplicial homology, any simplex would be continuously sent to a point, hence all homology groups modulo boundaries vanish. Is it true?2011-07-08
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    @Liudas: if there is only one homotopy class of closed loops passing through $x_0$, then how did you get $\pi_1(\mathbb{Q}, x_0) = \mathbb{Q}$?2011-07-08
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    @Liudas: it is not true that all homology groups of your space vanish. $H_0$, for example vanishes very, very rarely!2011-07-08
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    @Liudas: You should allways pick a basepoint for the fundamental group. It is true, that $\pi_1(X,x_0)$ and $\pi_1(X,x_1)$ are isomorphic, if $x_0$ and $x_1$ are elements of the same path component of $X$, but this isomorphism does depend on your choice of path between $x_0$ and $x_1$, except when $\pi_1(X,x_0)$ is abelian, and the isomorphism is therefor not a natural one.2011-07-08
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    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

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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.

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    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.2011-07-08
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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?