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what is the 0'th homology group of $\mathbb{Q}$ I mean $ H_{0}(\mathbb{Q})$?as the 0'th homology group is counting the path component of the space so it should be infinite direct sum of copies of $\mathbb{Z}$,and that exactly the place I feel doubt and I asked myself is it true for infinitely points?

so please help me and guide me ,thank you very much.

kpax
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    have a look at this – Math137 Jun 10 '14 at 09:09
  • @AymanHourieh: How is $\pi_0(\mathbb{Q}) = \mathbb{Q}$ wrong? It looks correct to me. – Najib Idrissi Jun 10 '14 at 12:56
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    @AymanHourieh: 1/ $\pi_0$ is a functor usually defined on topological spaces, not pointed spaces; 2/ $\mathbb{Q}$ is in particular a pointed space (being a group), so the identification isn't abusive; 3/ $\mathbb{Q}$ is an H-space and even a topological group, so $\pi_0$ does have more structure. In any case, there is a canonical bijection of pointed sets $\pi_0(\mathbb{Q}, q_0) \cong (\mathbb{Q}, q_0)$, basically the identity. – Najib Idrissi Jun 10 '14 at 13:10
  • @AymanHourieh: Yes, I got a bit carried away for my first point; it's also common to view $\pi_0$ as a pointed set. Point 3/ still stands though: $\pi_0(\mathbb{Q}, 0) = \mathbb{Q}$ in almost every possible way (not as topological spaces though!). – Najib Idrissi Jun 10 '14 at 13:20
  • @NajibIdrissi You have convinced me. I'll delete my comments in a few minutes. :) – Ayman Hourieh Jun 10 '14 at 13:22
  • @NajibIdrissi can $\pi_0$ not be given the compact open topology in which case I believe $\pi_0(\mathbb{Q},0)$ is actually homeomorphic to $\mathbb{Q}$ right? (Haven't worked out the details yet so I could be wrong.) – Dan Rust Jun 10 '14 at 13:43
  • @DanielRust: I think you're right. – Najib Idrissi Jun 10 '14 at 13:46

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From the definition of singular homology, the $n$th homology group of $\mathbb{Q}$ is trivial except for $H_0(\mathbb{Q};\mathbb{Z})$ which is countably infinitely freely generated abelian, that is $H_0(\mathbb{Q};\mathbb{Z})\cong\displaystyle\bigoplus_{n=1}^{\infty}\mathbb{Z}$.

(really the only cases where you can directly apply the definition of singular homology is when your space is totally disconnected).

Dan Rust
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