I'm trying to show that $H_1(\mathbb{R}, \mathbb{Q})$ is free abelian, this is another exercise in Hatcher.
I'm not sure but I thought I can use the exact sequence
$ \cdots 0 \xrightarrow{f} H_1(\mathbb{R}, \mathbb{Q}) \xrightarrow{g} H_0(\mathbb{Q}) \xrightarrow{h} H_0(\mathbb{R}) \xrightarrow{i} \cdots$
and then
$ 0 \xrightarrow{f} H_1(\mathbb{R}, \mathbb{Q}) \xrightarrow{g} H_0(\mathbb{Q}) \xrightarrow{h} \operatorname{ker} (i) \xrightarrow{i}0$
My idea is to use that an exact sequence $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow$ splits if $C$ is free. So I thought because $\operatorname{ker}(i)$ is a subgroup of $\mathbb{Z}$, it's free. If this is correct, all I need to finish is $H_0(\mathbb{Q})$. I think it should be a countable infinite direct sum of $\mathbb{Z}$'s but I don't know how to express that. How about $\mathbb{Z}^\mathbb{N}$?
Then I'd have $H_0(\mathbb{Q}) = H_1(\mathbb{Q}, \mathbb{R}) \oplus \operatorname{ker}(i)$ and therefore $H_1(\mathbb{Q}, \mathbb{R}) \cong H_0(\mathbb{Q}) / \operatorname{ker}(i) \cong H_0(\mathbb{Q})$.
Now I'm supposed to find a basis for that.
Can someone tell me if my calculations are correct and if yes give me a hint on how to find a basis for $\mathbb{Z}^\mathbb{N}$? Many thanks for your help!
Edit I just found out we have the following exact sequence: $0 \rightarrow H_1(\mathbb{R}, \mathbb{Q}) \xrightarrow{f} H_0(\mathbb{Q}) \xrightarrow{g} H_0(\mathbb{R}) \xrightarrow{h} 0$
Using the previous exercise, 16 (a), I know that $H_0(\mathbb{R}, \mathbb{Q}) = 0$ because $\mathbb{Q}$ meets every path-component of $\mathbb{R}$. I also have $H_0(\mathbb{R}) = \mathbb{Z}$.
Edit 2
OK, I think I have it:
$ H_0( \mathbb{Q} ) = \oplus_{q \in \mathbb{Q}} \mathbb{Z} $
$ g: \oplus_{q \in \mathbb{Q}} \mathbb{Z} \rightarrow \mathbb{Z}$
$ \operatorname{im}{g} = \mathbb{Z} \implies \operatorname{ker}{g} = \oplus_{q \in \mathbb{Q} - \ast} \mathbb{Z}$
Here $\mathbb{Q} - \ast$ should be the rationals minus a point.
$ H_1(\mathbb{R}, \mathbb{Q}) / 0 = H_1(\mathbb{R}, \mathbb{Q}) = \operatorname{im}{f} = \oplus_{q \in \mathbb{Q} - \ast} \mathbb{Z}$
Can you tell me if this is correct?
What is a basis for this?
Many thanks for your help!