$H_1(\mathbb{R}, \mathbb{Q})$ is free abelian 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!
 A: A subgroup of a free abelian group is free abelian, by a well known theorem of Kaplanski.
That fact and the long exact sequence do the trick.
Later. $\newcommand\QQ{\mathbb{Q}}\newcommand\RR{\mathbb{R}}\newcommand\ZZ{\mathbb{Z}}$Let's "compute" $H_1(\RR,\QQ)$. 
We know that the path-connected components of $\QQ$ are the points. It follows from the general description of $H_0$ that if we denote, for each $q\in\QQ$, $[q]_\QQ$ the homology class in $H_0(\QQ)$ of the point $q$ and $\ZZ[q]_\QQ$ the subgroup it generates in $H_0(\QQ)$, then there is an isomorphism $$H_0(\QQ)\cong\bigoplus_{q\in\QQ}\ZZ[q]_\QQ$$
We also know that $H_0(\RR)$ is free abelian of rank one, generated by the homology class of any point; moreover, we know that if $x$ and $y$ are points in $\RR$ and we denote $[x]_\RR$ and $[y]_\RR$ their homology classes in $H_0(\RR)$, then $[x]_\RR=[y]_\RR$.
Now, the map $g$ is induced by the inclusion $\QQ\to\RR$. In other words, for each $q\in\QQ$ we have $$g([q]_\QQ)=[q]_\RR.$$ In fact, since the set $\{[q]_\QQ:q\in\QQ\}$ is a basis of the abelian group $H_0(\QQ)$, this information completely determines the map $g$.
Next, and as you observed, there is a short exact sequence $$0\to H_1(\RR,\QQ)\xrightarrow fH_0(\QQ)\xrightarrow gH_0(\RR)$$ coming from the long exact sequence inhomology for the pair $(\RR,\QQ)$. We therefore obtain a description of $H_1(\RR,\QQ)$ as an abelian group isomorphic to the kernel of the map $g$.
Can you construct a basis of $\ker g$?
