TITLE: The relative homology group $H_{1}(\mathbb{R},\mathbb{Q})$
QUESTION [5 upvotes]: This problem is an extract of Hatcher's book.

Show that for the subspace $\mathbb{Q}\subset \mathbb{R}$, the relative homology group $H_{1}(\mathbb{R},\mathbb{Q})$ is free abelian and find a basis.

I have no idea (edit : for the first part of the question, I think I have to use the long exact sequence...). Thank you for help !

REPLY [2 votes]: Perhaps this is an easier way of seeing this, in a very similar way to Najib's answer.
Instead of using the LES on homology, use the LES on reduced homology:
$$H_1(\mathbb{R}) \rightarrow H_1(\mathbb{R}, \mathbb{Q}) \rightarrow \widetilde{H_0}(\mathbb{Q}) \rightarrow \widetilde{H_0}(\mathbb{R})$$
Note that, since $\mathbb{R}$ is contractible, $\widetilde{H_0}(\mathbb{R}) = 0$ and $H_1(\mathbb{R}) = 0$. We also know that $\widetilde{H_0}(\mathbb{Q}) \cong \bigoplus\limits_{n \in \mathbb{N}} \mathbb{Z}$.
So we have the following exact sequence:
$$0 \rightarrow H_1(\mathbb{R}, \mathbb{Q}) \rightarrow \bigoplus\limits_{n \in \mathbb{N}} \mathbb{Z} \rightarrow 0$$
By exactness, it follows that $H_1(\mathbb{R}, \mathbb{Q}) \cong \bigoplus\limits_{n \in \mathbb{N}} \mathbb{Z}$, so $H_1(\mathbb{R}, \mathbb{Q})$ is a free abelian group on countably infinitely many generators.