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.
Solution
Since \(\mathbb R\) is contractible, \[ H_1(\mathbb R)=0, \qquad H_0(\mathbb R)\cong\mathbb Z. \] The subspace \(\mathbb Q\) is totally disconnected, so each rational number is a path-component and \[ H_0(\mathbb Q)\cong\bigoplus_{q\in\mathbb Q}\mathbb Z[q]. \]
1. The long exact sequence gives\[ H_1(\mathbb R,\mathbb Q) \cong \ker\left(H_0(\mathbb Q)\to H_0(\mathbb R)\right). \]
The relevant segment is \[ 0=H_1(\mathbb R)\to H_1(\mathbb R,\mathbb Q) \to H_0(\mathbb Q)\to H_0(\mathbb R). \] Exactness gives the claimed identification.
2. Under the component bases, the map\[ H_0(\mathbb Q)\to H_0(\mathbb R)\cong\mathbb Z \]is the augmentation\[ \sum_q n_q[q]\longmapsto\sum_q n_q. \]
All points of \(\mathbb Q\) lie in the single path-component of \(\mathbb R\), so every generator \([q]\) maps to the same generator \(1\) of \(H_0(\mathbb R)\).
3. Fix \(0\in\mathbb Q\). Then\[ \mathcal B=\{[q]-[0]:q\in\mathbb Q,\ q\ne0\} \]is a basis of the augmentation kernel.
Each element lies in the kernel. Any finite sum \[ \sum_q n_q[q] \] with total coefficient zero equals \[ \sum_{q\ne0}n_q([q]-[0]), \] so \(\mathcal B\) spans. If a finite linear combination of elements of \(\mathcal B\) vanishes, comparison of the coefficient of each \([q]\) with \(q\ne0\) shows every coefficient is zero, so the set is independent.
4. Therefore\[ \boxed{H_1(\mathbb R,\mathbb Q) \cong\bigoplus_{q\in\mathbb Q\setminus\{0\}}\mathbb Z,} \]with basis represented by the classes \([q]-[0]\).
Combine <1>1–<1>3.