Compute the homology groups of the subspace of \(I \times I\) consisting of the four boundary edges plus all points in the interior whose first coordinate is rational.
Solution
Let \(X\) be the specified subspace of \(I\times I\). Thus \(X\) contains the whole boundary square and, for each rational \(q\in(0,1)\), the vertical segment \(\{q\}\times I\).
Define open subsets of \(X\) \[ U=X\cap\{y<3/4\}, \qquad V=X\cap\{y>1/4\}. \]
1. Both \(U\) and \(V\) are contractible.
The vertical homotopy \[ (x,y)\longmapsto(x,(1-t)y) \] deformation retracts \(U\) onto the bottom edge. It stays in \(X\) because an interior point of \(X\) either has rational first coordinate or lies on a vertical boundary edge, and the homotopy preserves the first coordinate. Similarly \(V\) deformation retracts onto the top edge.
2. The intersection \(U\cap V\) deformation retracts onto\[ Q=(\mathbb Q\cap I)\times\{1/2\}, \]so\[ \widetilde H_0(U\cap V) \cong \bigoplus_{q\in(\mathbb Q\cap I)\setminus\{0\}}\mathbb Z \]and\[ H_k(U\cap V)=0\quad(k>0). \]
On \(U\cap V\) the vertical contraction to height \(1/2\) is well-defined and stays in \(X\). At heights strictly between \(1/4\) and \(3/4\), points of \(X\) have rational first coordinate, except that the side edges have first coordinates \(0,1\), which are rational anyway. Hence the retract is precisely the copy of \(\mathbb Q\cap I\) displayed above.
Every path-component of \(\mathbb Q\cap I\) is a single point, so its \(H_0\) is free on the rational points and the reduced group is the augmentation kernel, with basis \([q]-[0]\) for \(q\ne0\). Any singular simplex of positive dimension in \(\mathbb Q\cap I\) has connected image and is therefore constant, so the positive-dimensional homology vanishes.
3. The space \(X\) is path connected, and\[ H_1(X)\cong\widetilde H_0(U\cap V). \]
Every point can be joined vertically to the top or bottom boundary and then along the boundary to any other point, so \(X\) is path connected.
The reduced Mayer–Vietoris sequence for \(X=U\cup V\) contains \[ 0=\widetilde H_1(U)\oplus\widetilde H_1(V) \to\widetilde H_1(X) \to\widetilde H_0(U\cap V) \to\widetilde H_0(U)\oplus\widetilde H_0(V)=0. \] Thus the middle arrow is an isomorphism.
4. All homology in dimensions at least two vanishes.
For \(k\ge2\), Mayer–Vietoris gives \[ \widetilde H_k(X)\cong\widetilde H_{k-1}(U\cap V)=0 \] by 2.
5. Therefore\[ \boxed{ H_0(X)\cong\mathbb Z, \qquad H_1(X)\cong \bigoplus_{q\in(\mathbb Q\cap I)\setminus\{0\}}\mathbb Z, \qquad H_k(X)=0\ (k\ge2). } \]
Combine <1>2–<1>4.