Show that \(S^1 \times S^1\) and \(S^1 \lor S^1 \lor S^2\) have isomorphic homology groups in all dimensions, but their universal covering spaces do not.
Solution
Set \[ T=S^1\times S^1, \qquad Y=S^1\vee S^1\vee S^2. \]
1. The spaces \(T\) and \(Y\) have isomorphic homology groups in every degree:\[ H_n(T)\cong H_n(Y)\cong \begin{cases} \mathbb Z,&n=0,\\ \mathbb Z^2,&n=1,\\ \mathbb Z,&n=2,\\ 0,&n>2. \end{cases} \]
For the torus this is the standard cellular chain computation with one \(0\)-cell, two \(1\)-cells, and one \(2\)-cell, all cellular boundary maps zero. For the wedge, reduced homology splits over a finite wedge: \[ \widetilde H_n(Y) \cong \widetilde H_n(S^1)\oplus\widetilde H_n(S^1)\oplus\widetilde H_n(S^2), \] which gives the same groups.
2. The universal cover of \(T\) is \(\mathbb R^2\), hence is contractible.
The quotient map \[ \mathbb R^2\longrightarrow\mathbb R^2/\mathbb Z^2\cong S^1\times S^1 \] is the universal covering map. Therefore its universal cover has trivial reduced homology.
3. The universal cover of \(Y\) has nonzero \(H_2\).
The fundamental group of \(Y\) is the free group \(F_2\) coming from the two circle summands. The universal cover of the \(1\)-skeleton \(S^1\vee S^1\) is the Cayley tree of \(F_2\). At every lift of the wedge point, a copy of the simply connected summand \(S^2\) lifts and is attached at that vertex. Hence the universal cover of \(Y\) is a tree with one \(2\)-sphere attached at each vertex.
Collapsing the tree to a point gives a homotopy equivalence to a wedge of one \(S^2\) for each vertex, so \[ H_2(\widetilde Y) \cong \bigoplus_{F_2}\mathbb Z, \] which is nonzero.
4. Therefore the universal covering spaces are not homeomorphic, and indeed not homotopy equivalent.
3. Homology is a homotopy invariant, so the universal covers cannot be homotopy equivalent or homeomorphic.