Use Mayer–Vietoris sequences to compute the singular homology groups of the 2-torus \(T=S^1\times S^1\). (You may use knowledge of the homology groups of the circle \(S^1\) in your calculations.)
Solution
Model \(T\) as the square with opposite sides identified, let \(x_0\in T\) be the image of an interior point of the square, let \(V\) be a small open disk around \(x_0\), and let \(U=T\setminus\{x_0\}\). \(U\) and \(V\) are open and cover \(T\).
1. \(U\simeq S^1\vee S^1\), \(V\simeq*\), and \(U\cap V\simeq S^1\).
\(V\) is a disk, and \(U\cap V=V\setminus\{x_0\}\) is a punctured disk. The punctured square retracts radially onto its boundary, which becomes the \(1\)-skeleton \(S^1\vee S^1\) of \(T\) after the side identifications.
2. The map \(H_1(U\cap V)\to H_1(U)\oplus H_1(V)\) in the Mayer–Vietoris sequence is zero.
Let \(a,b\) be the loops of \(S^1\vee S^1\simeq U\). A loop around \(x_0\) is homotopic in \(U\) to the boundary of the square, which represents \(aba^{-1}b^{-1}\) in \(\pi_1(U)\); its class in the abelianization \(H_1(U)\) is \(0\). Also \(H_1(V)=0\).
3. \(H_2(T)\cong\ZZ\) and \(H_1(T)\cong\ZZ^2\).
By step 1, \(H_2(U)=H_2(V)=0\), \(H_1(U\cap V)\cong\ZZ\), \(H_1(U)\oplus H_1(V)\cong\ZZ^2\), and \(U\), \(V\), \(U\cap V\) are path-connected. The Mayer–Vietoris sequence contains the exact segment \[0\to H_2(T)\xrightarrow{\delta}H_1(U\cap V)\xrightarrow{0}H_1(U)\oplus H_1(V)\to H_1(T)\xrightarrow{\partial}H_0(U\cap V)\xrightarrow{\phi}H_0(U)\oplus H_0(V),\] where the zero map is step 2 and \(\phi\) sends the generator to \((1,-1)\), so \(\phi\) is injective. Exactness makes \(\delta\) an isomorphism onto \(\ZZ\), and makes \(\partial=0\), so \(H_1(U)\oplus H_1(V)\cong\ZZ^2\to H_1(T)\) is an isomorphism.
4. Q.E.D.
\(T\) is path-connected, so \(H_0(T)\cong\ZZ\). For \(n\ge3\), \(H_n(U)\oplus H_n(V)=0\) and \(H_{n-1}(U\cap V)=0\), so \(H_n(T)=0\). With step 3, \[H_n(T)\cong\begin{cases}\ZZ,&n=0,\\\ZZ^2,&n=1,\\\ZZ,&n=2,\\0,&n\ge3.\end{cases}\]