Compute the (reduced) singular homology groups of the space \(X=S^1\times(S^1\vee S^1)\), which can be thought of as two copies of \(S^1\times S^1\) glued together along their copies of \(S^1\times\{x_0\}\). You may use your knowledge of the homology groups of \(T^2=S^1\times S^1\) in your calculations.
Solution
Write \(X=A\cup B\), where \(A\) and \(B\) are the two copies of \(T^2=S^1\times S^1\) and \(A\cap B=C=S^1\times\{x_0\}\cong S^1\). Let \(N\) be a small open neighborhood of \(x_0\) in \(S^1\vee S^1\) that deformation retracts onto \(x_0\). The open sets \(A\cup(S^1\times N)\) and \(B\cup(S^1\times N)\) deformation retract onto \(A\) and \(B\), with intersection \(S^1\times N\) deformation retracting onto \(C\), so the Mayer–Vietoris sequence applies to \(A\), \(B\), \(C\). Choose bases \((a,b)\) of \(H_1(A)\) and \((a',b')\) of \(H_1(B)\) with \(a\) and \(a'\) the classes of \(C\), and let \(c\) generate \(H_1(C)\).
1. The Mayer–Vietoris map \(\Phi=(i_*,-j_*)\colon H_1(C)\to H_1(A)\oplus H_1(B)\cong\ZZ^4\) is injective, with image the direct summand generated by \((a,0,-a',0)\).
\(\Phi(c)=(a,0,-a',0)\), which is a primitive vector of \(\ZZ^4\).
2. \(H_2(X)\cong\ZZ^2\).
The Mayer–Vietoris sequence contains the exact segment \(0=H_2(C)\to H_2(A)\oplus H_2(B)\cong\ZZ^2\to H_2(X)\to H_1(C)\xrightarrow{\Phi}\ZZ^4\). By step 1, \(\Phi\) is injective, so \(\ZZ^2\to H_2(X)\) is an isomorphism.
3. \(H_1(X)\cong\ZZ^3\).
The next segment is \(H_1(C)\xrightarrow{\Phi}H_1(A)\oplus H_1(B)\to H_1(X)\to H_0(C)\to H_0(A)\oplus H_0(B)\). \(A\), \(B\), \(C\) are connected, so the last map is \(\ZZ\to\ZZ^2\), \(1\mapsto(1,-1)\), which is injective. By exactness and step 1, \(H_1(X)\cong\ZZ^4/\langle(1,0,-1,0)\rangle\cong\ZZ^3\).
4. Q.E.D.
\(X\) is the union of the connected tori \(A\) and \(B\) with nonempty intersection, so it is connected and \(\widetilde H_0(X)=0\). For \(k\ge3\) the Mayer–Vietoris groups on both sides of \(H_k(X)\) vanish, because tori have no homology above degree \(2\) and the circle none above degree \(1\). With steps 2 and 3, \[\widetilde H_k(X)\cong\begin{cases}\ZZ^3,&k=1,\\\ZZ^2,&k=2,\\0,&\text{otherwise}.\end{cases}\]