Let \(X\) be the space obtained as the quotient of a disjoint union of a 2-sphere \(S^2\) and a torus \(T = S^1 \times S^1\) by identifying the equator in \(S^2\) with a circle \(S^1 \times \theset{p}\) in \(T\).
Compute the homology groups of \(X\).
Solution
Let \[ C\cong S^1 \] denote the circle obtained by identifying the equator of \(S^2\) with \[ S^1\times\{p\}\subset T^2. \]
1. Mayer–Vietoris may be applied using an open cover \(X=U\cup V\) with\[ U\simeq S^2, \qquad V\simeq T^2, \qquad U\cap V\simeq C\simeq S^1. \]
The equator in \(S^2\) and the circle \(S^1\times\{p\}\) in \(T^2\) both have annular neighborhoods. After the two circles are identified, enlarge the images of \(S^2\) and \(T^2\) slightly across these annuli. The resulting sets \(U\) and \(V\) are open, retract respectively onto the sphere and torus, and their intersection retracts onto the common circle \(C\). Thus their homology may be replaced by that of the displayed deformation retracts in the Mayer–Vietoris sequence.
2. The relevant homology groups are\[ H_k(S^2;\ZZ) \cong \begin{cases} \ZZ,&k=0,2,\\ 0,&\text{otherwise}, \end{cases} \]\[ H_k(T^2;\ZZ) \cong \begin{cases} \ZZ,&k=0,2,\\ \ZZ^2,&k=1,\\ 0,&\text{otherwise}, \end{cases} \]and\[ H_k(C;\ZZ) \cong \begin{cases} \ZZ,&k=0,1,\\ 0,&\text{otherwise}. \end{cases} \]
These are the standard integral homology groups of the sphere, torus, and circle.
3. The Mayer–Vietoris map\[ \Phi:H_1(C)\longrightarrow H_1(S^2)\oplus H_1(T^2) \]is injective and has cokernel isomorphic to \(\ZZ\).
Let \(a,b\) be the standard basis of \[ H_1(T^2;\ZZ)\cong\ZZ^2, \] with \(a\) represented by the factor circle \(S^1\times\{p\}\). The inclusion of the equator into \(S^2\) induces the zero map on \(H_1\) because \[ H_1(S^2)=0. \] The inclusion of \(C=S^1\times\{p\}\) into the torus sends a generator of \(H_1(C)\) to the primitive class \(a\). Hence, up to the Mayer–Vietoris sign convention, \[ \Phi:\ZZ\longrightarrow 0\oplus\ZZ^2, \qquad 1\longmapsto (0,a). \] This map is injective, and \[ \operatorname{coker}\Phi \cong \ZZ^2/\langle a\rangle \cong\ZZ. \]
4. One has\[ H_2(X;\ZZ)\cong\ZZ^2. \]
The degree-\(2\) portion of Mayer–Vietoris is \[ H_2(C) \longrightarrow H_2(S^2)\oplus H_2(T^2) \longrightarrow H_2(X) \longrightarrow H_1(C) \xrightarrow{\Phi} H_1(S^2)\oplus H_1(T^2). \] Using 2, this becomes \[ 0 \longrightarrow \ZZ^2 \longrightarrow H_2(X) \longrightarrow \ZZ \xrightarrow{\Phi} \ZZ^2. \] The map \(\Phi\) is injective by 3, so the image of \[ H_2(X)\longrightarrow\ZZ \] is zero by exactness. Therefore \[ \ZZ^2\longrightarrow H_2(X) \] is both injective and surjective, hence an isomorphism.
5. One has\[ H_1(X;\ZZ)\cong\ZZ. \]
The next part of Mayer–Vietoris is \[ H_1(C) \xrightarrow{\Phi} H_1(S^2)\oplus H_1(T^2) \longrightarrow H_1(X) \longrightarrow H_0(C) \longrightarrow H_0(S^2)\oplus H_0(T^2). \] All three spaces \(C,S^2,T^2\) are connected, so the last map is, up to sign, \[ \ZZ\longrightarrow\ZZ^2, \qquad 1\longmapsto(1,-1), \] and is injective. Exactness therefore gives \[ H_1(X)\cong\operatorname{coker}\Phi. \] By 3 this cokernel is \(\ZZ\).
6. The space \(X\) is connected, so\[ H_0(X;\ZZ)\cong\ZZ, \]and\[ H_k(X;\ZZ)=0 \qquad(k\ge3). \]
The images of \(S^2\) and \(T^2\) in \(X\) are connected and meet along the nonempty circle \(C\), so their union \(X\) is connected. Thus \(H_0(X)\cong\ZZ\).
4.In particular, the degree-\(3\) segment is\[ 0\longrightarrow0\longrightarrow H_3(X)\longrightarrow H_2(C)=0, \]so \(H_3(X)=0\), and the same vanishing argument applies in all higher degrees.
7. Therefore\[ H_k(X;\ZZ) \cong \begin{cases} \ZZ,&k=0,1,\\ \ZZ^2,&k=2,\\ 0,&k\ge3. \end{cases} \]
This collects <1>4–<1>6.