Use the Mayer-Vietoris sequence to calculate the integral homology of the following spaces.
\(\RP^2\), written as a Möbius strip \(M\) union a disk along their boundary.
\(T^2\), written as a punctured torus (homotopy-equivalent to \(S^1\vee S^1\)) union a disk along the boundary circle.
The Klein bottle, written as the twisted union of two cylinders \(S^1\times I\): on one overlap circle the gluing is the identity and on the other it is complex conjugation \(z\mapsto\bar z\).
\(S^1\cup_{z^n}B^2\), where the boundary circle of \(B^2\) is attached to \(S^1\) by \(z\mapsto z^n\).
Solution
All homology groups below have coefficients in \(\ZZ\).
1. For \(\RP^2=M\cup D^2\),\[ H_0(\RP^2)=\ZZ,\qquad H_1(\RP^2)=\ZZ/2,\qquad H_i(\RP^2)=0\quad(i\ge2). \]
Take collar neighborhoods so that the intersection deformation retracts to the common boundary circle \(S^1\). We have \[ H_1(S^1)\cong\ZZ,\qquad H_1(M)\cong\ZZ,\qquad H_1(D^2)=0. \] The boundary circle of a Möbius strip runs twice around its core, so the inclusion \[ H_1(S^1)\longrightarrow H_1(M) \] is multiplication by \(2\). Since all three spaces are connected, the map \[ H_0(S^1)\longrightarrow H_0(M)\oplus H_0(D^2) \] is injective. The relevant Mayer-Vietoris segment is therefore \[ 0\longrightarrow H_2(\RP^2)\longrightarrow\ZZ \xrightarrow{\times2}\ZZ \longrightarrow H_1(\RP^2)\longrightarrow0. \] Hence \(H_2=0\) and \(H_1\cong\ZZ/2\). Connectedness gives \(H_0\cong\ZZ\), and there is no homology above degree \(2\).
2. For the torus,\[ H_0(T^2)=\ZZ,\qquad H_1(T^2)=\ZZ^2,\qquad H_2(T^2)=\ZZ, \]and \(H_i(T^2)=0\) for \(i>2\).
Let \(P=T^2\setminus\operatorname{int}(D^2)\) be the punctured torus. Then \[ P\simeq S^1\vee S^1,\qquad H_1(P)\cong\ZZ^2,\qquad H_2(P)=0. \] Write \(T^2=P\cup D^2\), with intersection a collar of \(\partial P\cong S^1\). Under the deformation retraction \(P\simeq S^1\vee S^1\), the boundary loop represents the commutator \(aba^{-1}b^{-1}\), so its class vanishes in \[ H_1(P)=\pi_1(P)^{\mathrm{ab}}. \] Thus the Mayer-Vietoris map \[ H_1(S^1)\longrightarrow H_1(P)\oplus H_1(D^2) \] is zero. As in 1, the \(H_0\) map is injective. Hence exactness gives \[ 0\longrightarrow H_2(T^2)\longrightarrow\ZZ \xrightarrow{0}\ZZ^2 \longrightarrow H_1(T^2)\longrightarrow0, \] which yields \(H_2(T^2)\cong\ZZ\) and \(H_1(T^2)\cong\ZZ^2\).
3. For the Klein bottle \(K\),\[ H_0(K)=\ZZ,\qquad H_1(K)=\ZZ\oplus\ZZ/2, \qquad H_i(K)=0\quad(i\ge2). \]
Let \(A\) and \(B\) be the two cylinder pieces. Each deformation retracts to a circle, while \[ A\cap B\simeq S^1\sqcup S^1. \] Choose generators \(e_0,e_1\) for \[ H_1(A\cap B)\cong\ZZ^2 \] and generators \(a,b\) for \(H_1(A)\oplus H_1(B)\cong\ZZ^2\). Orient the first overlap so that both inclusions have degree \(+1\). On the second overlap the inclusion into \(A\) still has degree \(+1\), while the twisted gluing \(z\mapsto\bar z\) has degree \(-1\) into \(B\). With the Mayer-Vietoris sign convention \((i_*,-j_*)\), the \(H_1\) map is therefore represented, up to changing basis signs, by \[ \phi_1= \begin{pmatrix} 1&1\\ -1&1 \end{pmatrix}. \] Its determinant is \(2\), so \(\ker\phi_1=0\) and \[ \operatorname{coker}\phi_1\cong\ZZ/2. \]
On \(H_0\), both components of \(A\cap B\) map to the unique component of each cylinder, so \[ \phi_0:\ZZ^2\longrightarrow\ZZ^2, \qquad (x,y)\longmapsto(x+y,-x-y). \] Thus \(\ker\phi_0\cong\ZZ\) and \(\operatorname{coker}\phi_0\cong\ZZ\). The Mayer-Vietoris sequence gives \[ 0\to H_2(K)\to\ZZ^2\xrightarrow{\phi_1}\ZZ^2 \to H_1(K)\to\ZZ^2\xrightarrow{\phi_0}\ZZ^2 \to H_0(K)\to0. \] Hence \(H_2(K)=0\), \(H_0(K)\cong\ZZ\), and there is a short exact sequence \[ 0\longrightarrow\ZZ/2\longrightarrow H_1(K) \longrightarrow\ZZ\longrightarrow0. \] It splits because \(\ZZ\) is free, so \[ H_1(K)\cong\ZZ\oplus\ZZ/2. \]
4. Let\[ Y_n=S^1\cup_{z^n}B^2. \]Then\[ H_0(Y_n)=\ZZ,\qquad H_1(Y_n)=\operatorname{coker}(\times n)=\ZZ/n\ZZ,\qquad H_2(Y_n)=\ker(\times n), \]and \(H_i(Y_n)=0\) for \(i>2\). Equivalently,\[ (H_1,H_2)= \begin{cases} (\ZZ/|n|,0),&n\ne0,\\ (\ZZ,\ZZ),&n=0. \end{cases} \]
Decompose the attached disk into an outer annular collar and an inner disk. Let \(A\) be the original \(S^1\) together with the outer collar and let \(B\) be the inner disk together with a slightly overlapping collar. Then \[ A\simeq S^1,\qquad B\simeq *,\qquad A\cap B\simeq S^1. \] Under the deformation retraction \(A\simeq S^1\), the inclusion of \(A\cap B\) is the attaching map \(z\mapsto z^n\), hence induces multiplication by \(n\) on \(H_1\). The inclusion into \(B\) induces zero on \(H_1\). Since the \(H_0\) map is injective, Mayer-Vietoris gives \[ 0\longrightarrow H_2(Y_n)\longrightarrow\ZZ \xrightarrow{\times n}\ZZ \longrightarrow H_1(Y_n)\longrightarrow0. \] Taking kernel and cokernel yields the stated groups, including the previously omitted case \(n=0\).
5. Q.E.D.
The four Mayer-Vietoris computations in <1>1–<1>4 give all requested homology groups.