Express a Klein bottle as the union of two annuli.
Use the Mayer–Vietoris sequence and this decomposition to compute its homology.
Solution
Write the Klein bottle as the mapping torus of the reflection \[ r:S^1\longrightarrow S^1, \qquad r(z)=\overline z: \qquad K=(S^1\times[0,1])/((z,0)\sim(r(z),1)). \] Let \[ p:K\longrightarrow S^1, \qquad p([z,t])=e^{2\pi i t}, \] be the mapping-torus projection.
1. The Klein bottle is the union of two annuli \(U\) and \(V\) whose intersection is the disjoint union of two annuli.
Choose two closed arcs \(A,B\subset S^1\) such that \[ A\cup B=S^1, \qquad \operatorname{int}(A)\cup\operatorname{int}(B)=S^1, \] and \(A\cap B\) has exactly two connected components \(C_0,C_1\), each an arc. Set \[ U=p^{-1}(A), \qquad V=p^{-1}(B). \] The mapping torus is an \(S^1\)-bundle over \(S^1\) with monodromy \(r\). Over an arc the bundle is trivial, hence \[ U\cong S^1\times[0,1], \qquad V\cong S^1\times[0,1]. \] Thus \(U\) and \(V\) are annuli. Likewise \[ U\cap V=p^{-1}(C_0)\sqcup p^{-1}(C_1) \] is the disjoint union of two annuli. Because the interiors of \(U\) and \(V\) cover \(K\), this is an excisive pair and the Mayer–Vietoris sequence applies.
2. With suitable generators, the Mayer–Vietoris map\[ \alpha:H_1(U\cap V)\longrightarrow H_1(U)\oplus H_1(V) \]is represented by\[ \begin{pmatrix} 1&1\\ -1&1 \end{pmatrix}. \]
Each annulus deformation retracts onto a fiber circle, so \[ H_1(U\cap V)\cong\ZZ^2, \qquad H_1(U)\oplus H_1(V)\cong\ZZ^2. \] Choose fiber generators \(x_0,x_1\) on the two components of \(U\cap V\), and generators \(a,b\) for \(U,V\).
Choose the trivializations so that on the overlap component \(C_0\) away from the mapping-torus seam the transition function is the identity, while on the component \(C_1\) crossing the seam it is the reflection \(r\). The reflection has degree \(-1\) on \(S^1\). Thus, for the Mayer–Vietoris map \((i_*,-j_*)\), \[ \alpha(x_0)=(a,-b), \qquad \alpha(x_1)=(a,b), \] which gives the displayed matrix.
3. One has\[ H_2(K;\ZZ)=0 \qquad\text{and}\qquad \operatorname{coker}(\alpha)\cong\ZZ/2\ZZ. \]
The relevant part of Mayer–Vietoris begins \[ 0 \longrightarrow H_2(K) \longrightarrow \ZZ^2 \xrightarrow{\alpha} \ZZ^2. \] The matrix in 2 has determinant \(2\), so \(\alpha\) is injective over \(\ZZ\) and therefore \(H_2(K)=0\).
Its entries have greatest common divisor \(1\) and its determinant has absolute value \(2\). Hence its Smith normal form is \[ \operatorname{diag}(1,2), \] so \[ \operatorname{coker}(\alpha)\cong\ZZ/2\ZZ. \]
4. The map\[ \beta:H_0(U\cap V)\longrightarrow H_0(U)\oplus H_0(V) \]has kernel isomorphic to \(\ZZ\) and cokernel isomorphic to \(\ZZ\).
Both \(U\) and \(V\) are connected, while \(U\cap V\) has two components. Thus, under the standard generators, \[ \beta:\ZZ^2\longrightarrow\ZZ^2, \qquad \beta(m,n)=(m+n,-m-n). \] Therefore \[ \ker\beta=\{(m,-m):m\in\ZZ\}\cong\ZZ, \] and \[ \operatorname{coker}\beta\cong\ZZ. \]
5. The integral homology of the Klein bottle is\[ \boxed{ H_n(K;\ZZ)\cong \begin{cases} \ZZ,&n=0,\\ \ZZ\oplus\ZZ/2\ZZ,&n=1,\\ 0,&n\ge2. \end{cases}} \]
Exactness around \(H_1(K)\) gives \[ 0 \longrightarrow\operatorname{coker}(\alpha) \longrightarrow H_1(K) \longrightarrow\ker(\beta) \longrightarrow0. \] By 3 and 4 this is \[ 0\longrightarrow\ZZ/2\ZZ \longrightarrow H_1(K) \longrightarrow\ZZ \longrightarrow0. \] Since \(\ZZ\) is free, the sequence splits, so \[ H_1(K)\cong\ZZ\oplus\ZZ/2\ZZ. \] Also \(H_0(K)\cong\operatorname{coker}(\beta)\cong\ZZ\), 3 gives \(H_2(K)=0\), and Mayer–Vietoris gives \(H_n(K)=0\) for \(n\ge3\) because the annuli and their intersection have no homology in those degrees.