For the mapping torus \(T_f\) of a map \(f: X \to X\), we constructed in Example 2.48 a long exact sequence \(\cdots \to H_n(X) \xrightarrow{1-f_*} H_n(X) \to H_n(T_f) \to H_{n-1}(X) \to \cdots\). Use this to compute the homology of the mapping tori of the following maps:
A reflection \(S^2 \to S^2\).
A map \(S^2 \to S^2\) of degree 2.
The map \(S^1 \times S^1 \to S^1 \times S^1\) that is the identity on one factor and a reflection on the other.
The map \(S^1 \times S^1 \to S^1 \times S^1\) that is a reflection on each factor.
The map \(S^1 \times S^1 \to S^1 \times S^1\) that interchanges the two factors and then reflects one of the factors.
Solution
For the mapping torus \(T_f\), the Wang sequence yields short exact sequences \[ 0\to\operatorname{coker}(1-f_*:H_n(X)\to H_n(X)) \to H_n(T_f) \to\ker(1-f_*:H_{n-1}(X)\to H_{n-1}(X)) \to0. \] In the cases below the kernel term is free, so each sequence splits as an abstract group.
1. For a reflection \(f:S^2\to S^2\),\[ H_i(T_f)\cong \begin{cases} \mathbb Z,&i=0,1,\\ \mathbb Z_2,&i=2,\\ 0,&i\ge3. \end{cases} \]
On \(H_2(S^2)=\mathbb Z\), a reflection acts by \(-1\), so \(1-f_*\) is multiplication by \(2\). On \(H_0\) it is zero. The displayed groups follow immediately from the short exact sequences.
2. If \(f:S^2\to S^2\) has degree \(2\), then\[ H_i(T_f)\cong \begin{cases} \mathbb Z,&i=0,1,\\ 0,&i\ge2. \end{cases} \]
On \(H_2\), the map \(1-f_*\) is multiplication by \(-1\), hence an isomorphism. On \(H_0\) it is zero. Thus there is no homology contribution in degrees \(2\) or \(3\), while the kernel in degree \(0\) produces the usual \(\mathbb Z\) in \(H_1\).
Let \(T^2=S^1\times S^1\) and choose the standard basis of \(H_1(T^2)=\mathbb Z^2\).
3. If \(f\) is the identity on the first factor and a reflection on the second, then\[ H_i(T_f)\cong \begin{cases} \mathbb Z,&i=0,\\ \mathbb Z^2\oplus\mathbb Z_2,&i=1,\\ \mathbb Z\oplus\mathbb Z_2,&i=2,\\ 0,&i=3,\\ 0,&i>3. \end{cases} \]
On \(H_1\), \(f_*=\operatorname{diag}(1,-1)\), so \[ 1-f_*=\operatorname{diag}(0,2). \] Its kernel is \(\mathbb Z\) and its cokernel is \(\mathbb Z\oplus\mathbb Z_2\). On \(H_2(T^2)=\mathbb Z\), \(f_*\) is multiplication by the determinant \(-1\), so \(1-f_*=2\). Hence \[ H_2(T_f)\cong\mathbb Z_2\oplus\mathbb Z, \] while \[ H_1(T_f)\cong(\mathbb Z\oplus\mathbb Z_2)\oplus\mathbb Z. \] There is no \(H_3\) because multiplication by \(2\) on \(H_2\) has trivial kernel.
4. If \(f\) reflects both factors, then\[ H_i(T_f)\cong \begin{cases} \mathbb Z,&i=0,2,3,\\ \mathbb Z\oplus\mathbb Z_2\oplus\mathbb Z_2,&i=1,\\ 0,&i>3. \end{cases} \]
On \(H_1\), \(f_*=-I\), so \(1-f_*=2I\). Its kernel is zero and its cokernel is \((\mathbb Z_2)^2\). On \(H_2\), \(f_*\) is multiplication by \(\det(-I)=+1\), so \(1-f_*=0\). Thus \(H_3\cong\mathbb Z\), \(H_2\cong\mathbb Z\), and \[ H_1\cong(\mathbb Z_2)^2\oplus\mathbb Z. \]
5. If \(f\) interchanges the two factors and then reflects one factor, then\[ H_i(T_f)\cong \begin{cases} \mathbb Z,&i=0,2,3,\\ \mathbb Z\oplus\mathbb Z_2,&i=1,\\ 0,&i>3. \end{cases} \]
Up to changing the basis, the induced map on \(H_1\) has matrix \[ A= \begin{pmatrix} 0&-1\\ 1&0 \end{pmatrix}. \] Thus \[ I-A= \begin{pmatrix} 1&1\\ -1&1 \end{pmatrix}. \] This matrix has determinant \(2\) and Smith normal form \(\operatorname{diag}(1,2)\), so its kernel is zero and its cokernel is \(\mathbb Z_2\). Since \(\det A=1\), the induced map on \(H_2\) is the identity, hence \(1-f_*=0\) there. Therefore \(H_3\cong\mathbb Z\), \(H_2\cong\mathbb Z\), and \[ H_1\cong\mathbb Z_2\oplus\mathbb Z. \]