For the mapping torus $T_f$ of a map $f: X \to X$, we constructed an exact sequence of the form $$\cdots \longrightarrow H_n(X) \xrightarrow{ \ 1 – f_{\ast} \ } H_n(X) \longrightarrow H_n(T_f) \longrightarrow H_{n-1}(X) \longrightarrow \cdots.$$ Compute $H_n(T_f)$ for the map $S^1 \times S^1 \longrightarrow S^1 \times S^1$ that is the identity on the first factor and a reflection on the second factor.
Observe that $$H_k(S^1 \times S^1) = \begin{cases}
\mathbb{Z}, & k=0,2, \\
\mathbb{Z}^2, & k =1, \\
0, & \text{else}
\end{cases}$$
So we have realise the sequence to be $$H_3(T_f) \longrightarrow \mathbb{Z} \xrightarrow{ \ 1 – f_{\ast} \ } \mathbb{Z} \longrightarrow H_2(T_f) \longrightarrow \mathbb{Z}^2 \xrightarrow{ \ 1 – f_{\ast} \ } \mathbb{Z}^2 \longrightarrow H_1(T_f) \longrightarrow \mathbb{Z} \xrightarrow{1 – f_{\ast} \ } \mathbb{Z} $$ $$\longrightarrow H_0(T_f) \longrightarrow \mathbb{Z} \xrightarrow{1 – f_{\ast} } \mathbb{Z}.$$ My way of proceeding is then to break up this sequence into two sequences of the form $$H_3(T_f) \longrightarrow \mathbb{Z} \xrightarrow{ \ 1 – f_{\ast} \ } \mathbb{Z} \longrightarrow H_2(T_f) \longrightarrow \mathbb{Z} \xrightarrow{ \ 1 – f_{\ast} \ } \mathbb{Z} \longrightarrow H_1(T_f) \longrightarrow \mathbb{Z} \xrightarrow{1 – f_{\ast} \ } \mathbb{Z} $$ $$\longrightarrow H_0(T_f) \longrightarrow \mathbb{Z} \xrightarrow{1 – f_{\ast} } \mathbb{Z},$$ since the direct sum of $\mathbb{Z}$-modules is an exact sequence if each $\mathbb{Z}$-module is exact.
In the case that $f_{\ast}$ is simply the identity, $1- f_{\ast}$ is simply the zero map and the homology groups are given by $$H_k(T_f) = \begin{cases}
\mathbb{Z}, & 0 \leq k \leq 3, \\
0, & \text{otherwise}.
\end{cases}$$
While in the second case, where $f$ is a reflection, $\deg f = -1$ and the map $1 – f_{\ast}$ corresponds to multiplication by 2 except for the last map, which is simply the identity map for dimension reasons. Therefore, $$H_k(T_f) = \begin{cases}
\mathbb{Z}, & k=0,3, \\
\mathbb{Z}_2, & k=1,2,\\
0, & \text{otherwise}.
\end{cases}$$
I'm unsure of whether this is correct and am unsure as to how I can recombine the exact sequences to get one solution for $H_k(T_f)$.
Best Answer
I'm not 100% certain but I'm pretty sure this is the idea of how to solve it without breaking up the sequence as you've suggested.
Using your statement about $H_k(S^1 \times S^1)$, and your first long exact sequence, because $H_k(S^1 \times S^1) = 0$ for $k \geq 3$, the nontrivial part of the exact sequence becomes
$$0 \to H_3(T_f) \to \mathbb Z \xrightarrow[]{\mathbb 1 - f_*} \mathbb Z \to H_2(T_f) \to \mathbb Z^2 \xrightarrow[]{\mathbb 1 - f_*} \mathbb Z^2 \to H_1(T_f) \to \mathbb Z \xrightarrow{\mathbb1 - f_*} \mathbb Z \to H_0(T_f) \to 0,$$
where $\mathbb 1$ is the identity map. Since $T_f$ has one connected component, $H_0(T_f) = \mathbb Z$. The dimension-$2$ induced homology map is $1 \mapsto -1$. (One way of seeing this is by giving $S^1 \times S^1$ a simplicial structure as in Example 2.3 in Hatcher, and observing that $f_\sharp(U - L)$ is homologous to $-U + L$.) Therefore the left hand map $\mathbb 1 - f_* : \mathbb Z \to \mathbb Z$ is defined by $1 \mapsto 2$. Since this is injective, exactness gives us $\tilde H_3(T_f) = 0$.
The dimension-$0$ induced homology map $f_* : \mathbb Z \to \mathbb Z$ is the identity, so $\mathbb 1 - f_* = 0$ in the right hand map $\mathbb Z \to \mathbb Z$. Meanwhile, the dimension-$1$ induced homology map $f_* : \mathbb Z^2 \to \mathbb Z^2$ sends one generator to itself, and the other generator to its inverse, and so can be represented by the matrix $\left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right)$. Therefore $\mathbb 1 - f_* = \left( \begin{array}{cc} 0 & 0 \\ 0 & 2 \end{array}\right)$. By exactness, $\ker(\mathbb Z^2 \to H_1(T_f)) = \mathrm{im}\left( \begin{array}{cc} 0 & 0 \\ 0 & 2 \end{array}\right)$, which is generated by $2$, so $\mathbb Z^2 / \ker(\mathbb Z^2 \to H_1(T_f)) = \mathbb Z \oplus \mathbb Z_2$. This gives us a short exact sequence
$$0 \to \mathbb Z \oplus \mathbb Z_2 \to H_1(T_f) \to \mathbb Z \to 0.$$
Since $\mathbb Z$ is free, this exact sequence splits, meaning $H_1(T_f)$ is the direct sum of the first and third nontrivial terms. So we have $H_1(T_f) = \mathbb Z^2 \oplus \mathbb Z_2$.
Lastly, recall $\mathbb Z \to \mathbb Z$ on the left side of the first exact sequence is $1 \mapsto 2$. By exactness, this is also the kernel of $\mathbb Z \to H_2(T_f)$, so by the first isomorphism theorem, $$ \mathrm{im}(\mathbb Z \to H_2(T_f)) = \mathbb Z / 2\mathbb Z = \mathbb Z_2.$$ Also, the map $H_2(T_f) \to \mathbb Z^2$ has image generated by $(1,0)$, and again by the first isomorphism theorem and exactness, $$\mathbb Z = \mathrm{im}(H_2(T_f) \to \mathbb Z^2) = H_2(T_f)/\ker(H_2(T_f) \to \mathbb Z^2) = H_2(T_f) / \mathrm{im}(\mathbb Z \to H_2(T_f)) = H_2(T_f) / \mathbb Z_2.$$ By uniqueness of the decomposition of a finitely generated abelian group into cyclic subgroups, this gives us $H_2(T_f) = \mathbb Z \oplus \mathbb Z_2$. So, to summarize, $$ H_n(T_f) = \begin{cases} \mathbb Z & \textrm{if } n = 0, \\ \mathbb Z^2 \oplus \mathbb Z_2 & \textrm{if } n = 1, \\ \mathbb Z \oplus \mathbb Z_2 & \textrm{if } n = 2, \\ 0 & \textrm{if } n \geq 3. \end{cases} $$