= Solution
Use <integer> coefficients. The <mapping-torus homology from Mayer–Vietoris> works for every <continuous map> $f:X\to X$, without assuming that the <mapping torus> is a <fiber bundle>. Project to the time <circle> and choose two open arcs, one containing the seam and one away from it. The inverse image $U$ of the seam arc is a <mapping cylinder> neighborhood with a <deformation retraction> to its target copy of $X$; the inverse image $V$ of the other arc is a product with a <deformation retraction> to $X$. Their intersection has two components, each with a <deformation retraction> to $X$. Label the upper component first. With the convention that the second inclusion in the <Mayer–Vietoris sequence> has a minus sign, its <homology> map is
$$
H_q(X)\oplus H_q(X)\longrightarrow H_q(X)\oplus H_q(X),\qquad (a,b)\longmapsto(a+f_*b,-a-b).
$$
Indeed, the upper component includes into the seam neighborhood by the identity, while retracting the lower component across the seam applies $f$. Put $u=a+b$ and $v=b$, and replace target coordinates $(p,q)$ by $(p+q,-q)$. The map becomes $((f_*-\operatorname{id})v,u)$. Negating $v$ gives a <direct sum> of $\operatorname{id}-f_*$ and an <isomorphism>. Canceling the latter summands in the long <exact sequence> yields
$$
\boxed{\cdots\longrightarrow H_q(X)\xrightarrow{\operatorname{id}-f_*}H_q(X)\longrightarrow H_q(T_f)\longrightarrow H_{q-1}(X)\xrightarrow{\operatorname{id}-f_*}H_{q-1}(X)\longrightarrow\cdots.}
$$
In particular, there is a <short exact sequence>
$$
0\longrightarrow\operatorname{coker}(\operatorname{id}-f_*|_{H_q})\longrightarrow H_q(T_f)\longrightarrow\ker(\operatorname{id}-f_*|_{H_{q-1}})\longrightarrow0.
$$
No splitting of this <short exact sequence> is assumed at this stage.
For the <torus>, take the coordinate loops $c_1(t)=[(t,0)]$ and $c_2(t)=[(0,t)]$ as the <basis> of $H_1$. Their lifts to $\mathbb R^2$ have endpoint displacements $e_1,e_2$. Applying $A$ changes those displacements to its columns, so
$$
(f_A)_*[c_j]=\sum_i A_{ij}[c_i].
$$
This also follows from the identification of the <first homology group> with the <abelianization> of the <fundamental group>. The <Künneth theorem> gives $H_2(X)=\mathbb Z$, with generator the product <orientation class>. If $\alpha,\beta$ are the dual degree-one <cohomology> classes, then $\alpha\smile\beta$ evaluates to one on this generator, and
$$
f_A^*(\alpha\smile\beta)=(A_{11}\alpha+A_{12}\beta)\smile(A_{21}\alpha+A_{22}\beta)=\det(A)\,\alpha\smile\beta.
$$
Thus the degree-two <homology> map is multiplication by $\det A$, as in <homology of an integer torus endomorphism>. The claim that the <torus> map is a <homeomorphism> requires $A$ to be a <unimodular matrix>, equivalently an <integer> inverse or $\det A=\pm1$; nonsingularity over $\mathbb R$ alone would only give a <covering map> of finite degree. The particular <matrix> here is unimodular.
For this <matrix>, $\det A=1$ and
$$
I-A=\begin{pmatrix}1&-1\\1&1\end{pmatrix}.
$$
Its <kernel> is zero and its <Smith normal form> is $\operatorname{diag}(1,2)$, so its <cokernel> is $\mathbb Z/2$. The maps $\operatorname{id}-f_*$ on $H_0$ and $H_2$ are zero. Substituting into the <short exact sequences>, and splitting the degree-one extension because its quotient $\mathbb Z$ is free, gives
$$
\boxed{H_q(T_{f_A};\mathbb Z)\cong\begin{cases}\mathbb Z,&q=0,2,3,\\\mathbb Z\oplus\mathbb Z/2,&q=1,\\0,&\text{otherwise}.\end{cases}}
$$
Back to article page
