= Solution
The <torus> map acts on $H_1(T^2;\mathbb Z)$ by
$$
A=\begin{pmatrix}2&1\\-1&0\end{pmatrix}.
$$
Its <determinant> is one, so it is an orientation-preserving <homeomorphism> and acts as the identity on both $H_0(T^2)$ and $H_2(T^2)$.
To calculate the <homology> of its <mapping torus>, cover the time <circle> by two arcs. Each inverse image has the <homotopy type> of $T^2$, while their intersection has two components of that <homotopy type>. The map in the <Mayer–Vietoris sequence> can be written
$$
H_q(T^2)\oplus H_q(T^2)\longrightarrow H_q(T^2)\oplus H_q(T^2),
\qquad (a,b)\longmapsto(a+f_*b,-a-b).
$$
Adding the output coordinates and using the input coordinate $a+b$ separates an identity map from $f_*-I$. Canceling the identity summands gives the <Wang sequence>, and hence
$$
0\longrightarrow\operatorname{coker}(f_*-I\mid H_q(T^2))
\longrightarrow H_q(X)
\longrightarrow\ker(f_*-I\mid H_{q-1}(T^2))\longrightarrow0.
$$
In degree one,
$$
A-I=\begin{pmatrix}1&1\\-1&-1\end{pmatrix}.
$$
Its <image> and <kernel> are both generated by $(1,-1)$; its <cokernel> is $\mathbb Z$ since this image is primitive. Its <Smith normal form> is $\operatorname{diag}(1,0)$, so its <torsion subgroup> is zero. On the other two nonzero <homology groups> of the <torus>, $f_*-I=0$. The displayed <short exact sequences> split because their right-hand groups are free. Therefore
$$
\boxed{H_j(X;\mathbb Z)=
\begin{cases}
\mathbb Z,&j=0,3,\\
\mathbb Z^2,&j=1,2,\\
0,&j>3.
\end{cases}}
$$
In particular, the <mapping torus> contributes an additional degree-one class from the time <circle>, as well as the invariant degree-one class in the fiber.
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