homology group of $(T^2\times I)/\sim$ Let $T^2$ be the $2$-torus, $S_1\times S_1$, where $S_1=\{z\in \mathbb{C}:|z|=1\}\subset \mathbb{C}$. And the equivalence relation of $T^2\times I$ is defined as
$$ (z,\zeta,1)\sim (z,z^2\zeta,0)$$
where $I$ is the unit interval. Let $X$ be the quotient space of $T^2\times I$ by $\sim$, i.e. $X=(T^2\times I)/\sim$. I want to find the homology groups of $X$. The computation by myself is
$$H_n(X)=0 (n\geq 4),H_3(X)\cong \mathbb{Z},H_2(X)\cong \mathbb{Z}^2,H_1(X)\cong \mathbb{Z}^2\times \mathbb{Z}/2\mathbb{Z},H_0(X)\cong \mathbb{Z}$$
I want you to confirm my computation and teach me your method if you can. Thank you for your help.
 A: I think your answer is correct.  The space $X$ is a mapping torus, and there is a long exact sequence relating the homology of $X$ with the homology of $T^2$ and the gluing map.  See Hatcher, example 2.48 for an explanation of this sequence.  Since I suspect that this might be the method you've used, I'll be brief and highlight only the important parts.  
Let $f: T^2 \to T^2$ be the map $(z,\zeta) \mapsto (z, z^2 \zeta)$.  In this case, the long exact sequence takes the form $$\cdots \to H_n T^2 \xrightarrow{1-f_*} H_n T^2 \to H_n X \to H_{n-1} T^2 \to \cdots.$$
The homology groups of the torus $T^2$ are $\mathbb{Z}$, $\mathbb{Z}^2$, and $\mathbb{Z}$ in dimensions $0$, $1$, and $2$ respectively, so it remains to compute the effect of the map $f$ in homology.  
The induced map $f_*$ is the identity on $H_0$ since the degree of $f$ is $1$, and it is also the identity on $H_2$ since for example it preserves "volume".  If $a$ represents the $z$-copy of $S^1$ in $T^2 \cong S^1 \times S^1$ and $b$ the $\zeta$-copy, then on $H_1$, $f_*$ is the map $a \mapsto a$ and $b \mapsto 2a + b$.  
So in the long exact sequence, the map $1 - f_*$ is
\begin{align*}
\cdots \to H_0 T^2 &\xrightarrow{0} H_0 T^2 \to \cdots \\
\cdots \to H_1 T^2 &\xrightarrow{\substack{a \mapsto 0  \\ b \mapsto 2a}} H_1 T^2 \to \cdots \\
\cdots \to H_2 T^2 &\xrightarrow{0} H_2 T^2 \to \cdots.
\end{align*}
Computing the kernels and cokernels of these maps, we find that $H_0 X \cong H_0 T^2 \cong \mathbb{Z}$ and $H_3 X \cong H_2 T^2 \cong \mathbb{Z}$.  Moreover, $H_1$ and $H_2$ sit in short exact sequences
\begin{align*}
0 \to \mathbb{Z} \oplus \mathbb{Z}/2 \to &H_1 X \to \mathbb{Z} \to 0 \\
0 \to \mathbb{Z} \to &H_2 X \to \mathbb{Z} \to 0.
\end{align*}
Since $\mathbb{Z}$ is projective, there are no nontrivial extensions, so $H_1 X \cong \mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z}/2$ and $H_2 X \cong \mathbb{Z} \oplus \mathbb{Z}$.  Of course, since $X$ is 3-dimensional, $H_{>3} X = 0$.  
