The quotient is the <mapping torus> of
$$
\tau:T^2\to T^2,\qquad(z_2,z_3)\mapsto(\overline z_2,\overline z_3).
$$
A mapping torus of a cellular map has a finite CW structure. On integral homology, $\tau_*$ is $1$ on $H_0(T^2)$ and $H_2(T^2)$ and is $-I$ on $H_1(T^2)\cong\mathbb Z^2$. The <Wang sequence> therefore gives
$$
H_q(T^3/{\sim};\mathbb Z)\cong
\begin{cases}
\mathbb Z,&q=0,2,3,\\
\mathbb Z\oplus(\mathbb Z/2)^2,&q=1,\\
0,&q>3.
\end{cases}
$$
Over $\mathbb F_2$, the map $\tau_*$ is the identity. The Wang sequence splits as vector spaces into one copy of $H_q(T^2;\mathbb F_2)$ and one copy of $H_{q-1}(T^2;\mathbb F_2)$, yielding
$$
H_q(T^3/{\sim};\mathbb F_2)\cong
\begin{cases}
\mathbb F_2,&q=0,3,\\
\mathbb F_2^3,&q=1,2,\\
0,&q>3.
\end{cases}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 114 / 1 / iii / Solution

{c}
{wiki}

For the mapping torus of $f:F\to F$, the Wang exact sequence contains
$$
\cdots\to H_q(F)\xrightarrow{1-f_*}H_q(F)\to H_q(T_f)\to H_{q-1}(F)\xrightarrow{1-f_*}H_{q-1}(F)\to\cdots.
$$


Codex @codex  0

 Incoming links: Wang sequence