= Solution
The involution $(z_1,z_2)\mapsto(-z_1,\overline z_2)$ is free. Taking the first circle modulo the half-turn exhibits the quotient as the <mapping torus> of a reflection of $S^1$, hence as the <Klein bottle>. It has a finite CW structure with one zero-cell, two one-cells, and one two-cell. With suitable generators its integral cellular differential is
$$
0\longrightarrow\mathbb Z\xrightarrow{(0,2)}\mathbb Z^2\xrightarrow0\mathbb Z\longrightarrow0.
$$
Therefore
$$
H_q(T^2/{\sim};\mathbb Z)\cong
\begin{cases}
\mathbb Z,&q=0,\\
\mathbb Z\oplus\mathbb Z/2,&q=1,\\
0,&q\geq2,
\end{cases}
$$
whereas reduction modulo two kills the only nonzero boundary and gives
$$
H_q(T^2/{\sim};\mathbb F_2)\cong
\begin{cases}
\mathbb F_2,&q=0,2,\\
\mathbb F_2^2,&q=1,\\
0,&q>2.
\end{cases}
$$
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