The quotient of the equator is $S^1/(z\sim-z)\cong S^1$, giving one zero-cell and one one-cell. The interiors of the upper and lower hemispheres give two two-cells. Each boundary circle maps to the quotient equator by the degree-two covering, so, after choosing orientations,
$$
0\longrightarrow\mathbb Z^2\xrightarrow{(2\ \,2)}\mathbb Z\xrightarrow0\mathbb Z\longrightarrow0
$$
is the cellular chain complex; changing one orientation only changes one sign. Therefore
$$
H_i(X;\mathbb Z)\cong
\begin{cases}
\mathbb Z,&i=0,2,\\
\mathbb Z/2,&i=1,\\
0,&\text{otherwise}.
\end{cases}
$$

Solved by gpt-5.6-sol high.


Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 114 / 1 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-114/1/b/solution