Write $F=K(\mathbb Z/4,1)$, $E=K(\mathbb Z,2)$, and $B=K(\mathbb Z,2)$. In total degree at most two, the $E_2$ page of the mod-two <Serre spectral sequence> has
$$
E_2^{0,0}=E_2^{0,1}=E_2^{0,2}=E_2^{2,0}=\mathbb F_2,
$$
and $E_2^{1,0}=E_2^{1,1}=0$. A periodic free resolution of the cyclic group gives $H_1(F;\mathbb Z)=\mathbb Z/4$ and $H_2(F;\mathbb Z)=0$. Hence $H^1(F;\mathbb F_2)\cong\operatorname{Hom}(\mathbb Z/4,\mathbb F_2)$, while the <universal coefficient theorem for cohomology> gives $H^2(F;\mathbb F_2)\cong\operatorname{Ext}(\mathbb Z/4,\mathbb F_2)$; both are one-dimensional.

The edge map $H^2(B;\mathbb F_2)\to H^2(E;\mathbb F_2)$ is induced by multiplication by $4$ and is therefore zero modulo two. Consequently
$$
d_2:E_2^{0,1}\longrightarrow E_2^{2,0}
$$
is an isomorphism. The differential out of $E_2^{0,2}$ is zero, because the total space has a one-dimensional $H^2$ which must survive in filtration zero. Thus for every $r\geq3$ and $p+q\leq2$,
$$
E_r^{0,0}=E_r^{0,2}=\mathbb F_2
$$
and all other groups in that range vanish.

It follows that
$$
H^0(F;\mathbb F_2)=H^1(F;\mathbb F_2)=H^2(F;\mathbb F_2)=\mathbb F_2.
$$
The surviving filtration-zero class is the restriction of the degree-two class of $E$, so
$$
\operatorname{im}H^2(f;\mathbb F_2)=H^2(F;\mathbb F_2).
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 165 / 4 / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-165/4/i/solution