Apply the rational <Serre spectral sequence> to the path-loop fibration of $K(\mathbb Z,3)$. Its fiber is
$$
\Omega K(\mathbb Z,3)\simeq K(\mathbb Z,2)\simeq\mathbb{CP}^{\infty},
$$
whose rational <cohomology ring> is $\mathbb Q[c]$ with $|c|=2$. Since the path space is contractible, $c$ must transgress to a nonzero class $x\in H^3(K(\mathbb Z,3);\mathbb Q)$. Multiplicativity gives
$$
d_3(c^m)=m c^{m-1}x.
$$
Over $\mathbb Q$ these differentials pair and kill every positive-degree class except $x$, while graded commutativity gives $x^2=0$. Hence
$$
H^i(K(\mathbb Z,3);\mathbb Q)\cong
\begin{cases}
\mathbb Q,&i=0,3,\\
0,&\text{otherwise}.
\end{cases}
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 165 / 3 / ii / Solution

Use the <path-loop fibration>
$$
\Omega S^4\longrightarrow PS^4\longrightarrow S^4
$$
and its rational <Serre spectral sequence>. The total space is contractible, while $H^*(S^4;\mathbb Q)$ is $\mathbb Q$ in degrees $0$ and $4$ and zero otherwise. The only possible nonzero differential is
$$
d_4:E_4^{0,q}\longrightarrow E_4^{4,q-3}.
$$
Convergence to the cohomology of a point first forces $d_4:H^3(\Omega S^4;\mathbb Q)\to H^4(S^4;\mathbb Q)$ to be an isomorphism, and then inductively forces an isomorphism from each nonzero vertical group to the group three degrees below it in the other column. Therefore
$$
H^i(\Omega S^4;\mathbb Q)\cong
\begin{cases}
\mathbb Q,&i=0,3,6,9,\ldots,\\
0,&\text{otherwise}.
\end{cases}
$$

Solved by gpt-5.6-sol high.

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 165 / 3 / i / Solution

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.

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

{c}
{wiki}

For a fibration $F\to E\to B$ with suitable connectivity, the cohomological Serre spectral sequence has
$$
E_2^{p,q}=H^p(B;H^q(F;R))
\Longrightarrow H^{p+q}(E;R),
$$
with differential $d_r:E_r^{p,q}\to E_r^{p+r,q-r+1}$.


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