Products of <Eilenberg–MacLane spaces> satisfy
$$
K(\mathbb Z/4,3)\times K(\mathbb Z/5,3)
\simeq K(\mathbb Z/4\oplus\mathbb Z/5,3)
\cong K(\mathbb Z/20,3),
$$
where the last isomorphism uses the <Chinese remainder theorem>. This space is $2$-connected, so the <Hurewicz theorem> identifies
$$
H_3(-;\mathbb Z)\cong\pi_3(-)\cong\mathbb Z/20.
$$
The assumed surjection on $H_3$ is an isomorphism because the group is finite. Hence $f$ is an isomorphism on $\pi_3$; all other homotopy groups of the source and target vanish. Thus $f$ is a weak homotopy equivalence, and the <Whitehead theorem> for <Kan complexes> makes it a <homotopy equivalence>.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 165 / 2 / vi / Solution

The <loop-space shift of homotopy groups> gives
$$
\pi_i(\Omega S^4)\cong\pi_{i+1}(S^4).
$$
Thus $\Omega S^4$ is $2$-connected and its first nonzero homotopy group is
$$
\pi_3(\Omega S^4)\cong\pi_4(S^4)\cong\mathbb Z.
$$
The <Hurewicz theorem> now gives
$$
H_3(\Omega S^4;\mathbb Z)\cong\mathbb Z,
$$
and every positive integral homology group below degree $3$ vanishes. The smallest requested degree is therefore $i=3$.

Solved by gpt-5.6-sol high.

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

{c}
{wiki}

If a path-connected space is $(n-1)$-connected for $n\geq2$, then its first potentially nonzero homotopy group maps isomorphically to homology:
$$
\pi_n(X)\xrightarrow{\sim}H_n(X;\mathbb Z).
$$


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 Incoming links: Hurewicz theorem