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

Use the path-loop fibrationand its rational Serre spectral sequence. The total space is contractible, while $H^{\ast }(S^4;\mathbb{Q})$ is $\mathbb{Q}$ in degrees $0$ and $4$ and zero otherwise. The only possible nonzero differential isConvergence 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

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