Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 115 / 1

A smooth map $i:N\to M$ is a smooth embedding when it is an injective immersion and a homeomorphism onto its image with the subspace topology. The unit sphere is the inverse image of the regular value $1$ under $x\mapsto |x{|}^2$, so the preimage theorem makes it a smooth submanifold of ${\mathbb{R}}^{n+1}$ and its inclusion an immersion. It is injective, and a continuous injection from the compact sphere into the Hausdorff Euclidean space is a homeomorphism onto its image. Thus the inclusion is an embedding.

Products of spheres can be embedded by iterated spinning. Start with a round $S^{n_1}\subset {\mathbb{R}}^{n_1+1}$ translated into the half-space whose last coordinate $r$ is positive. If a compact $m$-manifold $N$ is embedded bythenis an injective immersion; compactness again makes it an embedding. Iterating with $q=n_2,\dots ,n_k$ embeds $S^{n_1}\times \cdots \times S^{n_k}$ in ${\mathbb{R}}^{n+1}$, where $n={\sum }_i{n}_i$.

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The de Rham cohomology isThe Poincare lemma says that every closed positive-degree differential form is locally exact, and is exact on every star-shaped open subset of Euclidean space.

Let $n>1$ and let $\alpha$ be a closed one-form on $M$. The hypothesis gives $\alpha =dg$ on $M\setminus B$. Choose a slightly larger coordinate ball $B^{\prime }$ around $B$. The Poincare lemma gives $\alpha =dh$ on $B^{\prime }$. The annulus $B^{\prime }\setminus B$ is connected when $n>1$, so $d(g-h)=0$ there and $g-h$ is constant. Adjusting $h$ by this constant makes $g$ and $h$ agree on the overlap, and they glue to a global primitive of $\alpha$. Hence $H^1(M)=0$. For $n=1$ the claim fails: remove a closed proper interval from $S^1$. Its complement is an interval and has vanishing first de Rham cohomology, whereas $H^1(S^1)\cong \mathbb{R}$.

Finally choose a nowhere-vanishing $n$-form $\omega$ on $S^n$ and a coordinate $t$ on the finite interval $I$. Every $(n+1)$-form on $S^n\times I$ is $a(x,t)\omega \wedge dt$. Fix $t_0\in I$ and putSince $d_{S^n}A\wedge \omega =0$ for dimensional reasons,Every top-degree form is exact, so $H^{n+1}(S^n\times I)=0$ without using the de Rham theorem.

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