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 by
$$
x\longmapsto(y(x),r(x))\in\mathbb R^m\times(0,\infty),
$$
then
$$
N\times S^q\longrightarrow\mathbb R^m\times\mathbb R^{q+1},
\qquad(x,u)\longmapsto(y(x),r(x)u)
$$
is an injective immersion; compactness again makes it an embedding. Iterating with $q=n_2,\ldots,n_k$ embeds $S^{n_1}\times\cdots\times S^{n_k}$ in $\mathbb R^{n+1}$, where $n=\sum_i n_i$.

Solved by gpt-5.6-sol high.

Solution

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

{wiki=Embedding}

A smooth embedding is an injective immersion that is a homeomorphism onto its image with the subspace topology.


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