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It seems obviously true, but how does one actually show this? Or what tools does one use? I only know the basics of homotopy theory and homology.

Can I use invariance of domain somehow? If $S^n$ embeds, then so does a neighborhood of it in $R^{n+1}$?

Elle Najt
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  • The image would have to be both open and compact in $\Bbb R^n$? – Ted Shifrin May 18 '14 at 19:11
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    By invariance of domain, the image would be open. The image already is compact thus closed, and you'd get a clopen subset of $\Bbb R^n$ of which there are only two, $\emptyset$ and $\Bbb R^n$ itself. But the sphere is homeomorphic to neither. – Olivier Bégassat May 18 '14 at 19:12
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    @OlivierBégassat Why does invariance of domain imply that the imagine of $S^n$ is open? – Seth May 18 '14 at 19:27
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    @Seth because an embedding of the sphere restricts to an embedding of $S^n\setminus\lbrace p\rbrace(\simeq \Bbb R^n)$ for any point $p$, and such a map is open by invariance of domain. – Olivier Bégassat May 18 '14 at 19:27
  • @OlivierBégassat thanks, so I guess any compact $n$-manifold embedded in $\mathbb{R}^n$ is embedded as an open set then? – Seth May 18 '14 at 19:33
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    @Seth yes, from which it follows that there are no such embeddings. – Olivier Bégassat May 18 '14 at 19:35
  • @user8268 Are you sure that result doesn't use invariance of domain? How do you prove it? – Elle Najt May 18 '14 at 19:55
  • @user54092 sorry, when I see 'embedding' I automatically understand 'smooth embedding'; for topological embeddings one needs invariance of domain – user8268 May 18 '14 at 20:02

2 Answers2

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To add to Ted Shifrin's hint, one can actually show a stronger result via the invariance of domain theorem:

If $M$ is a compact $n$-manifold and $N$ is a connected $n$-manifold, then an embedding $h : M \to N$ must be surjective, hence a homeomorphism.

This is corollary 2B.4 in Hatcher's Algebraic Topology. To prove it, note that $h(M)$ must be closed since $M$ is compact and $N$ Hausdorff. $h(M)$ is also open by the invariance of domain. Since $N$ is connected, it follows that $h(M) = N$.

Your question now follows since $S^n$ and $\Bbb R^n$ are not homeomorphic.

Ayman Hourieh
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The image of $S^n$ would have to be compact (as $S^n$ is compact) and open, but $\mathbb R^n$ is connected and not compact.

user8268
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