Why doesn't $S^n$ embed into $R^n$?

Question

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 $R^{n+1}$?

Answer 1

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 $h(M) = N$.

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

Answer 2

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.