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Are there any "nice" functions that can take a point from the surface of an n-sphere and map it to a the surface of an (n+1)-sphere?

By "nice", I mean it should be continuous, one-to-one (but not necessarily onto), and cover lots of surface area (not just $f(x) = (x, 0)$).

  • The last condition is not "nice" in any reasonable sense; it is very close to requiring that the map is "space-filling" (http://en.wikipedia.org/wiki/Space-filling_curve) and as Will's answer shows below all such maps are very poorly behaved, in particular they fail to be differentiable. – Qiaochu Yuan Mar 30 '14 at 02:18

2 Answers2

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No. Sard's Theorem. Short version, the image of $\mathbb S^n$ has measure zero in $\mathbb S^{n+1}$ unless the mapping is highly non-differentiable.

Will Jagy
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  • Well, OP only asks for continuity for niceness. But he also asks for invertibility, which is almost certainly a dealbreaker. –  Mar 29 '14 at 06:08
  • Whoops, I meant one-to-one, not invertible. Edited. – Aaron Voelker Mar 29 '14 at 06:36
  • @AaronVoelker Well, that's the reason it would be a dealbreaker. Once you're doing stuff with positive measure you're close to being able to cover the whole space anyway. (Think of the space-filling curves; those are never injective.) –  Mar 29 '14 at 07:08
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First of all, I do not think this question has anything to do with algebraic topology per se. It appears you are asking for examples topological embeddings $S^{n}\to S^{n+1}$ whose images have positive $n+1$-dimensional Lebesgue measure (this is how I read yours "covers lots of surface area"). You can find some very readable constructions of Jordan curves $J$ in the plane which have positive 2-dimensional Lebesgue measure for instance here. (Note that registering and reading on line at Jstor is free.) Applying the inverse to the stereographic projection, you obtain examples in $S^2$. Multiplying curves $J$ as above by $[0,1]^{n-1}$ and adding "flat top and bottom" you get examples of $n$-dimensional topological spheres in $R^{n+1}$ which have positive $n+1$-dimensional Lebesgue measure.

Moishe Kohan
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