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I have read in this post Clarify: "$S^0$, $S^1$ and $S^3$ are the only spheres which are also groups" that there is a bijection between $S^1$ and the sphere $S^n$, $n \geq 1$ and this was confirmed in the answer!

What is an explicit bijection $f : S^1 \rightarrow S^n$ ?

Thanks!

Mira
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  • Both are clearly uncountable, and contained in $\mathbb R^n$ for some $n$, so they must have cardinality $\aleph_1$. So, you get a bijection. – Sntn Mar 13 '21 at 15:53
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    @MathIsNice1729 $\aleph_1$ is not necessarily the cardinality of $\Bbb R$. (continuum hypothesis) – Kenny Lau Mar 13 '21 at 15:54
  • @KennyLau Continuum hypothesis is pretty well-accepted though, so I guess we should be able to use it. – Sntn Mar 13 '21 at 15:55
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    @MathIsNice1729 Continuum hypothesis is pretty well-accepted though That is really not the case! You have results with or without it. No more than that. – mathcounterexamples.net Mar 13 '21 at 15:56
  • @mathcounterexamples.net What do you mean? I know that it's not provable under ZFC but in my (rather limited) experience, I've always seen that it's assumed to be true in mainstream mathematics. Isn't it? – Sntn Mar 13 '21 at 15:58
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    Please define what mainstream mathematics is! The continuum hypothesis is not assumed in mathematics as a usual axiom. – mathcounterexamples.net Mar 13 '21 at 15:59
  • @mathcounterexamples.net Look, I get your point. It's not provable, so we get two different answers. But just like axiom of choice, I've seen that it's usually assumed in most textbooks/research work. – Sntn Mar 13 '21 at 16:01
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    I've seen that it's usually assumed in most textbooks/research work really not what I experience. If you can quote some usual math books (analysis, calculus...) where it is assumed that would convince me. What classical math theorem assumes continuum hypothesis? – mathcounterexamples.net Mar 13 '21 at 16:02
  • @mathcounterexamples.net Now that I think about it, it was dumb of me to say that. You're right. – Sntn Mar 13 '21 at 16:18

1 Answers1

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There is no point seeking an explicit bijection. It will not have any special property. It will not be close to continuous.

$S^1$ and $S^n$ are in bijection because they have the same cardinality, and here is why:

If you remove a point from each of them, then you end up with $\Bbb R$ and $\Bbb R^n$ because of the stereographic projection.

Now, $\Bbb R$ clearly injects into $\Bbb R^n$, so by the Schröder–Bernstein theorem, we only need to inject $\Bbb R^n$ back into $\Bbb R$: given $(x_1, \cdots, x_n) \in \Bbb R^n$, let $0.a_{i1}a_{i2} \cdots$ be the binary expansion of $\frac1{1+\exp(x_i)}$ (this is the sigmoid function, an injection $\Bbb R \to (0, 1)$). Then say that the result is $0.a_{11} a_{21} \cdots a_{n1} a_{12} a_{22} \cdots a_{n2} \cdots$ as ternary. This different base is taken to avoid things like $0.012022222\ldots_3 = 0.121_3$. Then this gives an injection $\Bbb R^n \to \Bbb R$, so we are done.

Kenny Lau
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