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I was reading about inversion of sphere. Wikipedia defines it as:

Let $f: S^2\to R^3$ be the standard embedding; then there is a regular homotopy of immersions $f_t\colon S^2\to R^3$ such that $f_0 = f $ and $f_1 = −f.$

I would like to know if such an inversion is impossible via embeddings. Thanks.

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    $\begingroup$ It is impossible. By the generalized Jordan separation theorem, an embedding of the sphere separates $\mathbb{R}^3$ into two regions, and so we can define an orientation of $S^2$ by the field of outward pointing unit normals (i.e. they point towards the unbounded component of the complement). Given a homotopy through embeddings of your original embedding, this orientation will vary continuously and thus will not change. So it's not possible to end up with a reversed orientation. $\endgroup$ – Pedro May 30 '16 at 19:17
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    $\begingroup$ @Pedro That sounds like an answer to me ... $\endgroup$ – Neal May 30 '16 at 19:41
  • $\begingroup$ Indeed, not only must your $f_t$ self-intersect, it is a theorem that at least one $t$ has some quadruple point ($x$ in the image such that $f_t^{-1}(x)$ is four points). $\endgroup$ – user98602 May 30 '16 at 19:51
  • $\begingroup$ @Neal It's kind of sketchy, but I suppose I can post it $\endgroup$ – Pedro May 30 '16 at 21:54
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(Reposting my comment which I suppose is an answer)

It is impossible. By the generalized Jordan separation theorem, an embedding of the sphere separates $\mathbb{R}^3$ into two regions, and so we can define an orientation of $S^2$ by the field of outward pointing unit normals (i.e. they point towards the unbounded component of the complement). Given a homotopy through embeddings of your original embedding, this orientation will vary continuously and thus will not change. So it's not possible to end up with a reversed orientation.

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