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By the generalized Jordan theorem any continuous injective map $S^{n-1} \hookrightarrow R^n$ splits $R^n$ into two regions, one being bounded (interior) and the other one unbounded (exterior). It must exist a stronger statement that the winding number for the points of the interior with respect to this sphere is always $\pm 1$. Anybody can tell me a reference how this theorem is called and who must be given credit for?

In fact a friend of mine asked me a different question. Is it true that any continuous injective map $R^n\hookrightarrow R^n$ is always open? I expect the answer is yes. It can be easily reduced to the previous one, but perhaps there is a simpler proof?

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For your first question the credit goes to Brouwer. It's sometimes called the Jordan-Brouwer Separation Theorem. In this context you do not need a sphere, your $n-1$-manifold can be any compact connected boundaryless submanifold of $\mathbb R^n$. I believe both appear in the Guillemin and Pollack textbook, and Bredon's Topology and Geometry textbook.

Your last question has the yes answer expected. The proof is very much similar. I'm not sure if I would call either proof easier than the other. This one tends to be called the open mapping theorem and at this level of generality you would need Bredon's book, but most intro algebraic topology texts that cover singular homology cover this.

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  • $\begingroup$ Hi Ryan. I will check the references, but I am not convinced by your answer as the theorem in question only states that the complement consists of two regions - unbounded and bounded. It does not imply that the winding number is one for the points of the interior region. $\endgroup$
    – Victor
    Commented Feb 22, 2016 at 20:34
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    $\begingroup$ The proof in the Guillemin and Pollack textbook provides this. Technically they are using mod-2 degrees in their proof but that is only because integer-valued degrees are not defined at that point. But if you re-do their proof using integer-valued degrees it gives you exactly what you want. $\endgroup$
    – Ryan Budney
    Commented Feb 22, 2016 at 20:37
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A statement that is both stronger and simpler to state is: For compact subsets of $\mathbb{R}^n$, the number of connected components of the complement is a topologic invariant. This is an immediate consequence of the Alexander duality (See e.g. Spanier's Algebraic Topology Ch. 6, Theorem 16). There is also an elementary proof, I believe due to Leray, by means of the Browder degree; you can find a very clear exposition of it in Louis Nirenberg's Topics in Nonlinear Functional Analysis.

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  • $\begingroup$ Pietro, thank you for your answer. I don't think that the Alexander duality helps here to determine the winding number. $\endgroup$
    – Victor
    Commented Feb 23, 2016 at 0:16
  • $\begingroup$ I am sorry, actually you are right. The geometrical interpretation of the Alexander duality is through the linking number, and the winding number in our case is its particular occurrence. $\endgroup$
    – Victor
    Commented Feb 23, 2016 at 3:25

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