Does there exist a continuous open function $f:B^n\to B^n$ which is not injective? (Here $B^n\subseteq\mathbb{R}^n$ is the open unit ball)
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What is an open function? – Vincenzo Tibullo Jul 29 '12 at 16:12
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@enzotib: It's a function that maps open sets to open sets. (See http://en.wikipedia.org/wiki/Open_and_closed_maps.) – ruakh Jul 29 '12 at 16:13
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I mean that $f$ is such that the image of every open set is still an open set. – Mizar Jul 29 '12 at 16:13
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In two (or more) dimensions it is easy to construct such a map, by homotopy. Homotop the ball into something which looks like a sausage, then move the two ends of the 'sausage' on top of each other. – Jul 29 '12 at 16:17
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Ohh, right. Sorry. I missed the open part. So careless of me. – Jul 29 '12 at 16:18
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1Do you want the function to be surjective? Otherwise the question is not much fun. – hmakholm left over Monica Jul 29 '12 at 16:20
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3is this map works? $re^{ix}\mapsto re^{2ix}$? – Myshkin Jul 29 '12 at 16:27
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@Patience: yup, now use this to do it for all $n \geq 2$... – t.b. Jul 29 '12 at 16:29
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@Thomas: could you explain better what do you mean? Your map seems to be injective, as the 'ends' of the 'sausage' are on its boundary. – Mizar Jul 29 '12 at 16:30
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@Patience: ok, thank you, that clearly works! – Mizar Jul 29 '12 at 16:32
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3@carizio: Please don't delete your question just because you get a satisfactory answer. Doing so deprives the answerer of the chance to earn reputation, and deprives future vistors to the site of the chance to learn from the question and its answer. – hmakholm left over Monica Jul 29 '12 at 16:54