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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?2012-07-29
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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](http://en.wikipedia.org/wiki/Open_and_closed_maps).)2012-07-29
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    I mean that $f$ is such that the image of every open set is still an open set.2012-07-29
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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.2012-07-29
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    Ohh, right. Sorry. I missed the open part. So careless of me.2012-07-29
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    Do you want the function to be surjective? Otherwise the question is not much fun.2012-07-29
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    is this map works? $re^{ix}\mapsto re^{2ix}$?2012-07-29
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    @Patience: yup, now use this to do it for all $n \geq 2$...2012-07-29
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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.2012-07-29
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    @Patience: ok, thank you, that clearly works!2012-07-29
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    @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.2012-07-29

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Hint for your problem: $re^{ix}\mapsto re^{2ix}$.