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)
Open mapping of the unit ball into itself
2
$\begingroup$
general-topology
-
0What is an open function? – 2012-07-29
-
0@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
-
0I mean that $f$ is such that the image of every open set is still an open set. – 2012-07-29
-
0In 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
-
0Ohh, right. Sorry. I missed the open part. So careless of me. – 2012-07-29
-
1Do you want the function to be surjective? Otherwise the question is not much fun. – 2012-07-29
-
3is this map works? $re^{ix}\mapsto re^{2ix}$? – 2012-07-29
-
0@Patience: yup, now use this to do it for all $n \geq 2$... – 2012-07-29
-
0@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
-
0@Patience: ok, thank you, that clearly works! – 2012-07-29
-
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. – 2012-07-29
1 Answers
5
Hint for your problem: $re^{ix}\mapsto re^{2ix}$.