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Let $X,Y$ be topological spaces, and let $f\colon X \to Y$ be a continuous function. Further suppose that there exist an open and dense subset $U$ of $X$, such that $f\vert_{U} \colon U \to Y$ is an open mapping.

Is it then true that $f$ itself is open? How is the situation if $X = \mathbb{R}^{n}$ and $Y=\mathbb{R}^{m}$?

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    $\begingroup$ $X = [0,\infty),\; Y = \mathbb{R};\; f = \operatorname{id}$ $\endgroup$ – Daniel Fischer Jul 22 '14 at 16:01
  • $\begingroup$ Thanks for the quick counterexample. $\endgroup$ – Sebastian Jul 22 '14 at 16:05
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    $\begingroup$ $X = \Bbb{R}$, $U = \Bbb{R} \setminus \{0\}$, $f(x) = x^2$ gives a counter example for the case of a function defined on a complete $\Bbb{R}^n$. $\endgroup$ – PhoemueX Jul 22 '14 at 22:04

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