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Let $m,n \geq1$, $U$ is a nonempty open subset of $\mathbb{R}^m$, $f:U \rightarrow \mathbb{R}^n$ is a surjection, if $f$ is continuous on $U$, then $m \geq n$?

Is this right?

David Chan
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    Well, there are things like space filling curves (see, e.g. http://math.stackexchange.com/questions/290148/are-there-any-continuous-functions-from-the-real-line-onto-the-complex-plane) – lulu Sep 29 '15 at 02:11
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    this the third link that pops out when you google continuous surjection: http://math.stackexchange.com/questions/357052/does-there-exists-a-continuous-surjection-from-mathbbr-to-mathbbr2 – Giovanni Sep 29 '15 at 02:16
  • While the linked questions only address the case where $m=1$ and $n=2$, the case of $m=1$ and arbitrary $n$ is very similar, and the case of general $m$ follows from the $m=1$ case by composing with any continuous surjection $\mathbb{R}^m\to\mathbb{R}$. – Eric Wofsey Sep 29 '15 at 02:19

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