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There should exist such a function, but I cannot think of any example. Onto continuous functions mapping $(0,1)$ to $\mathbb R$ are easy to find.

Edit: Sorry - I mentioned Tietze extension theorem proved the existence of such a function - that was wrong. I mixed it up with a different question.

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    do you mean onto?2012-03-25
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    Exactly. Sorry. It's now fixed.2012-03-25
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    @Poly Maybe you want $(0,1)$ into $\mathbb R$. You can use $\tan \pi z$ for example.2012-03-25
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    Pedro Tamaroff, You should wrote $\cot\pi z$2014-03-22

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No such function exists. $[0,1]$ is compact, $\mathbb{R}$ is not compact, and the continuous image of a compact space is compact.

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    You are right. I don't know why I forgot that... In fact I am looking for a continuous onto function mapping R to R^2. I was trying to find some functions to compose with space filling curve [0,1] to $[0,1]^2$2012-03-25
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No such function exists from $\mathbb{R}$ to $\mathbb{R}^2$. If you omit some lines from $\mathbb{R}$, it's not connected anymore, but $\mathbb{R}^2$ stays connected.

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    I'm not sure why you're talking about functions $f: \mathbb{R} \rightarrow mathbb{R}^2$, but what you're saying is incorrect: there *are* surjective continuous functions $f$. There are just not *bijective* continuous functions.2012-05-28
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    I'm not sure what you mean by lines in $\mathbb R$, but in any case, while it's certainly true that the continuous image of a connected set is connected, the converse doesn't hold: sometimes the continuous image of a disconnected set is also connected :)2012-05-28
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    It's possible he read Polymorpher's comment to Chris Eagle's answer, where Polymorpher said that he is looking for a map $\mathbb{R} \to \mathbb{R}^2$. But the statement is still incorrect.2012-05-28