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$\newcommand{\dim}{\operatorname{dim}}\newcommand{\im}{\operatorname{im}}$ In linear algebra it's a classic exercise to demonstrate that you cannot build a surjective linear map from a lower rank domain (call it $V$) to a higher rank one (call it $W$) (for example $f:\mathbb{R}\rightarrow \mathbb{R^2} $ - here $V$ is $\mathbb{R}$ and $W$ is $\mathbb{R^2}$). A simple demonstration is to start from the fact that: $$\dim V = \dim \ker f + \dim \im f.$$

But now I'm thinking about whether you can build a surjective function (not linear map) from $V$ to $W$. Obviously the theorem above does not apply any more.

Let's limit to a particular case. $V$ is $\mathbb{R}$ and $W$ is $\mathbb{R^2}$ (or $\mathbb{C}$ if you prefer... for this particular example it's the same thing). Considering the fact that $\mathbb{R}$ and $\mathbb{R^2}$ have the same cardinal ($2^{\aleph_0}$), it would result (even though it's a bit counterintuitive) that you could build a bijective function from the first one to the other. But this is as far as I got. I fail to be able to give an example.

Any idea how could I go on to find an example? Or to demonstrate it is not possible?

jgon
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zozo
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Try looking at space filling curves (link to wiki included). These are examples of continuous surjective functions from the unit interval to the unit square. I'm pretty sure these can be extended to continuous surjective functions from $\Bbb{R}$ to $\Bbb{R^2}$ by tiling the plane in an appropriate way.

Also see the answers to the related stack exchange question here.

jgon
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  • The space filling curves looks interesting (reading it now), but the SE question linked is the other way around (higher rank to lower, problem being injectivity) – zozo Nov 08 '17 at 02:30
  • The linked se question is about a bijection, so the order of the spaces doesn't matter, since bijections are invertible. – jgon Nov 08 '17 at 02:31
  • Sorry... missread it. I read injection instead of bijection :). – zozo Nov 08 '17 at 04:00