Does there exist any continuous bijective function $f:[0,1] \times [0,1] \to [0,1] $ , where $[0,1]$ is equipped with usual Euclidean metric of $\mathbb R$ and $[0,1] \times [0,1]$ is equipped with the usual Euclidean metric of $\mathbb R^2$ ? What if we require the continuous mapping to be only injective or only surjective ?

  • $\begingroup$ Only surjective is pretty easy, just take the projection in one of the coordinates. $\endgroup$ – 115465 Mar 17 '15 at 13:16
  • $\begingroup$ @GEdgar continuous and injective should be "No" due to invariance of domain, right? (restrict to $(0,1)^2$ first, such that the theorem applies). Or by GTET such a continuous injective map would be a homeomorphism onto its image. $\endgroup$ – kahen Mar 17 '15 at 13:18
  • $\begingroup$ @Léo : yeah I figured out the surjective one. $\endgroup$ – user217921 Mar 17 '15 at 13:20
  • $\begingroup$ In order: No. No, yes. (Corrected, thanks to kahen.) $\endgroup$ – GEdgar Mar 17 '15 at 13:20
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    $\begingroup$ Hint: removing finitely many points from $[0,1]^2$ leaves a connected space. $\endgroup$ – kahen Mar 17 '15 at 13:25

There does not exist a continuous bijective function $f:[0,1] \times [0,1] \to [0,1]$. Suppose such function exist. Consider the set $A= ([0,1] \times [0,1]) \setminus f^{-1}(\frac{1}{2})$(Instead of $\frac{1}{2}$ you can take any point in the interior of $[-1,1]$). Then the function $f$ restricted to $A$ is also continuous. It is easy to see that $A$ is connected, but $f(A)$ is not connected. Since continuous image of connected set is connected, thus there does not exist such function. Now image of the injective continuous function is a non-degenerate interval in $[-1, 1]$ (it is continuous image of a connected set). Take any interior point in that interval and argue as above.


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