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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 ?

  • Only surjective is pretty easy, just take the projection in one of the coordinates. – Ugo Iaba Mar 17 '15 at 13:16
  • @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. – kahen Mar 17 '15 at 13:18
  • @Léo : yeah I figured out the surjective one. –  Mar 17 '15 at 13:20
  • In order: No. No, yes. (Corrected, thanks to kahen.) – GEdgar Mar 17 '15 at 13:20
  • @kahen , GEdgar : Can you please give the proofs of the first two . I see that proving the non-existence of injective continuous function is enough , but I don't know invariance of domain and am not willing to apply it . Thanks –  Mar 17 '15 at 13:23
  • Hint: if $f$ is continuous and injective from compact space to Hausdorff space, then the inverse map is also continuous. – GEdgar Mar 17 '15 at 13:24
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    Hint: removing finitely many points from $[0,1]^2$ leaves a connected space. – kahen Mar 17 '15 at 13:25
  • @GEdgar : But any metric space is Hausdorff ; are you saying that any continuous injective map from a compact metric space to a metric space has continuous inverse ? –  Mar 17 '15 at 13:28
  • @SaunDev: Yes. See if you can prove it. – GEdgar Mar 17 '15 at 13:36
  • @GEdgar : Yeah , I have been able to prove it , but how do I apply it here ? –  Mar 17 '15 at 13:42

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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.

mrka
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