Let $f\colon \mathbb{R}^n \to \mathbb{R}^{n-1}$ be a continuous function and $K\subset \mathbb{R}^n$ a subset of positive Lebesgue measure. Is it possible that $f$ is one-to-one on $K$?

If $K$ contains a (nonempty) open set this is impossible because of the invariance of domain theorem. But can we say anything for arbitrary (measurable) sets?

  • $\begingroup$ This smells like Baire's category theorem and absolute continuity to me, but that's just a first impression $\endgroup$ – Ben Grossmann Nov 16 '15 at 14:33

Let $n = 2$, $C \subseteq \mathbf R$ a fat cantor set, $f \colon C \times C \to C$ a homeomorphism. As $C \times C \subseteq \mathbf R^2$ is closed, there is a continuous extension $F \colon \mathbf R^2 \to [0,1]$ by the Tietze extension theorem. Now $\lambda(C \times C) > 0$ and $F|_{C \times C} = f$ is one-to-one.

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    $\begingroup$ Maybe it's worth mentioning that every two Cantor spaces are homeomorphic, with some reference. $\endgroup$ – Silvia Ghinassi Nov 16 '15 at 14:49
  • $\begingroup$ Ok, so this reduces the problem to finding a homeomorphism from $C\times C \to C$. Is this well known? I think the reference to homeomorphic cantor spaces won't suffice for me, as I don't know why $C\times C$ is a Cantor space... $\endgroup$ – KoliG Nov 16 '15 at 14:58
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    $\begingroup$ $C$ is known to be homeomorphic to $2^{\mathbf N}$ (the countable infinite product of the space $2 = \{0,1\}$ [with the discrete topology] with itself), hence $C^2 \cong 2^{\mathbf N} \times 2^{\mathbf N}$, but this is homeomorphic to $2^{\mathbf N}$ by mapping the first factor to the even, the second to the odd coordinates. $\endgroup$ – martini Nov 16 '15 at 15:02
  • $\begingroup$ Thanks! So continuous is obviously not enough to keep dimension with a one-to-one mapping. I will ask a new question for more regular mappings. There's just one typo in your answer, n=2 ;) $\endgroup$ – KoliG Nov 16 '15 at 15:06

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