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Let $\phi:\Bbb D^2\to\Bbb R^n$ be a continuous mapping of the 2-disc $\Bbb D^2$ that is injective on the boundary $\partial\Bbb D^2=\Bbb S^1$. Does its image contain an embedded disc with the same boundary? That is, is there an injective map $\psi:\Bbb D^2\to\Bbb R^n$ with $\phi(x)=\psi(x)$ for $x\in\partial \Bbb D^2$ and $\mathrm{im}(\psi)\subseteq\mathrm{im}(\phi)$? Does this depend on $n$? Is it at least true if $\phi$ is a PL map?

This is meant to be an analogy to how every curve between two points contains a non-self-crossing curve between the same points (at least I believe that this is true).

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  • $\begingroup$ If I understand you correctly, this is the Schoenflies theorem. $\endgroup$
    – Oleg Eroshkin
    Nov 12 at 0:40
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    $\begingroup$ @Oleg For $n=2$ you mean? This seems plausible. I would be most interested in $n=3,4$. $\endgroup$
    – M. Winter
    Nov 12 at 0:41
  • $\begingroup$ @ThomasKojar Hm, I do not immediately see the relation of my question to the Schoenflies problem for $n\ge 3$. Can you elaborate? I still think that the horned sphere (even a less wild version) might be a counterexample in 3D. $\endgroup$
    – M. Winter
    Nov 12 at 0:50
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    $\begingroup$ I missed that your map goes to $\mathbb{R}^n$. The answer is no for $n=3$ - attach a disk (necessary with self-intersections) to a non-trivial knot. I don't know the answer for $n>3$. $\endgroup$
    – Oleg Eroshkin
    Nov 12 at 1:22
  • $\begingroup$ @Oleg Yes, that should suffice as a counterexample. Thank you. It also yields a counterexample for $n\ge 4$ because we can just make $\phi$ map your counterexample to a 3-dimensional subspace, and since $\psi$ must be contained in $\phi$ we are back at the 3-dimensional problem. $\endgroup$
    – M. Winter
    Nov 12 at 1:25

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The answer is "yes" for $n=2$ (that is the Schoenflies theorem). For $n>2$ the answer is "no". Consider a non-trivial knot $K$ in $\mathbb{R}^3$, and a continuous injective map $\partial\mathbb{D}^2\mapsto K$. Extend that map to a map $\mathbb{D}^2\to \mathbb{R}^3$. The resulting disk has a knot as a boundary, so it must have a self-intersection. As pointed out by M. Winter, that gives a counter-example for all $n\geq3$.

Another way to construct counter-examples is to consider an embedded disk with a single point of a self-intersection. For $n\ge 4$ one may consider a disk with a transversal self-intersection. For $n=3$ one may consider a disk that touches itself -- fold a piece of paper and pinch two halves together at one point.

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