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Let $U\subseteq\Bbb R^n$ be an open set and $\gamma\subset U$ a closed null-homotopic curve in $U$ (i.e. it can be contracted to a point). Then is there an embedded disc $D\subset U$ with boundary $\gamma$?

This is a follow up on this question, from which I know the answer for

  • $n=2$: Yes, using the Schoenflies theorem.
  • $n=3$: No if $\gamma$ is a non-trivial knot. Is it true if $\gamma$ is the unknot?

I am mostly interested in $n=4$ and everything being piecewise linear, but a more general answer is welcome as well.

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  • $\begingroup$ The answer is no for $n=4$. Take a disk with a transversal self-intersection. Then a sufficiently small neighborhood of it will be a counter-example. $\endgroup$
    – Oleg Eroshkin
    Nov 12 at 22:48
  • $\begingroup$ @Oleg Hm, I don't immediately see why this is a counterexample. Can you elaborate? $\endgroup$
    – M. Winter
    Nov 12 at 23:56
  • $\begingroup$ If $D_1$ is a transversely self-intersecting disk and $U$ is a small enough neighborhood of $D_1$, then any disk in $U$ attached to the $\gamma=\partial D_1$ must also have a self-intersection. $\endgroup$
    – Oleg Eroshkin
    Nov 13 at 12:31
  • $\begingroup$ @Oleg I am not sure this is true. Couldn't I push the self-intersection to and then beyond the disc's boundary, modifying the disc only in an arbitrarily small neighbourhood? Having said that, shouldn't something like this work starting from every null-homotopy (make all intersections transversal and push them to the boundary)? I think your comment is right if $\gamma$ lies in the boundary of $U$. $\endgroup$
    – M. Winter
    Nov 13 at 21:17
  • $\begingroup$ Yes, you are right. My construction does not work. And it seems that your idea works for any smooth $\gamma$. $\endgroup$
    – Oleg Eroshkin
    Nov 14 at 2:08

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