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Nov 12 at 21:47 vote accept M. Winter
Nov 12 at 17:46 answer added Oleg Eroshkin timeline score: 6
Nov 12 at 2:05 comment added M. Winter @OlegEroshkin I don't need any projections. Just use that $\mathrm{im}(\psi)\subseteq\mathrm{im}(\phi)$. So whatever dimension $\phi$ "lives in" is forced on $\psi$. And you can choose $\phi$ essentially 3-dimensional.
Nov 12 at 2:01 comment added Oleg Eroshkin I don't follow you. The fact that a projection on $\mathbb{R}^3$ is not injective says nothing about the map to $\mathbb{R}^4$.
Nov 12 at 1:25 comment added M. Winter @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.
Nov 12 at 1:22 comment added Oleg Eroshkin 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 $n>3$.
Nov 12 at 0:50 comment added M. Winter @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.
Nov 12 at 0:41 comment added M. Winter @Oleg For $n=2$ you mean? This seems plausible. I would be most interested in $n=3,4$.
Nov 12 at 0:40 comment added Oleg Eroshkin If I understand you correctly, this is the Schoenflies theorem.
Nov 11 at 23:46 history asked M. Winter CC BY-SA 4.0