In Rolfsen's knots and links, he shows that, as a consequence of the unknotting theorem, that if you connect sum two knots and get the unknot, they both had to be unknotted. Does the same statement hold for smooth two knots? I know the smooth unknotting conjecture is still open, so one can't appeal to the same proof, but perhaps someone has shown it's true in another way or constructed a counterexample. Any references would be appreciated.
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$\begingroup$ I would suspect not. Analogously, it is not known that a connect summand of the 4-sphere is standard (the smooth 4D Schoenflies problem). This question is a sort of equivariant version of the Schoenflies problem, since a 2-fold branched cover is a homotopy 4-sphere. $\endgroup$– Ian AgolCommented Sep 24, 2021 at 18:11
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$\begingroup$ @IanAgol could you possibly say more? Is the problem that when you glue two four balls together along the S^3 boundary, you don't get an S^4? $\endgroup$– Daniel H. HartmanCommented Sep 24, 2021 at 18:16
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3$\begingroup$ You are asking about 2-knots, so 2-spheres in S^4? If the fundamental group is Z, then the 2-fold branched cover is a homotopy 4-sphere. So a summand of the unknot would have 2-fold branched cover either being a counter example to Schoenflies or a 4-sphere with an exotic involution. $\endgroup$– Ian AgolCommented Sep 24, 2021 at 18:22
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1$\begingroup$ Couldn't you use the proof using wild knots and the Eilenberg swindle in higher dimensions to prove non-cancellation in the topological category? Then the only thing you would have to worry about are knots that are smoothly knotted but not topologically knotted. $\endgroup$– LindaCommented Sep 24, 2021 at 19:44
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1$\begingroup$ Okay, but your first question was if the same argument works. Technically the argument is perfectly valid, but it starts with the unknown unknotting conjecture in dimension 4, i.e. it is a valid implication but we just don't know the starting point. I don't know of any attacks on the monoid structure for 2-knots that don't start with the unknotting conjecture. That said, there are a lot of invariants of the group completion via concordance. $\endgroup$– Ryan BudneyCommented Sep 24, 2021 at 19:48
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