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Suppose you have an arc in a ball with ends fixed on a boundary. In other words, we have an isolated knot on one of the arcs of a tangle. If we glue the ends we get an ordinary knot and clearly isotopic isolated knots, when ends are connected, give the same knot. But is this a bijection, i.e. can we have different knots of the first kind that give the same knot after the ends are glued?

This seems like it should be obvious, but I did try to google and ask people in my department - neither helped.

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Two long knots are equivalent if and only if their closures are equivalent. This is because any Reidemeister move on the knot can be realized by Reidemeister moves which avoid a single point, which you can take to be the point at infinity. This is a combinatorial proof. I think there's a smooth proof also. –  Daniel Moskovich Mar 21 '13 at 13:51
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When you close the knot you lose the sense of orientation given by the open ends of the knot. So your "isolated knots" up to isotopy is equivalent to "oriented closed knots" up to isotopy. –  Ryan Budney Mar 21 '13 at 17:24
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FYI Daniel there's much more than a smooth proof. If you're careful about the argument you get a proof that the space of closed knots $Emb(S^1,S^n)$ has the homotopy-type of $SO_{n+1} \times_{SO_{n-1}} K_{n,1}$ where $K_{n,1}$ is the space of knots with fixed endpoints in the ball $D^n$, and the $\times_{SO_{n-1}}$ indicates you take the product then mod out by the diagonal action of $SO_{n-1}$. The above question is asking how the two spaces $\pi_0$ are connected in the $n=3$ case. –  Ryan Budney Mar 21 '13 at 17:38
    
There's a similar argument relating the homotopy-type of the above spaces to the space of embeddings of $S^1$ in $\mathbb R^n$ but this one is a step more elaborate. –  Ryan Budney Mar 21 '13 at 17:39
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