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Let a knot in $\mathbb{R}^n$ be an embedding of $f: S^1 \to \mathbb{R}^n$ under the relation that two knots $f,g$ are equivalent if there is a 'non-crossing' homotopy of maps from $f$ to $g$ (i.e. $f_t(x): S_1\times I \to \mathbb{R}^n$ is injective for each $t \in [0,1]$.) Is this definition equivalent to simply defining the equivalence classes to simply be the isomorphism classes of the complement of the knot in $\mathbb{R}^n$?

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No. Without assuming that the embeddings are locally flat or smooth or something, the isotopy extension theorem doesn't hold, and there's no reason isotopic knots should have homeomorphic complements. For instance, the trefoil (or ANY smooth knot) is continuously isotopic to the unknot: just shrink the knotted part down to nothing.

When you assume that they're smooth, the answer is (essentially) yes. For the moment, restrict to knots in $S^3$. For then "smoothly isotopic embeddings" is equivalent to "there is an orientation-preserving diffeomorphism $S^3 \to S^3$ taking one embedding to the other". See my answer here. So in particular isotopic knots have diffeomorphic complements. It is a difficult theorem, the Gordon-Luecke theorem, that if the complements are orientation-preserving diffeomorphic, then the knots are equivalent.

Other dimensions are silly. It is the Schoenflies theorem that knots $S^1 \hookrightarrow S^2$ are the unknot, and it is reasonably elementary to prove that every smooth embedding $S^1 \hookrightarrow S^n$, $n \geq 4$, is isotopic to the unknot.

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