# Equivalence of two definitions of a knot

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 $\mathbb{R}^n$?

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.