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Maybe this question is too trivial for a research site but there are so many notions of equivalence of knots that I am lost in literature. The question that interests me is the following.

A knot is a continuous embedding $\gamma: S^1 \rightarrow S^3$.

Two knots $\gamma_0$ and $\gamma_1$ in $S^3$ are homotopically equivalent if there is a homotopy $\gamma_t: S^1 \times [0,1] \rightarrow S^3$.

If I understand correctly, if the knots are homotopically equivalent then their complements are homeomorphic (we can use isotopy extension theorem).

Is the converse true or not?

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    $\begingroup$ Ah, so you are not assuming that the homotopy is through knot embeddings? Note that knots are not just continuous maps, in most peoples' minds, but continuous embeddings. Your answer to this will have a big impact on what people can say about your question. $\endgroup$
    – David Roberts ♦
    Commented Jun 10, 2014 at 4:59
  • $\begingroup$ @David Roberts Thank you! Of course, I want to be as others.) $\endgroup$
    – Olga
    Commented Jun 10, 2014 at 5:10
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    $\begingroup$ Your notion of "homotopically equivalent" is too weak. Even if you require that $\gamma_t$ is an embedding for each $t$, you'll still have all knots homotopically equivalent to the unknot. The homotopy consists of "pulling the knot tight" so that the knotted part shrinks down to a point while the rest of the knot approaches an unknot. I think this observation is called "Alexander's trick". $\endgroup$
    – Andreas Blass
    Commented Jun 10, 2014 at 6:07
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    $\begingroup$ The right notion of equivalence is ambient isotopy (en.wikipedia.org/wiki/Ambient_isotopy). Any other had better be equivalent to this one. $\endgroup$
    – Paul Siegel
    Commented Jun 10, 2014 at 7:37
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    $\begingroup$ To the best of my knowledge, the usual definition of knot equivalence is ambient isotopy, and this is equivalent (modulo orientation issues) to saying that there's a self-homeomorphism of $S^3$ that sends one knot to the other. $\endgroup$
    – Andreas Blass
    Commented Jun 10, 2014 at 15:41

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This is really just a long comment and figure. It summarizes the comments above and addresses the theme of the question for a natural class of knots.

In their book "Knots,'' Burde and Zieschang define a knot as an embedded subset of $S^3$ homeomorphic to $S^1$. A general notion of equivalence for two knots $K_1,K_2$ is if they are equivalent under ambient isotopy, i.e. there is a isotopy I:$S^3 \times [0,1] \rightarrow S^3$, $I(y,t)=h_t(y)$ with $h_t$ a homeomorphism for all $t$ and $I(x,0)=x$ and $h_1(K_1)=K_2$.

A further common restriction is that a knot is tame, i.e. the knot is equivalent to a finite sided embedded polygon in $S^3$.

A famous (and non-trivial) theorem of Gordon and Luecke states: Two tame knots are equivalent if and only if their complements are homeomorphic.

Edit: As pointed out in the comments, Gordon and Luecke's notion of equivalence is that two knots $K_1$ and $K_2$ are equivalent if and only if $K_1$ or its mirror image is ambient isotopic to $K_2$.

As pointed out in the comments, for your definitions, a knot complement is not well defined (see also "Knots" figure 1.1). Consider the trefoil knot embedded in the standard way as in the figure on right. The notice we can put a ball around the part of the knot that looks like the figure on the left and contract that ball to a point via a homotopy of $S^3$ that is also an isotopy of the knot but not the ambient space. If one used this notion of equivalence, the trefoil is equivalent to the unknot. However, by a whole host of knot invariants (Alexander polynomial, fundamental group, tri-colorability, etc.), the trefoil is a non-trivial knot.

Trefoil: taken from wiki commons

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    $\begingroup$ One needs to say something about the orientation. Gordon and Luecke consider a simpler equivalence notion, not through isotopy. $\endgroup$
    – Włodzimierz Holsztyński
    Commented Jun 11, 2014 at 6:09

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