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The definition of an element of the fundamental group of a space $X$ based at point $p$ is $$f:[0,1]\rightarrow X,\quad f(0)=p=f(1),$$ defined up to homotopy. This homotopy allows self-intersection, so that there is no sense in thinking about images of loops as knots. Suppose we were to constrain the homotopy class by disallowing self-intersection. Then for each element of the ordinary fundamental group, we would have a copy of every possible knot.

My feeling is that this would get us nowhere, but I can't help but ask if anyone has thought about this and if there might be an interesting application. For instance, how about the complement of a sphere that has some infinite set of knots drilled out of it?

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    $\begingroup$ If you want the set you obtain to be a group, you need to allow «self-intersections» (although you used this term without defining it properly, so I can only guess what you mean by this). $\endgroup$
    – Loïc Teyssier
    Commented Dec 1, 2017 at 14:35
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    $\begingroup$ If you just take $X= S^1$ (a circle), then your "group" would have only two elements: go around the circle in one direction, or in the other direction. You could not compose these elements, and there would be no neutral element. $\endgroup$
    – Goldstern
    Commented Dec 1, 2017 at 14:38
  • $\begingroup$ And if $X$ is a single point, the "group" would be empty. shudder $\endgroup$
    – Goldstern
    Commented Dec 1, 2017 at 14:39
  • $\begingroup$ @Goldstern Good point. $\endgroup$
    – j0equ1nn
    Commented Dec 1, 2017 at 14:47
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    $\begingroup$ I think the reasonable thing would be to look at the space of embeddings of $S^1$ up to isotopy. This is reasonable in dimension $3$, but for higher dimensions this is probably just the space of homotopy classes of maps from $S^1$ into $X$ by general position arguments. I do not see how you could define compositions on this space. $\endgroup$
    – Thomas Rot
    Commented Dec 1, 2017 at 16:07

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Homotopy avoiding self-intersections does not preserve knottedness: Grab a knot made of a non-stretchable string of zero thickness and pull it outwards as shown below: The location where the knot occurs gets smaller and smaller, becoming a single point at the very end of the homotopy. No self-intersection occurs during this homotopy, transforming a non-trivial knot into a trivial one. Reverse this homotopy and an unknotted simple closed curve "miraculously" and instantly becomes knotted.

Perhaps you had in mind an isotopy of the ambient space, not just of the knot, but then this would be nothing else but the usual knot theory.

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    $\begingroup$ for this to work you need a topologically tame segment... I don't know what happens for very wild knots, without any tame arc. $\endgroup$
    – Daniele Zuddas
    Commented Dec 1, 2017 at 17:54
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    $\begingroup$ @DanieleZuddas: Now THAT'S a good question! I believe it still can be done, though not so easily. $\endgroup$
    – Wlodek Kuperberg
    Commented Dec 1, 2017 at 18:02
  • $\begingroup$ I acknowledge my sloppiness in properly defining the situation I wanted, but I feel the notion I described could be communicated by a proper change in definition (next time I have a chance I will edit the question to express it correctly). In the meantime, I think that the commenters have a better sense than I do of how to define what I meant and perhaps could still think about the idea. $\endgroup$
    – j0equ1nn
    Commented Dec 1, 2017 at 18:24

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