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Imagine that I have a closed or open piece of string that I move continuously through space without self-intersection. Formally, this means that I have a homotopy $\phi_t:[0,1]\times X\to\Bbb R^3$ where $X=[0,1]$ or $X=S^1$ and so that $\phi_t$ is a homeomorphism onto its image for all $t\in[0,1]$.

Question: Does this homotopy extend to a homotopy of the whole space? That is, is there a homotopy $\bar\phi_t:[0,1]\times \Bbb R^3\to\Bbb R^3$ so that $\bar\phi_t$ is a homeomorphism and $\bar\phi_t|_X=\phi_t$ for all $t\in[0,1]$?

Is this a standard result with a name? It strikes me as something belonging to knot theory, but I am not familar enough with this subject to recognize the result.

My main question is stated as above. But I do also wonder how general such an extension property can hold (if at all). For example, I believe it fails for $X=S^2$ (Alexander's horned sphere maybe?).

M. Winter
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    What you describe sounds more like isotopy than homotopy and there are isotopy extension theorems under various assumptions. There is also a homotopy extension criterion. – Conifold Oct 07 '21 at 18:18
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    In your definition you are making the "standard mistake" when defining your homotopy (actually, isotopy, as correctly noted by Conifold): Under such definition one can "untie" any given tame knot in $R^3$. There are several ways to avoid this trouble, for instance, use PL maps. But, with the definition that you gave, in general, such a homotopy does not extend to an ambient isotopy. https://math.stackexchange.com/questions/1336275/which-two-knots-are-isotopic-but-not-ambient-isotopic – Moishe Kohan Oct 10 '21 at 23:11
  • @MoisheKohan This was a helpful link which highlights some flaws in my thinking. Thank you. – M. Winter Oct 11 '21 at 11:14

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