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In the lecture note of my knot theory course, there is a very short explanation why we need ambient isotopy instead of just isotopy. It uses the following figure to illustrate without any explanation. I don't understand why it's OK to make a knot become unknot when it's isotopy, but not OK when it's ambient isotopy. Can anyone give an intuitive explanation?

Reference

isotopic

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    $\begingroup$ It would be good to add the reference of the picture. $\endgroup$
    – Javi
    Commented Sep 9, 2020 at 12:28
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    $\begingroup$ This is relevant $\endgroup$
    – Cronus
    Commented Sep 25, 2021 at 7:23

1 Answer 1

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The idea of the diagram is to demonstrate that isotopy between two embeddings is an incorrect notion of knot equivalence. Note that a knot is an embedding:

$k:S^1\to \mathbb{R}^3$ (or more conveniently $\mathbb{S^3}$)

Given two knots (embeddings) $k_0$ and $k_1$, we may construct an isotopy of embeddings:

$k_t:S^1\times [0,1] \to S^3$

such that for each $t\in [0,1]$, $k_t$ is an embedding.

For any tame knot we may construct an isotopy that intuitively "pulls" the knotted portion of an knot down to a point (as pictured). For any $t<1$ the knot is not self-intersecting, and for $t=1$ the embedding is that of the unknot and thus not self-intersecting. Therefore, the map described is an isotopy of embeddings. However, this map is not differentiable (smooth) around $t=1$. To avoid making all tame knots equivalent, a more sensitive measure of equivalence is adopted in which the isotopy of embeddings is smooth.

To reconcile this with the standard definition of knot equivalence via ambient isotopy, it should be noted that any smooth isotopy of embeddings lifts to an ambient isotopy.

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