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My understanding is that if the embeddings $f_0,f_1$ are tame knots then

$H(t,\theta) = (1-t)f_0(\theta) + t f_1(\theta)$

is a homotopy between them, thus all tame knots are homotopic. Is this the case?

To provide some context, I'm writing a paper on introductory knot theory and I want to use this to motivate why isotopy (where we additionally require $H(t,\theta)$ to be an embedding for all $t$) is used to define equivalence of tame knots. My supervisor thinks the above statement is wrong but after a quick exchange of emails I'm none the wiser as to how it is wrong. I'm meeting with her on monday but it would be nice to get this straightened out before then.

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up vote 3 down vote accepted

This follows from the fact that $\pi_{1}(\mathbb{R}^3) = 0$. In fact, any two knots are also isotopic to the un-knot by shrinking the knotted parts of the knots to a point (an embedding for all $t$). The correct idea is an ambient isotopy, since when you shrink a portion of Euclidean space to a point you do not have an embedding of Euclidean space into itself.

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Thanks, I think I understand. Is there any reason why ambient isotopy is used instead of requiring $H(t,\theta)$ to be a tame knot for all $t$? It seems this would be a less convoluted way of doing things, though i suppose it doesn't extend to wild knots. – remus Apr 18 '13 at 14:26
If you shrink knotted parts to a point, you should remain tame for all $t$. As you shrink the knotted portion, you remain equivalent to a polygonal knot where the lines representing the knotted portion shrink. In the limit, you are equivalent to the unknot, which is very tame! – Isaac Solomon Apr 18 '13 at 15:46

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