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I want to show (although Artin gave an ad hoc proof) that if two braids \beta and \beta' are isotopic as braids, then they are equivalent as tangles. I'd like to use the homotopy extension property (HEP) to show that there is an isotopy H of the tangle ball B^3 which is the identity on the boundary of B^3 and at one end (ie. H_0(x)=x all x in B^3), and throws \beta onto \beta' at the other; ie. H_1(\beta)=\beta' (sometimes called an "ambient" isotopy). The problem is that the HEP gives me a homotopy, whereas I want an isotopy, ie. H_t is a homeomorphism B^3 --> B^3 for all t in I=[0,1].

Can someone please help?

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You need a relative version of the Isotopy Extension Theorem. There is a proof of the non-relative version in Hirsch's "Differential Topology", pp. 177-180. Perhaps more relevant to your question is the Braid Isotopy Extension Theorem, apparently due to Artin, which is Theorem 1.11 in this book.

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  • $\begingroup$ I'm glad I hit "refresh" before posting this answer :-) $\endgroup$
    – Peter Samuelson
    Commented Aug 20, 2012 at 21:42
  • $\begingroup$ @Peter: Unlucky for you I am on the other side of the Atlantic than usual! $\endgroup$
    – Mark Grant
    Commented Aug 20, 2012 at 22:03

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