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From Hirsch's Differential Topology, p. 180.

The first of the isotopy extension theorems says;

Let $A\subset M$ be a compact submanifold and $F:V\times I \rightarrow S^{3}$ an isotopy of $A$. If either $F(V\times I) \subset \delta M$ or $F(V\times I) \subset M - \delta M$, then $F$ extends to a diffeotopy of $M$ having compact support.

What I don't understand is why this does not imply that every isotopy of knots is automatically an ambient isotopy. Since knots are compact and $S^{3}$ has no boundary.

Please help, very confused!

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    $\begingroup$ Hatcher's Differential Topology ?? What's that? FYI, it does imply isotopic knots are ambient isotopic, that's the point. Perhaps you mean a book by Milnor or Guillemin and Pollack by a similar name? $\endgroup$
    – Ryan Budney
    Nov 2, 2011 at 0:59
  • $\begingroup$ Sorry, the book's by Morris Hirsch. $\endgroup$
    – Dilitante
    Nov 2, 2011 at 1:01
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    $\begingroup$ A remark that might be helpful: for smooth knots, ambient isotopy and isotopy are the same. However in the topological category, these notions differ. For example, you could pull a trefoil tight until the knotted part vanishes to a point. This is a topological isotopy, but not a smooth isotopy. $\endgroup$
    – Cheerful Parsnip
    Nov 2, 2011 at 1:11

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