So what this is saying is that ensuring sufficient (smooth) structure obviates any need explicitly to involve the rest of the exterior space. Which I could sort of see from the start, but failed to appreciate the significance of involving the fibre/fiber. Overall, excellently instructive problem & discussion! WFL On 4/6/20, Andy Latto <andy.latto@pobox.com> wrote:
On Fri, Apr 3, 2020 at 12:38 PM Fred Lunnon <fred.lunnon@gmail.com> wrote:
What theorem were you employing to deduce that such an extension exists?
I found the theorem that confirms my intuitions; its the Isotopy Extension Theorem, which only works for manifolds in the smooth category (Alexander's Horned Sphere is a counterexample in the topological category). http://math.ucr.edu/~res/math260s10/isotopyextension.pdf
Isotopy is what I was calling "homotopy injective on fibers"; a homotopy I X N -> M is an isotopy if the restriction {t} X N -> M is an embedding.
The theorem says that if N is a smooth compact manifold M is a smooth manifold, and f: N X I -> M is a smooth isotopy, that it extends to an ambient isotopy, that is an isotopy f* : M x I -> M. Applying this to the isotopy that "unfolds" the boundary of the standard embedding of the mobius strip to the geometric circle, and then restricting the endpoint of this map to the original mobius strip, we have the desired embedding of the mobius strip with a geometric circle as an edge.
The isotopic extension property (in the case where N is the circle and M is R^3, which is the same case we used here) is what you need to show that the fundamental group of the knot complement is a knot invariant.
Andy
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