Can we homotope away a (non-trivial) knot.?

[Bacle]

The answer is pretty clearly yes, but , embarrassingly, I cannot find a good

argument to this-effect. Any hints, please.? (I know the answer is no for

isotopies, i.e., we cannot isotope-away a knot.). I have thought of embedding

the knot in S^3 , and avoiding the point-at-infinity, but this is not working too well.

Thanks.

 

[quasar987]

What you are after here is a mathematical proof of the obvious fact that any knot can be unknotted if we allow self intersections... that is so say, by an ambient homotopy.

So, we start with an complicated embedding p:S^1--> R^3 and the standard embedding j:S^1-->R^3. Call K_1:=j(S^1) and K_2:=p(S^1)

Then we can define a straight-line homotopy H:K_1 x I -->R^3, $H(x,t) = x + t(p(j^{-1}(x))-x)$ between K_1 and K_2. But since K_1 and K_2 are compact and embedded, there exists an open epsilon-tubular neigborhood T_1 of K_1 and T_2 of K_2. This allows us to extend the straight-line homotopy H to T_1 in an obvious way. (Namely, every point of T_1 can be uniquely written in the form x+a*n(x) where n(x) is the unit normal to K_1 at x and |a|<epsilon and analogously for K_2. Define H':T_1 x I -->R^3 by H(x+a*n(x),t)=[x+a*n(x)] + t([y+a*n(y)]-[x+a*n(x)]), where $y = p(j^{-1}(x))$.)

Finally, patch H to the whole of R^3 by a bump function: Let $\lambda:\mathbb{R}^3\rightarrow [0,1]$ be continuous with value 0 on R^3 - T_1 and value 1 on K_1 (see Urysohn's Lemma) and define H'':R^3 x I --> R^3 by parts by:

$H''(x,t)=H'(x,\lambda(x)t)$ if x is in T_1
$H''(x,t)=x$ if x is in R^3 - T_1