I'm not very familiar with the Whitehead manifold. But: With 4, or as few as 2, tori linked around (the interior of) a larger one T, clearly a meridian of T lying in R^3 - T (or S^3 - T) cannot be shrunk to a point in the complement of the linked tori. Yet with only 1 torus T_1 self-linked around a larger one T, then -- surprisingly -- a meridian M of T *can* be shrunk to a point in the complement of T_1. (Not through homeomorphisms, but through a continuous family of continuous maps f_t : S^1 -> R^3 - T_1, 0 <= t <= 1, where f_0 is a homeomorphism of S^1 onto M. There must be at least one t, 0 < t < 1, for which f_t is not one-to-one.) Note: The meridian M and the core circle of T_1 form what's known as the Whitehead link, with linking number = 0. (For a useful image, see fig. 16 at < http://www.math.cornell.edu/~mec/2008-2009/HoHonLeung/page2_knots.htm >. It's a symmetrical link: the two closed curves can be interchanged.) --Dan On 2012-12-08, at 1:32 PM, Andy Latto wrote:
On Sat, Dec 8, 2012 at 4:23 PM, Dan Asimov <dasimov@earthlink.net> wrote:
I had a suspicion I knew that space [the intersection of nested tori, each pulled around the previous one so as to link itself], from a previous encounter with it.
Sure enough, it turns out that it's the *complement* of the amazing "Whitehead contractible manifold" W in the 3-sphere.
When you nest 4 linked tori inside each torus at each step, you've proved that the complement of the intersection is not simply connected, hence not contractible. Why doesn't the same proof work in the case where you place a single self-linked torus inside the torus each time?