So I wanted to know what is this glueing map h : T^2 - D^2 —> T^2 - D^2 which for the trefoil knot K determines its complement S^3 - K via S^3 - K = (T^2 - D^2) x [0,1] / (x,1) ~ (h(x),1) ??? I finally figured it out, and it's kind of pretty. (No doubt this was known long, long ago to knot theorists.) Consider the hexagonal torus: a regular hexagon with each pair of its opposite sides identified by a translation across the hexagon. Now remove a small disk D^2 centered at the center of the hexagon. Finally, the glueing map h just takes this holey hexagonal torus and rotates it 120º into itself. That's it. That's S^3 - K when K is the trefoil. If you're partial to knots in R^3 instead, just remove another point from that. (Something similar happens with all (p,q) torus knot complements. —Dan ----- There's an interesting feature of many knots called "fibring", and such knots are called "fibred knots". A knot can be thought of as living in R^3 but it can be more useful to think of it as living in the 3-sphere S^3 (which is topologically just R^3 with a point added at infinity). To distinguish knots — since all knots are topologically circles — people have studied not the knot itself but what becomes of S^3 after the knot is removed. S^3 - K is called the knot complement. For many knots K in S^3, it turns out that there is a lovely surjective map π : S^3 - K —> S^1 to the circle, such that the inverse image of each point of S^1 is always a certain M (with boundary) in a continuous manner. If we removed one of these surfaces, say π^(-1)(1), what remained of S^3 - K would be topologically M x (0, 1), the product of the surface with an open interval. If we like we can glue on the missing end-surfaces to get M x [0, 1]. Then knot complement is completely determined by the way the end-surfaces are glued back together: S^3 - K = M x [0,1] / (x,1) ~ (h(x),0) where h is some self-homeomorphism h : M —> M of M. ----