A Topological Puzzle

The proof simply consists in computing the fundamental group of the complement X of W in R3 and checking that the class of r in π1(X) is not 1. The first step means finding a presentation of π1(X). This could quickly be achieved using the Wirtinger presentation, but I prefer to give a more self-contained proof, using only van Kampen’s theorem. Let W1 be the wire obtained from W by adding to it the two straight bars P and Q (see Figure 2) and W2 the wire obtained by adding to W the two straight bars R and S. Clearly W = W1 ∩ W2. We call W0 the union of W1 and W2 obtained by attaching to W the four bars P , Q, R, and S. Let S3 be the one point compactification of R3. We think of the extra point as the unique point at infinity and we choose it as the base point for the various fundamental groups that are going to appear. This simplifies the drawing and the deformation of loops: instead of beginning and ending at the same point of R3, we let them come from an infinite distance (from an arbitrary direction) and return to infinity in a possibly different direction. Let X = S3 \ W and Xi = S3 \ Wi for i = 0, 1, 2, so that X = X1 ∪ X2 and X0 = X1 ∩ X2. By van Kampen’s theorem (see, for instance, [1, chap. 4, sec. 2]) π1(X) is the amalgamated product of π1(X1) and π1(X2) over π1(X0). It is not difficult to see that there exists a deformation of S3 that transforms W1 into a wedge of plane (in particular, unknotted) circles (more precisely, solid tori), like the wire in Figure 3. Our first task is to determine the fundamental group of its complement. The fundamental group of the complement—call it Y1—of a single unknotted circle is infinite cyclic, generated by a loop that crosses the disc spanned by the circle only once. This follows, for instance, from the well-known decomposition of S3 into two solid tori (see [1, chap. 4, sec. 6]) showing that the complement of an unknotted solid torus in S3 is homeomorphic to a solid torus. Assume, by induction, that the fundamental group of the complement Yn of a wedge of n circles is the free group of rank n. Then, given