Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 2 / 21F (Algebraic Topology)

Regard the 3-sphere asIt is the union of the two closed solid toriExplicit homeomorphisms from the two copies in the question areandOn their boundaries both maps send $(u,v)\in S^1\times S^1$ to $(u,v)/\sqrt{2}$. They therefore descend through the quotient to a continuous bijection $S\to S^3$. Since $S$ is compact and $S^3$ is Hausdorff, this bijection is a homeomorphism.

Writing the square coordinates modulo one, the curve has the parametrizationIt is therefore the $(2,3)$ torus knot, or trefoil, on the common boundary of the two solid tori. Applying the Seifert-van Kampen theorem to the complements on the two sides gives one generator $a$ for the meridian of one solid torus and one generator $b$ for the other; the boundary slope identifies two turns of the first with three turns of the second. HenceThis group is nonabelian: imposing the additional relation $a^2=b^3=1$ gives a quotientwhich maps onto the nonabelian group $S_3$ by sending $a$ to a transposition and $b$ to a three-cycle. An abelian group cannot have a nonabelian quotient.