Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 1 / 21F (Algebraic Topology)

The van Kampen theorem says: if $X=U\cup V$ with $U,V$ open and path connected, $U\cap V$ path connected, and a base point in the intersection, then ${\pi }_1(X)$ is the free product ${\pi }_1(U)\ast {\pi }_1(V)$ modulo the normal subgroup generated by identifying the two images of every element of ${\pi }_1(U\cap V)$. Equivalently it is the pushout of the two inclusion-induced fundamental-group maps.

Choose open neighborhoods of the sphere and circle in the wedge, each enlarged by a small contractible neighborhood of the attaching point in the other piece. They deformation retraction to $S^2$ and $S^1$ and have contractible intersection. Van Kampen therefore givesThe generator traverses the circle once; the sphere contributes no fundamental-group generator.

A geometric circle is an unknot. Compactify ${\mathbb{R}}^3$ to $S^3$, and carry the circle to $\{(z_1,z_2)\in {\mathbb{C}}^2:|z_1{|}^2+|z_2{|}^2=1,z_2=0\}$. Its complement has coordinates $z_1\in \Delta$ and $z_2=\sqrt{1-|z_1{|}^2}{e}^{i\theta }$, making it homeomorphic to $\Delta \times S^1$. It deformation retracts to $S^1$. Removing the additional point at infinity does not change the fundamental group: van Kampen for a small ball about that point uses a simply connected punctured ball, homotopy equivalent to $S^2$. ThusA meridian linking the circle once generates it. This uses “circle” in its geometric sense; the complement of an arbitrary knotted embedded copy of $S^1$ need not have this group.