Let \(X\) be the space obtained by deleting three distinct points from \(\RR^2\). Compute \(\pi_1(X)\).
Solution
Let the deleted points be \(p_1,p_2,p_3\), and choose pairwise disjoint closed discs \(D_i\) centered at \(p_i\).
1. \(X\) deformation retracts onto a graph \(G\) consisting of the three circles \(\partial D_i\) joined to one common basepoint by pairwise disjoint arcs.
Choose a closed disc \(D\) containing the \(D_i\) in its interior. Radial contraction outside \(D\) retracts \(X\) onto \(D\setminus\{p_1,p_2,p_3\}\). Pushing each punctured disc \(D_i\setminus\{p_i\}\) radially onto \(\partial D_i\) gives a deformation retraction onto the compact planar surface \(M=D\setminus\bigcup_{i=1}^3\operatorname{int}(D_i)\). Choose three pairwise disjoint arcs from one point of \(\partial D\) to the three inner boundary circles. Cutting \(M\) along these arcs yields a disc whose boundary consists of \(\partial D\), two copies of each arc, and the three circles \(\partial D_i\). This disc deformation retracts onto its boundary minus the open outer arc coming from \(\partial D\), and regluing gives a deformation retraction of \(M\) onto \(G\).
2. \(G\) is homotopy equivalent to \(S^1\vee S^1\vee S^1\).
The joining arcs form a tree. Collapsing that tree to the common basepoint is a homotopy equivalence of graphs and leaves the three circles \(\partial D_i\) as three wedge summands.
3. Q.E.D.
By steps 1 and 2, \(X\simeq S^1\vee S^1\vee S^1\). By the Seifert–van Kampen theorem, \[\pi_1(X)\cong\pi_1(S^1\vee S^1\vee S^1)\cong\ZZ*\ZZ*\ZZ=\boxed{F_3},\] the free group on three generators.