Consider the covering map indicated in Figure 60.3 of the text. Here, \(p\) wraps \(A_1\) around \(A\) twice and wraps \(B_1\) around \(B\) twice; \(p\) maps \(A_0\) and \(B_0\) homeomorphically onto \(A\) and \(B\), respectively. Use this covering space to show that the fundamental group of the figure eight is not abelian.
Solution
Let \(X=A\vee B\) be the figure eight, based at \(x_0\). Let \(f\) traverse \(A\) once counterclockwise and let \(g\) traverse \(B\) once counterclockwise.
Use the covering in Figure 60.3 and begin lifts at the central point \(e_0\). The lift \(\widetilde f\) runs along the half of \(A_1\) from \(e_0\) to the vertex \(e_2\). From \(e_2\), the loop \(g\) lifts to the loop \(B_0\), so the lift of \(f*g\) beginning at \(e_0\) ends at \(e_2\).
Similarly, the lift \(\widetilde g\) runs along the half of \(B_1\) from \(e_0\) to the vertex \(e_1\). From \(e_1\), the loop \(f\) lifts to the loop \(A_0\), so the lift of \(g*f\) beginning at \(e_0\) ends at \(e_1\).
The endpoints \(e_1\) and \(e_2\) are distinct. By the homotopy-lifting criterion, path-homotopic loops based at \(x_0\) have lifts beginning at \(e_0\) with the same endpoint. Therefore \[ f*g\not\simeq_p g*f. \] Consequently \[ [f][g]\ne[g][f] \] in \(\pi_1(X,x_0)\). Hence the fundamental group of the figure eight is not abelian.