Generalize the proof of Theorem 54.5 to show that the fundamental group of the torus is isomorphic to the group \(\mathbb{Z} \times \mathbb{Z}\).
Solution
Write the torus as \[ T=S^1\times S^1. \] Consider the covering map \[ p:\mathbb R^2\longrightarrow T, \qquad p(s,t)=\bigl(e^{2\pi i s},e^{2\pi i t}\bigr). \] It is the product of the standard coverings \(\mathbb R\to S^1\), hence is a covering map. The total space \(\mathbb R^2\) is simply connected.
Fix the basepoint \((1,1)\in T\) and its lift \((0,0)\). For a loop \(f\) in \(T\) based at \((1,1)\), let \(\widetilde f\) be its unique lift beginning at \((0,0)\). Its endpoint lies in \[ p^{-1}(1,1)=\mathbb Z^2. \] Define \[ \Phi:\pi_1(T,(1,1))\to\mathbb Z^2, \qquad \Phi([f])=\widetilde f(1). \] Exactly as in the proof for \(S^1\), the homotopy-lifting theorem shows that \(\Phi\) is well defined, and lifting concatenations shows \[ \Phi([f]*[g])=\Phi([f])+\Phi([g]). \] It is surjective: for \((m,n)\in\mathbb Z^2\), the path \[ \widetilde f(t)=(mt,nt) \] projects to a loop whose image under \(\Phi\) is \((m,n)\).
It is injective: if \(\Phi([f])=(0,0)\), then \(\widetilde f\) is a loop in the simply connected space \(\mathbb R^2\), so it is nullhomotopic there. Projecting the nullhomotopy by \(p\) shows that \(f\) is nullhomotopic in \(T\).
Therefore \[ \boxed{\pi_1(S^1\times S^1,(1,1))\cong\mathbb Z\times\mathbb Z}. \]