Let \(T = S^1 \times S^1\) be the torus; let \(x_0 = b_0 \times b_0\).
- Prove the following.
Theorem. Every isomorphism of \(\pi_1(T, x_0)\) with itself is induced by a homeomorphism of \(T\) with itself that maps \(x_0\) to \(x_0\).
[Hint: Let \(p: \mathbb{R}^2 \to T\) be the usual covering map. If \(A\) is a \(2 \times 2\) matrix with integer entries, the linear map \(T_A: \mathbb{R}^2 \to \mathbb{R}^2\) with matrix \(A\) induces a continuous map \(f: T \to T\). Furthermore, \(f\) is a homeomorphism if \(A\) is invertible over the integers.]
- Prove the following.
Theorem. If \(E\) is a covering space of \(T\), then \(E\) is homeomorphic either to \(\mathbb{R}^2\), or to \(S^1 \times \mathbb{R}\), or to \(T\).
[Hint: You may use the following result from algebra: if \(F\) is a free abelian group of rank 2 and \(N\) is a nontrivial subgroup, then there is a basis \(a_1, a_2\) for \(F\) such that either (1) \(ma_1\) is a basis for \(N\), for some positive integer \(m\), or (2) \(ma_1, na_2\) is a basis for \(N\), where \(m\) and \(n\) are positive integers.]
Solution
Identify \[ \pi_1(T,x_0)\cong\mathbb Z^2. \] Every automorphism is therefore represented by a matrix \[ A\in GL_2(\mathbb Z). \] Let \(L_A:\mathbb R^2\to\mathbb R^2\) be the linear map with matrix \(A\). Since \(A(\mathbb Z^2)=\mathbb Z^2\), the map descends through the quotient \(\mathbb R^2/\mathbb Z^2=T\) to \[ f_A:T\to T,\qquad [v]\mapsto[Av]. \] It fixes the basepoint \([0]\). Since \(A^{-1}\in GL_2(\mathbb Z)\), \(f_{A^{-1}}\) is its inverse, so \(f_A\) is a homeomorphism. On fundamental groups, \((f_A)_*=A\). Hence every automorphism of \(\pi_1(T)\) is induced by a basepoint-preserving homeomorphism.
Assume the covering \(p:E\to T\) is connected, and put \[ H=p_*\pi_1(E,e_0)\le\mathbb Z^2. \] By the subgroup classification quoted in the hint, after changing the basis of \(\mathbb Z^2\) by an automorphism, exactly one of the following occurs:
- \(H=0\).
- \(H=\langle(m,0)\rangle\) for some \(m>0\).
- \(H=\langle(m,0),(0,n)\rangle\) for some \(m,n>0\).
By part (a), the basis change is induced by a homeomorphism of the base torus, so it does not change the homeomorphism type of the total covering space. The standard coverings realizing these three subgroups are exactly those in E-XFMJA: \[ \mathbb R^2\to T,\qquad S^1\times\mathbb R\to T,\qquad T\to T. \] The classification theorem for connected covering spaces says two connected coverings corresponding to the same subgroup (up to the chosen basepoint equivalence) are equivalent. Therefore \[ \boxed{E\cong\mathbb R^2,\quad S^1\times\mathbb R,\quad\text{or}\quad T.} \]