 Solution

Choose ${\hat{x}}_i\in p_i^{-1}(x_0)$. By the classification of connected covering spaces and the degree of a connected covering, each cover corresponds to an index-two subgroupAn index-two subgroup is the kernel of a nonzero homomorphism ${\mathbb{Z}}^2\to \mathbb{Z}/2\mathbb{Z}$. There are three:The group $G{L}_2({\mathbb{F}}_2)$ acts transitively on the three nonzero linear functionals on ${\mathbb{F}}_2^2$, and its elementary matrices lift to the integral general linear group. Hence there is $A\in G{L}_2(\mathbb{Z})$ such that

Take $\psi =f_A$. Since ${\psi }_{\ast }(H_1)=H_2$, the lifting criterion for a covering space gives a based liftsuch thatApplying the same argument to $A^{-1}$ gives a lift in the reverse direction. The composites are based lifts of the appropriate identity maps; uniqueness of based lifts makes them identities. Thus $\varphi$ and $\psi$ are homeomorphisms, and the required square commutes. This proves the equivalence of connected double covers of the torus.