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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.

A covering space is a map $p:\tilde{X}\to X$ such that every $x\in X$ has an open neighbourhood $U$ for whichwith every restriction $p:U_{\lambda }\to U$ a homeomorphism.

Let $N=⟨⟨a⟩⟩$ in
${\pi }_1(Y)=⟨a,b\mid a^2{b}^{-3}⟩$. ThenThe subgroup is normal of index three, so the degree of a connected covering shows that every fibre $p^{-1}(y)$ has exactly three points.

The cell complex of a covering from a coset graph gives an explicit model. Take vertices $v_0,v_1,v_2$ with indices modulo three; put an $a$-loop $A_i$ at every $v_i$, and a directed $b$-edge $B_i:v_i\to v_{i+1}$. Attach a 2-cell at each $v_i$ alongMap every $v_i$ to $y_0$, every $A_i$ to the $a$-cell, every $B_i$ to the $b$-cell, and each 2-cell homeomorphically to the unique 2-cell of $Y$. Its attaching word is the lift of $a^2{b}^{-3}$ beginning at $v_i$. This defines the required connected three-sheeted covering.