 Solution

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.