Let $\tilde{X}=\{(x.y)\in \mathcal{R}^2; \text{x or y is an integer}\}$
Let X=$\{(z_1, z_2) \in S^1\times S^1; z_1=1$ or $z_2=1\}$
and let $p:\tilde{X}\rightarrow X$ be defined by $p(x,y)=(exp(2\pi ix), exp(2\pi iy))$.
Show that $p:\tilde{X} \rightarrow X$ is a covering map.
My attempt: Let $(z_1, z_2) \in S^1 \times S^1$. Let U be an open set in $S^1 \times S^1$ such that $(z_1, z_2) \in U$. Now $(z_1, z_2)=((exp(2\pi ix_0), exp(2\pi iy_0))$ for some $(x_0, y_0)\in R^2$. and then i have no idea how to proceed.
We have the universal cover $\mathbb R\longrightarrow S^1$ which gives the product covering map $p:\mathbb R\times \mathbb R\longrightarrow S^1\times S^1$. Now the space $ X$ you give is just the wedge sum $S^1\vee S^1$ and the restriction $p^{-1}(S^1\vee S^1)=\tilde X\longrightarrow S^1\vee S^1=X$ is also a covering map.