I can slightly confused about the difference between covering space and evenly covered. Does there exists a difference between a definition called "evenly covered space" and "covering space"? According to wikipedia:
The map $\pi$ is called the covering map, the space $X$ is often called the base space of the covering, and the space $C$ is called the total space of the covering. For any point $x$ in the base the inverse image of $x$ in $C$ is necessarily a discrete space, called the fiber over $x$.
The special open neighborhoods $U$ of $x$ given in the definition are called evenly covered neighborhoods
Then thing I am confused about is the following. Take $x,y \in X, x \neq y$, and take an evenly covered neighborhood $U_x, U_y$ of $x,y$ respectively. Then, we have discrete preimage $\pi^{-1}(U_x) = \{V_\lambda\}_{\lambda \in \Lambda}$ and $\pi^{-1}(U_y) = \{V'_{\alpha}\}_{\alpha \in \mathcal A}$. My question is, does there exist a bijection between $\Lambda$ and $\mathcal A$? The question can be written as, for every evenly covered neighborhood $U, V \subset X$, must there exist a bijection between their preimages?
Best Answer
Consider for example the $2$ point space $\{x,y\}$ with the discrete topology. The space is covered by $\{x,y_1,y_2\}$ (discrete topology) where $p(x)=x$ and $p(y_i)=y$.
Now $x\in U_x=\{x\}$ and $y\in V_y=\{y\}$. Now $p^{-1}(U_x)=\{x\}$ and $p^{-1}(U_y)=\{y_1\}\cup\{y_2\}$.