When explaining the degree of a covering map, my lecture notes say:
Let $p: \bar Y \to Y$ be a covering map. Let $V \subset Y$ be a well-covered set (in the sense that its preimage consists of a disjoint union of open sets each of which is mapped homeomorphically to $V$ by $p$).
Let $y \in V$. For each of the open sets $U_\alpha$ making up $p^{-1}(V)$ as in the definition of covering map, $U_\alpha \cap p^{-1}(y)$ consists of a single point.
Why does each $U_\alpha$ have to contain a point of $p^{-1}(y)$? Can it not be the case where a $U_\alpha$ does not contain a point of $p^{-1}(y)$?
Best Answer
The definition of a covering map will help shed some light on this.
For the map $p$ to be a covering space projection, for every point $x$ in the base space, there must exist a neighbourhood $V$ of that point such that the preimage of that neighbourhood is a disjoint union of sets homeomorphic to $V$ under $p$, and in particular, restriction of $p$ to one of these disjoint components is a bijection.
In particular, the intersection of the fiber of $X$ with the restricted preimage of the neighbourhood is a single point because $p$ restricted to this set is a bijection. That is, surjectivity implies that a preimage of the point lies in the restricted preimage, and injectivity implies the uniqueness of this point.
To put it more succinctly; By the definition of a covering map, $p|_{U_{\alpha}}$ is a homeomorphism, hence a bijection, hence a surjection. And so for all $y$ in $V$, there exists a $z$ in $U_{\alpha}$ such that $p|_{U_{\alpha}}(z)=y$. Hence, $p(z)=y$.