There is a staement in Hatcher's book:If $p:\tilde{X} \to X$ is a covering space,then the cardinality of the set $p^{-1}(x)$ is locally constant over $X$.
Does it mean that for any $x\in X$ there exits a neighborhood $U_x$ of $x$ such that $|p^{-1}(U)|$ is a constant? How to check it?
The statement implies that each $x \in X$ has an open neighborhood $U$ such that $|p^{-1}(y)|$ is constant for all $y \in U$.
To see why, let $U$ be an open neighborhood of $x \in X$ such that $p^{−1}(U) = \bigsqcup_{i\in I} V_i$ is a union of disjoint open sets in $\widetilde X$, each of which is mapped homeomorphically onto $U$ by $p$. Then each point $y \in U$ has exactly one preimage in each $V_i$. It follows that $|p^{-1}(y)| = |I|$.