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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?

math112358
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  • It means that the function $n:X\to \text{cardinals}$, $n(x)=\lvert {y\in \tilde X,:, p(y)=x}\lvert$ is locally constant, i.e. that for all $c\in X$ there is an open set $U\ni x$ such that $n(x)=c$ for all $x\in U$. –  Oct 03 '19 at 13:21

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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 $|p^{-1}(y)| = |I|$.

Ayman Hourieh
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  • If $X$ is a connected space,how to show that the cardinality of $p^{-1}(x)$ is constant for all $x\in X$? – math112358 Oct 04 '19 at 10:59
  • Any local property is global on a connected space. See this question for a proof: https://math.stackexchange.com/q/44850/4583 – Ayman Hourieh Oct 04 '19 at 11:31