Proposition.
Let $p:E\rightarrow B$ a covering map. If $U\subset B$ is evenly covered by p and $U'\subset U$ is open, $U'$ is evenly covered by p.
Proof.
Let $\{V_\alpha\}$ be the partition of $p^{-1}(U)$ into pieces and $p|_{V_\alpha}:V_\alpha\rightarrow U$, a homeomorphism. Let $V_\alpha=(p|_{V_\alpha})^{-1}(U)$ are disjoint, so $(p|_{V_\alpha})^{-1}(U')$ are disjoint and $\bigcup (p|_{V_\alpha})^{-1}(U')=p^{-1}(U')$. Let $W_\alpha=(p|_{V_\alpha})^{-1}(U')$. $p|_{W_\alpha}:W_\alpha\rightarrow U'$ is a restriction of $p|_{V_\alpha}$, which is a homeomorphism, so it is a homeomorphism. This concludes $U'$ is evenly covered by $p$, and $\{W_\alpha\}$ is the partition of $p^{-1}(U')$ into pieces.
End of proof.
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예전에는 챗 지피티한테 위상수학 물어보면 정신 못차렸는데 이제는 이런 간단한 증명 정도는 구글도 해준다..

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