(Purdue Jan ’07) Let \(p:E\to B\) be a covering map. Prove that \(p\) takes open sets to open sets.
Solution
Let \(O\subset E\) be open. For each \(e\in O\), choose an evenly covered neighborhood \(U_e\) of \(p(e)\), and let \(V_e\) be the sheet over \(U_e\) containing \(e\). Then \[ O\cap V_e \] is open in \(V_e\), and the restriction \(p|_{V_e}:V_e\to U_e\) is a homeomorphism. Hence \[ p(O\cap V_e) \] is open in \(U_e\), and therefore open in \(B\). Finally, \[ p(O)=\bigcup_{e\in O}p(O\cap V_e), \] a union of open subsets of \(B\). Thus \(p\) is an open map.