Let \(p: X \to Y\) be an open map. Show that if \(A\) is open in \(X\), then the map \(q: A \to p(A)\) obtained by restricting \(p\) is an open map.
Solution
Let \(U\subseteq A\) be open in the subspace \(A\). Since \(A\) itself is open in \(X\), every subspace-open set \(U\) is actually open in \(X\). Because \(p\) is an open map, \(p(U)\) is open in \(Y\). Therefore \[ p(U)\cap p(A)=p(U) \] is open in the subspace \(p(A)\). Hence the restriction \[ q=p|_A:A\to p(A) \] is open.