Let \(f: X \to Y\) be a continuous open map. Show that if \(X\) satisfies the first or the second countability axiom, then \(f(X)\) satisfies the same axiom.
Solution
Replace \(Y\) by \(f(X)\), so \(f\) is surjective.
Suppose \(X\) is first countable. Fix \(y\in Y\) and choose \(x\in f^{-1}(y)\). Let \(B_1,B_2,\dots\) be a countable local basis at \(x\). Since \(f\) is open, each \(f(B_n)\) is an open neighborhood of \(y\). If \(V\) is any neighborhood of \(y\), then \(f^{-1}(V)\) is a neighborhood of \(x\), so some \(B_n\subset f^{-1}(V)\). Hence \(f(B_n)\subset V\). Thus \(Y\) is first countable.
Suppose \(X\) is second countable with basis \(\mathcal B=\{B_n:n\ge1\}\). Then \(\{f(B_n):n\ge1\}\) is a countable collection of open sets. It is a basis for \(Y\): if \(y\in V\) with \(V\) open, choose \(x\in f^{-1}(y)\). Since \(f^{-1}(V)\) is open, some basis element \(B_n\) satisfies \[ x\in B_n\subset f^{-1}(V). \] Then \[ y\in f(B_n)\subset V. \] So \(Y\) is second countable.