Discussion
Let X be a second countable space and \(f : X \to Y\) be continuous open surjective map. Then :
Answer: A
Y is second countable.
Let \(\{B_{n}\}\) be a countable base for X. Since \(f\) is open, each \(f(B_{n})\) is open in Y. Given an open set \(U \subseteq Y\) and \(y \in U\), pick \(x\) with \(f(x) = y\) (surjectivity); \(f^{-1}(U)\) is open (continuity) and contains \(x\), so some \(B_{n}\) has \(x \in B_{n} \subseteq f^{-1}(U)\). Then \(y \in f(B_{n}) \subseteq U\). So \(\{f(B_{n})\}\) is a countable base for Y.
A second countable space is also first countable and separable, so the options denying either property are wrong.
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