Is every open surjection also closed? This confuses me. It is mentioned that this is not true, but I can't see why.
Suppose that $f: X \to Y$ is an open surjection between topological spaces. Then for every closed $C \subset X$ the set $Y \setminus f(X \setminus C)$ is closed. Does the equality $Y \setminus f(X \setminus C) = f(C)$ hold? Let's see.
Let $y \in Y$ be some arbitrary element. Then $y \not\in f(X \setminus C)$ iff $f^{-1}(y) \cap (X \setminus C) = \varnothing$ iff $f^{-1}(y) \subset C$ iff $f^{-1}(y) = \varnothing$ or $y \in f(C)$, so for a surjection openness and closedness must be equivalent, right? Something must elude me here, but embarassingly I don't understand what it is.