Lemma 5.26.2. Let $f : X \to Y$ be a continuous map of topological spaces. Assume $f$ is surjective and $f(E) \not= Y$ for all proper closed subsets $E \subset X$. Then for $U \subset X$ open the subset $f(U)$ is contained in the closure of $Y \setminus f(X \setminus U)$.
Proof. Pick $y \in f(U)$ and let $V \subset Y$ be any open neighbourhood of $y$. We will show that $V$ intersects $Y \setminus f(X \setminus U)$. Note that $W = U \cap f^{-1}(V)$ is a nonempty open subset of $X$, hence $f(X \setminus W) \not= Y$. Take $y' \in Y$, $y' \not\in f(X \setminus W)$. It is elementary to show that $y' \in V$ and $y' \in Y \setminus f(X \setminus U)$. $\square$
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