Lemma 5.6.4. Let $f : X \to Y$ be surjective, open, continuous map of topological spaces. Let $T \subset Y$ be a subset. Then

$f^{-1}(\overline{T}) = \overline{f^{-1}(T)}$,

$T \subset Y$ is closed if and only if $f^{-1}(T)$ is closed,

$T \subset Y$ is open if and only if $f^{-1}(T)$ is open, and

$T \subset Y$ is locally closed if and only if $f^{-1}(T)$ is locally closed.

In particular we see that $f$ is submersive.

**Proof.**
It is clear that $\overline{f^{-1}(T)} \subset f^{-1}(\overline{T})$. If $x \in X$, and $x \not\in \overline{f^{-1}(T)}$, then there exists an open neighbourhood $x \in U \subset X$ with $U \cap f^{-1}(T) = \emptyset $. Since $f$ is open we see that $f(U)$ is an open neighbourhood of $f(x)$ not meeting $T$. Hence $x \not\in f^{-1}(\overline{T})$. This proves (1). Part (2) is an easy consequence of (1). Part (3) is obvious from the fact that $f$ is open and surjective. For (4), if $f^{-1}(T)$ is locally closed, then $f^{-1}(T) \subset \overline{f^{-1}(T)} = f^{-1}(\overline{T})$ is open, and hence by (3) applied to the map $f^{-1}(\overline{T}) \to \overline{T}$ we see that $T$ is open in $\overline{T}$, i.e., $T$ is locally closed.
$\square$

## Comments (2)

Comment #3561 by Laurent Moret-Bailly on

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