Lemma 5.6.5. Let $f : X \to Y$ be surjective, closed, continuous map of topological spaces. Let $T \subset Y$ be a subset. Then

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

$T \subset Y$ 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 $\overline{f^{-1}(T)} \subset f^{-1}(\overline{T})$. Then $T \subset f(\overline{f^{-1}(T)}) \subset \overline{T}$ is a closed subset, hence we get (1). Part (2) is obvious from the fact that $f$ is closed and surjective. Part (3) follows from (2) applied to the complement of $T$. For (4), if $f^{-1}(T)$ is locally closed, then $f^{-1}(T) \subset \overline{f^{-1}(T)}$ is open. Since the map $\overline{f^{-1}(T)} \to \overline{T}$ is surjective by (1) we can apply part (3) to the map $\overline{f^{-1}(T)} \to \overline{T}$ induced by $f$ to conclude that $T$ is open in $\overline{T}$, i.e., $T$ is locally closed.
$\square$

## Comments (2)

Comment #3562 by Laurent Moret-Bailly on

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