Prove that $f$ is closed if and only if: exist $V \subset Y$ open set such that $y \in V$ and $f^{-1}(y)\in V$.

Question

Let $f: X \longrightarrow Y$ be a continuous function between two topological spaces. Prove that $f$ is closed if and only if: for every $U\subset X$ and for every $y\in Y$ such that $f^{-1}(y)\in U$, there exists an open set $V \subset Y$ such that $y \in V$ and $f^{-1}(y)\in V$.

Answer 1

The correct formulation should be:

$f$ is closed iff for all $y \in Y$ and for every open set $U \subseteq X$ such that $f^{-1}[\{y\}] \subseteq U$, there exists some open set $V \subseteq Y$ such that $y \in V$ and $f^{-1}[V] \subseteq U$.

This is true regardless of continuity of $f$ (a map can be closed without being continuous).

Note that this is a sort of dual to continuity, using fibres (inverse images of singletons).

For normal continuity we have that for every $x \in X$ and every open neighbourhood $V$ of the forward image $f(x)$ we have some open $U$ that contains $x$ and such that $f[U] \subseteq V$.

In this criterion for closedness we take an open neighbourhood of the fibre of $y$ (so going back instead of forward) and ask for a neighbourhood of that point whose inverse image stays inside the given neighbourhood, instead of the forward image.

It's a useful criterion to think about closed maps in this way, especially if you're working with so-called perfect maps (closed, continuous, surjective and fibres are compact) and preservation of properties under those maps.

That's where I first learnt about it.

If $f$ is closed, then given any open set $U$ in $X$ with $f^{-1}\big[ \{ y \} \big] \subseteq U$, we see that $X \setminus U$ is closed in $X$, and so $f \big[X\setminus U \big]$ is closed in $Y$, and we can define $V \colon= Y \setminus f \big[ X \setminus U \big]$, which is then open.

Then $y \in V$, because if $y \not\in V$, this would mean $y \in f \big[X\setminus U \big]$, and so $y=f(p)$ for some $p \in X \setminus U$, but this $p$ cannot be in $f^{-1}\big[ \{y\} \big]$ since $p \notin U$.

Also $f^{-1}[V] \subseteq U$: if $p \in f^{-1}[V]$, then $f(p) \in V$, so $f(p) \notin f \big[X \setminus U \big]$, so $p$ cannot be in $X \setminus U$, so must be in $U$, as required.

If $f$ obeys the condition as stated, $f$ is closed:

Suppose $C$ is closed in $X$ and let $y \notin f[C]$. The latter means that $f^{-1} \big[ \{ y \}\big] \subseteq U$, where $U \colon= X\setminus C$, and so we have an open set $V$ of $Y$ such that $y \in V$ and $$ f^{-1}[V] \subseteq U = X \setminus C,$$ which says that $V$ misses $f[C]$ entirely. So $y$ is an interior point of $Y \setminus f[C]$, making the latter set open and thus $f[C]$ closed.