# Necessary and sufficient condition for $f$ to be a closed mapping

Let $$f:X\to Y$$ be a mapping from the topological space $$X$$ to $$Y$$. Show that the existence of neighborhood $$V_y$$ of $$\forall y \in Y$$ such that $$f^{-1}(V_y)\subset U$$ for any open $$U: f^{-1}(y)\in U$$ is necessary and sufficient condition for $$f$$ to be a closed mapping.

I'm fairly new to topology, so even the statement of the problem was quite tough to understand for me.

I started by trying to prove the sufficient condition:

• Let $$f:X\to Y$$ be a mapping from the topological space $$X$$ to $$Y$$. Assume there exists a neighborhood $$V_y$$ of $$\forall y \in Y$$ such that $$f^{-1}(V_y)\subset U$$ for any open $$U: f^{-1}(y)\in U$$. Now, we know that $$f(y)=U$$ and $$V_y \subseteq f(U)$$. This implies that $$f(U)$$ is an open set since $$y$$ lies in $$f(U)$$ with some neighborhood. Then, $$f$$ sends open sets to open sets $$(\because U\overset f{\mapsto} f(U) \text{ is open to open})$$.

I do not know if I am right so far, but even if I am, I do not see any use of that piece of observation.

For the second direction (assuming $$f$$ is closed), I do not have idea on how to use the definition of the closed mapping (if it's really needed).

• I'm afraid you are not right so far. Closed maps need not be open maps. A mistake appears when you write $f(y)=U$. Here $y\in Y$, so $f(y)$ is not even defined. Also it does not follow that $V_y\subseteq f(U)$. For instance the map $f$ might not be surjective (consider the case in which $f$ is the inclusion of a closed subspace). Commented Oct 30, 2020 at 18:13
• What you are trying to show is that $C\subseteq X$ closed $\Rightarrow$ $f(C)\subseteq Y$ closed. So take $C$ and use the asusmptions to show that $Y\setminus f(C)$ is open. Do this by showing that if $y\in Y\setminus f(C)$, then $y$ has a neighbourhood contained inside $Y\setminus f(C)$. (Hint: there is an obvious open set of $X$ containing $f^{-1}(y)$) Commented Oct 30, 2020 at 18:18
• What you’ve set out to prove there is sufficiency, not necessity: you’re trying to prove that the given condition is sufficient to ensure that $f$ is closed. Commented Oct 30, 2020 at 18:18
• @Tyrone Thank you for your hints. Let me see if I can take it from there.
– user775075
Commented Oct 30, 2020 at 18:21
• @Tyrone So, I need to show that $$\forall y \in Y \setminus f(C), ~ \exists W_y: W_y \subseteq Y\setminus f(C)$$ Well, there is an open set $U: U \ni f^{-1}(y)$. But how can I show that there exists such $W_y$?
– user775075
Commented Oct 30, 2020 at 18:39

Assume that whenever $$y\in Y$$ and $$U\subseteq X$$ is an open set with $$f^{-1}(y)\subseteq U$$, then there exists an open set $$V_y\subseteq Y$$ with $$y\in V_y$$ such that $$f^{-1}(V_y)\subseteq U$$.
Suppose $$C\subseteq X$$ is closed. We will show that $$f(C)$$ is closed by showing that its complement $$Y\setminus f(C)$$ is open. So let $$y\in Y\setminus f(C)$$. Then $$f^{-1}(y)\subseteq X\setminus C$$. Note that $$X\setminus C$$ is open, so by assumption there is $$V_y\subseteq Y$$ such that $$f^{-1}(V_y)\subseteq X\setminus C$$. Necessary $$V_y\subseteq Y\setminus f(C)$$. We conclude that $$Y\setminus f(C)$$ is open, and so that $$f$$ is a closed map.
Now assume that $$f:X\rightarrow Y$$ is a closed map. Let $$y\in Y$$ and suppose that $$f^{-1}(y)\subseteq U$$, where $$U\subseteq X$$ is open. Then $$X\setminus U$$ is closed, and thus also $$f(X\setminus U)$$, since $$f$$ is a closed map. Put $$V_y=Y\setminus f(X\setminus U)$$ and notice that $$f^{-1}(V_y)\subseteq U$$.