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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).

Thanks in advance for your help!

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  • $\begingroup$ 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). $\endgroup$
    – Tyrone
    Commented Oct 30, 2020 at 18:13
  • $\begingroup$ 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 $f^{-1}(y)$) $\endgroup$
    – Tyrone
    Commented Oct 30, 2020 at 18:18
  • $\begingroup$ 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. $\endgroup$
    – Brian M. Scott
    Commented Oct 30, 2020 at 18:18
  • $\begingroup$ @Tyrone Thank you for your hints. Let me see if I can take it from there. $\endgroup$
    – user775075
    Commented Oct 30, 2020 at 18:21
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    $\begingroup$ @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 $W_y$? $\endgroup$
    – user775075
    Commented Oct 30, 2020 at 18:39

1 Answer 1

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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 $f^{-1}(V_y)\subseteq U$.

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  • $\begingroup$ Thank you very much! $\endgroup$
    – user775075
    Commented Nov 2, 2020 at 3:50

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