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 $f^{-1}(y)\in V$.
Prove that $f$ is closed if and only if: exist $V \subset Y$ open set such that $y \in V$ and $f^{-1}(y)\in V$.
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general-topology
applications
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0$f^{-1}(y)$ is a subset of $X$, so I guess you mean $f^{-1}(y)\subset U$. Now $V$ is a subset of $Y$, so how is $f^{-1}(y)\subset V$? – 2017-01-16
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0$f^{-1}(y) \in V$ makes no sense, $f^{-1}(y)$, more properly denoted as $f^{-1}[\{y\}]$, is a subset of $X$, not a member of some subset of $Y$. I gave a correct formulation in my answer. – 2017-01-17
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The correct formulation should be : $f$ is closed iff for all $y \in Y$ and for every $U \subseteq X$ open such that $f^{-1}[\{y\}] \subseteq U$, there exists some open $V \subseteq Y$ 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 $U$ with $f^{-1}[\{y\}] \subseteq U$, we see that $X \setminus U$ is closed in $X$ and so $f[X\setminus U]$ is closed in $Y$, and we can define $V = Y \setminus f[X \setminus U]$, which is then open.
Then $y \in V$, because if $y \notin V$, this would mean $y \in f[X\setminus U]$ so $y=f(p)$ for some $p \in X \setminus U$, but this cannot be ($p \in f^{-1}[\{y\}]$ and $p \notin U$). Also $f^{-1}[V] \subseteq U$: if $p \in f^{-1}[V]$, then $f(p) \in V$, so $f(p) \notin f[X \setminus U]$, 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 $f^{-1}[\{y\}] \subseteq U := X\setminus C$ and so we have $V$ an open neighbourhood $V$ of $y$ such that $f^{-1}[V] \subseteq X \setminus C$, 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.