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Let $X$ and $Y$ be two topological spaces and $f:X \longrightarrow Y$ continuous, surjective and closed (i.e. it send closed sets in $X$ to closed sets in $Y$). How can I prove that if $X$ is normal, that $Y$ is normal too? So far I have the following:

Let $E,F$ be closed, disjoint sets in $Y$. Since $f$ is continuous, $f^{-1}(E)$ and $f^{-1}(F)$ are closed too, and since $X$ is normal, there are two disjoint open sets $U,V \subset X$, such that $f^{-1}(E) \subset U$ and $f^{-1}(F) \subset V$.

Now let be $U' := f(U^c)^c$ and $V' := f(V^c)^c$. Then $U',V'$ are open (as $f$ is closed), and it's easy to show that $U' \cap V' = \emptyset$. Now I need to show that $E \subset U'$ and $F \subset V'$.

Let be $y \in E \Longrightarrow \exists x \in f^{-1}(E): f(x) = y$ (since $f$ is surjective). Now I had to approaches

$(i)$ Then $x \in U \Longrightarrow x \in V^c \Longrightarrow y = f(x) \in f(V^c) = V'^c$. And then I'm stuck.

$(ii)$ Then $x \in U \Longrightarrow x \notin U^c \Longrightarrow y = f(x) \notin f(U^c) \Longrightarrow y \in f(U^c)^c = U'$. But that one is wrong, , I noticed, since $x \notin U^c \Longrightarrow y = f(x) \notin f(U^c)$ is not necessarily right.

I think the approach should be right, but I can't conclude. Any hints?

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  • $\begingroup$ You mean by that $U', V'$ are open. $\endgroup$ – Asaf Karagila Jan 24 '16 at 0:17
  • $\begingroup$ Yes, sorry, they're open. Otherwise the proof would be finished. $\endgroup$ – K.A. Jan 24 '16 at 0:18
  • $\begingroup$ See also here. $\endgroup$ – Jendrik Stelzner Jan 24 '16 at 0:34
  • $\begingroup$ How we can show $U' \cap V' $ is empty $\endgroup$ – Believer Apr 28 '19 at 11:45
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Let $y\in E$, and suppose that $y\notin U'$. Then $y\in Y\setminus U'=f[X\setminus U]$, so there is an $x\in X\setminus U$ such that $f(x)=y$. But then $x\in f^{-1}[E]\subseteq U$, which is a contradiction.

I suspect that what you were missing is the fact that $f^{-1}[E]$ contains every point of $X$ that maps to $E$.

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For every $x \notin U$ we have in particular $x \notin f^{-1}(E)$ and therefore $f(x) \notin E$. So $f(U^c) \cap E = \emptyset$ and therefore $E \subseteq f(U^c)^c = U'$. Similarly $V \subseteq f(V^c)^c$.

Notice that we need the surjectivity to conclude that $U' \cap V' = \emptyset$.

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  • $\begingroup$ How to prove that $U^/ /cap V^/ = /phi$? $\endgroup$ – reflexive Aug 15 '18 at 4:35

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