This question already has an answer here:

I have this exercise:

Let $p: X \rightarrow Y$ be a closed continuous surjective map. Show that if X is normal, then so is Y. [Hint: If U is an open set containing $p^{-1}(\{y\})$, show there is a neighborhood W of y such that $p^{-1}(W)\subset U$.]

my attempt:

If I assume the hint is true. Let A, B be to closed sets disjoint in Y. $p^{-1}(A), p^{-1}(B)$ are two closed sets in X, the are also disjoint by elementary set-theory. Let x be an element of A or B, then the hints says that there is $W_x$, open, such that $p^{-1}(W_x)\subset p^{-1}(A)$, or $p^{-1}(W_x)\subset p^{-1}(B) $. Now, if $x \in A, x' \in B$, then $W_x \cap W_{x'}=\emptyset$, if not, if y is in both sets, then $p^{-1}(\{y\})$, intersects both $p^{-1}(A)$ and $p^{-1}(B)$, which it can't. Is this solution correct?

But now comes the real problem, and that is proving what the hint told us. So assume that we have a point Y, and that $p^{-1}(\{y\})\subset U$, U open in X. How do I go about in finding the open neighborhood W in Y? I still have not used that p is a closed map, so it seems I should use this.

Can you please help me?


marked as duplicate by user642796 Oct 21 '15 at 7:43

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


Your attempt to apply the hint is incorrect, I’m afraid. Let’s stick with an $x\in A$ for a moment: the hint does not say that there is an open nbhd $W_x$ of $x$ such that $p^{-1}[W_x]\subseteq p^{-1}[A]$, because $p^{-1}[A]$ is not (in general) an open set containing $p^{-1}[\{x\}]$. Notice also that you never used the hypothesis that $X$ is normal. What you should do is use the normality of $X$ to say that there are disjoint open sets $U$ and $V$ in $X$ such that $p^{-1}[A]\subseteq U$ and $p^{-1}[B]\subseteq V$. Now you can apply the hint: each $y\in A$ has an open nbhd $W_y$ such that $p^{-1}[W_y]\subseteq U$, and each $y\in B$ has an open nbhd $W_y$ such that $p^{-1}[W_y]\subseteq V$. Let $G=\bigcup_{y\in A}W_y$ and $H=\bigcup_{y\in B}W_y$; it’s not hard now to check that $G$ and $H$ are disjoint open nbhds of $A$ and $B$, respectively, and hence that $Y$ is normal.

To prove the hint you’ll need to use the fact that $p$ is a closed map. You have a point $y\in Y$ and an open set $U\subseteq X$ such that $p^{-1}[\{y\}]\subseteq U$, and you want to find an open nbhd $W$ of $y$ such that $p^{-1}[W]\subseteq U$. Here’s a further HINT: consider the set $Y\setminus p[X\setminus U]$.


Not the answer you're looking for? Browse other questions tagged or ask your own question.