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$p : X \to Y$ is continuous closed and surjective, and $X$ is a normal space. Show $Y$ is normal.

There is a hint, which I'm trying to prove: show that if $U$ is open in $X$ and $p^{-1}(\{y\}) \subset U$, $y \in Y$, then there is a neighbourhood W of y s.t. $p^{-1}(W) \subset U$.

I have a candidate for W namely $W=Y\setminus p(X \setminus U)$. I did prove that this W is open, and that $p^{-1}(W) \subset U$, but I don't see how $y \in W$...I think this would require injectivity of p...

I have also shown that $y \in p(U)$ and that $W \subset p(U)$, so if also $W \supset p(U)$, then $y \in W$.

Can anyone help me? Thank you in advance.

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2 Answers 2

I'm actually not so sure how helpful the hint you've been given is. I think the following might get you to the proof quite quickly.

Hint: Note that essentially by de Morgan's Laws, normality of a topological space $X$ is equivalent to the following: given open $U , V \subseteq X$ such that $U \cup V = X$ there are closed $E \subseteq U$ and $F \subseteq V$ such that $E \cup F = X$.

As for the hint you have been given, you have given the correct set $W$. Note that as $p^{-1} \{ y \} \subseteq U$, then $p(x) \neq y$ for all $x \in X \setminus U$, which implies that $y \notin p [ X \setminus U ]$, or, equivalently, $y \in Y \setminus p [ X \setminus U ] = W$.

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Thanks. I failed to use that $p^{-1}(\{y\}) \subset U$. –  fuente Jan 12 '13 at 15:16
Deleted. Misprint. –  fuente Jan 12 '13 at 15:18
@fuente: It was no problem to flesh out the last remaining detail for the hint. –  Arthur Fischer Jan 12 '13 at 15:21
No, apparently not :) –  fuente Jan 12 '13 at 15:27

Why go through all that hassle, when you can do as follows:

Let $p: X \rightarrow Y$ be a closed, continuous surjection. Now let $A,B$ be two disjoint closed subsets of $Y$. Because $X$ is normal, we can separate the closed disjoint sets $p^{-1}(A), p^{-1}(B)$ in $X$ by respective neighborhoods $U_1, U_2$. Now choose neighborhoods $V_1$ of $A$, and $V_2$ of $B$ s.t. $p^{-1}(V_1) \subset U_1$, and $p^{-1}(V_2) \subset U_2$. Then it follows that $V_1, V_2$ are disjoint. Hence, $Y$ is normal.

Note that in general, a continuous image of a normal space is not necessarily normal.

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This in fact means that the image of a Hausdorff space under a closed, continuous surjection is Hausdorff. This can be proved easily: just replace your closed subsets with points instead. –  Libertron Jan 12 '13 at 19:34
does it work same for images of T4 space under closed continuous map is T4 –  math Feb 21 '13 at 17:01
Yes, I believe it does. This sounds like a good exercise. Haha, actually a $T_4$-space is normal and Hausdorff, so of course it should work! –  Libertron Apr 11 '13 at 21:24

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