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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 $y$ such that $p^{-1}(W) \subset U$.

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?

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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, $y \in Y \setminus p [ X \setminus U ] = W$.

I would be tempted to attack this problem in a alightly different manner, noting 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 $E \cup F = X$.

So let $U,V \subseteq Y$ be open sets such that $U \cup V = Y$. Then by continuity of $f$, $f^{-1}[U], f^{-1}[V]$ are open subsets of $X$, and $f^{-1}[U] \cup f^{-1}[V] = X$. As $X$ is normal the condition above implies that there are closed $E \subseteq f^{-1}[U]$ and $F \subseteq f^{-1}[V]$ such that $E \cup F = X$. It is easy to check that $f[E] \subseteq U$ and $f[F] \subseteq V$. As $f$ is a closed mapping, then $f[E],f[F]$ are closed subsets of $Y$, and by the surjectivity of $f$ it follows that $f[E] \cup f[F] = Y$. Thus $f[E],f[F]$ are as required.

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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
I think dREaM means why can we take neighborhoods $V_i$ such that $p^{-1}(V_i)\subset U_i$, and I think it requires an additional argument, because of examples like the following: Take $U_1=[0,1)$, $U_2=(0,1]$, $Y=[0,1]$. We have obvious inclusions $\eta_i:U_i\to Y$, so take $X=U_1\sqcup U_2$, and $p=\eta_1\sqcup\eta_2$. Finally take $A=\{0\}$ and $B=\{1\}$, then everything else in your proof works just fine, except no such $V_i$ could be chosen, even if $Y$ is normal, and (I think) $p$ is a closed. – Ruian Chen Aug 25 '15 at 6:53
No, my $p$ is clearly not closed. I guess I am just trying to point out that, unlike Arthur Fischer's argument that brings open sets around, the argument that deals with closed sets requires some caution, as one no longer has $p(U_1\cap U_2)=p(U_1)\cap p(U_2)$. – Ruian Chen Aug 25 '15 at 7:58

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