Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$p : X \to Y$ 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?

share|improve this question

2 Answers 2

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.

share|improve this answer

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.

share|improve this answer
    
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
1  
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 at 6:53
1  
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 at 7:58

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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