Image of a normal space under a closed and continuous map is normal

Question

$p : X \to Y$ 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?

Answer 1

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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I would be tempted to attack this problem in a slightly 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.

Answer 2

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.

Answer 3

Here is a complete proof:

First, we will show that singletons in $Y$ are closed. Take any arbitrary $y \in Y$. Since $p$ is surjective $p^{-1}(y)$ is not empty. Let $x \in p^{-1}(y)$, then $$p(\{x\})=\{y\}.$$ The set $\{x\}$ is closed by normality, and $p$ is closed. Thus, $\{y\}$ is closed, as desired.

Next, we will show that, for every pair of disjoint closed sets $C_1$ and $C_2$ in $Y$, there exist disjoint open sets $O_1 \supset C_1$ and $O_2 \supset C_2$.

By continuity of $p$, $p^{-1}(C_1)$ and $p^{-1}(C_2)$ are closed. They are also disjoint, and so there exists an open set $U_1 \supset p^{-1}(C_1)$ disjoint from $p^{-1}(C_2)$ by normality of $X$. Moreover, there exists an open set $V$ such that $$p^{-1}(C_1) \subset V \subset \bar{V} \subset U$$ again by the normality of $X$. The sets $X-V$ and $\bar{V}$ are closed and so are $p(X-V)$ and $p(\bar{V})$ since $p$ is closed. Consequently, $O_1:=Y-p(X-V)$ and $O_2:=Y-p(\bar{V})$ are open. We will show that $$\begin{cases} C_1 \subset O_1\\ C_2 \subset O_2\\O_1 \cap O_2 =\emptyset \end{cases}.$$ Note that, $p^{-1}(A)$ and $B$ being disjoint implies that $A$ and $p(B)$ are disjoint. Using this, we have $$p^{-1}(C_1) \cap (X-V)=\emptyset \Longrightarrow C_1 \cap p_1(X-V)=\emptyset \Longrightarrow C_1\subset O_1,$$ and $$p^{-1}(C_2) \cap \bar{V}=\emptyset \Longrightarrow C_2 \cap p(\bar{V})=\emptyset \Longrightarrow C_2 \subset O_2.$$ Finally, \begin{align}O_2&=Y-p(\bar{V})=p(X)-p(\bar{V}) \subset p(X)-p(V) \subset p(X-V) \\ \Longrightarrow O_2 &\cap\left( Y-p(X-V)\right) =O_2 \cap O_1=\emptyset.\end{align}