Validity in choosing open sets in proof

Question

I was taking a look at this post of the problem $f:X \rightarrow Y$ is continuous, closed surjective, and $X$ is normal, show $Y$ is normal. Here is a link to the post Image of a normal space under a closed and continuous map is normal

In particular, this concise solution provided by a user: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.

Several people mentioned in the comments, for what reason are we able to choose the neighborhoods $V_1,V_2$ in this proof? The user did not seems to address the situation, and I still do not understand it.

So, for what reason are $V_1,V_2$ able to be chosen in the manner in which they are in this user's proof?

My guess is that he is using the alternate definition of normality:For any closed set $A$ and open set $U$ containing it, there is another open set $V$ containing $A$ with $\bar V \subset U$.If not, can someone explain this step of the proof?

Answer 1

Let $W_1=X\setminus f^{-1}[f[X\setminus U_1]]$; $X\setminus U_1$ is closed, so $f[X\setminus U_1]$ is closed, and hence $W_1$ is open. Moreover, $f^{-1}[A]\cap(X\setminus U_1)=\varnothing$, so $A\cap f[X\setminus U_1]=\varnothing$, and therefore $f^{-1}[A]\cap f^{-1}[f[X\setminus U_1]]=\varnothing$, i.e., $f^{-1}[A]\subseteq W_1$. Next,

$$X\setminus W_1=f^{-1}[f[X\setminus U_1]]\supseteq X\setminus U_1\,,$$

so $W_1\subseteq U_1$. Similarly, if $W_2=X\setminus f^{-1}[f[X\setminus U_2]]$, then $W_2$ is an open nbhd of $f^{-1}[B]$ contained in $U_2$. Clearly $W_1\cap W_2=\varnothing$.

Let $V_i=f[W_i]$ for $i=1,2$. $X\setminus U_i$ is closed for $i=1,2$, so

$$V_i=f[X\setminus f^{-1}[f[X\setminus U_i]]]=Y\setminus f[X\setminus U_i]$$

is open for $i=1,2$. Finally, $W_i=f^{-1}[f[W_i]]=f^{-1}[V_i]$ for $i=1,2$, so

$$f^{-1}[V_1\cap V_2]=f^{-1}[V_1]\cap f^{-1}[V_2]=W_1\cap W_2=\varnothing\,,$$

and it follows that $V_1\cap V_2=\varnothing$.

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Say that a subset $A$ of $X$ is saturated if $f^{-1}[f[A]]=A$, i.e., if $A$ is a union of fibres of $f$. What I’ve done is shrink the open sets $U_1$ and $U_2$ to saturated open nbhds $W_1$ and $W_2$ of $f^{-1}[A]$ and $f^{-1}[B]$, respectively, and used the fact that since $f$ is closed, the image under $f$ of a saturated open set $W$ is open: in that special case

$$f[W]=Y\setminus f[X\setminus W]\,,$$

the complement of a closed set.

Answer 2

If $f: X \to Y$ is a closed map, then it obeys a sort of "reverse continuity property"

$$\forall B \subseteq Y: \forall U \subseteq X \text{ open with } f^{-1}[B] \subset U: \exists V \subseteq Y \text{ open such that } B \subseteq V \text{ and } f^{-1}[V] \subseteq U\tag{1}$$

Proof: Define $V=Y\setminus f[X\setminus U]$ which is open precisely because $f$ is closed. Let's check that it works:

$B \subseteq V$: suppose $y \in B$ and $y \notin V$. The latter implies that $y \in f[X\setminus U]$, so $y=f(x)$ for some $x \in X\setminus U$ so $y=f(x) \notin U$,while $y \in B$ implies $x \in f^{-1}[B]$ so $x \in U$. This is a contradiction, so $B \subseteq V$ holds.

$f^{-1}[V] \subseteq U$: suppose $x \in f^{-1}[V]$ and $x \notin U$. The former tells us $f(x) \in V$ and the latter that $f(x) \in f[X\setminus U] = Y\setminus V$, a blatant contradiction. QED

Note that ontoness of $f$ is not even used, just it being closed.

Now if $X$ is normal and $f: X \to Y$ is both closed, continuous and onto (so $f[X]=Y$) we know $Y$ is normal: if $C,D$ are closed and disjoint in $Y$ the same holds for $f^{-1}[C]$ and $f^{-1}[D]$ in $X$; here continuity is used to conclude the sets are closed, disjointness holds for any map. Then apply $(1)$ to the open (disjoint) sets in $X$ (say $U_A, U_B$) that separate them and we get $V_A,V_B$ in $Y$ open so that

$$C \subseteq V_C, D \subseteq V_D, f^{-1}[V_C] \subseteq U_C, f^{-1}[V_D] \subseteq U_D\tag{2}$$

Now $V_C$ and $V_D$ are as required: if they intersected in some $y$, let $x \in X$ be such that $f(x)=y$ (here we finally use the surjectivity!) but then $x \in f^{-1}[V_C]$ so $x \in U_C$ and similarly $x \in U_D$, contradiction by how these sets were chosen..

This way of writing up the proof makes where we use which assumption on $f$ a bit clearer IMO.