Chapter 11, Theorem 5.2 (2) of James Dugundji Topology

Question

Let $p:X \to Y$ be a perfect map. Then: $(2)$ If $X$ is regular, so also is $Y$.

Dugundji’s proof:

Given $y\in U$ in $Y$, there is by 1.5(b) an open $V \subseteq X$ with $p^{-1}(y) \subseteq V\subseteq \overline{V} \subseteq p^{-1} (U)$. Since $p$ is a closed map, we find a nbhd $W \supseteq y$ with $p^{-1}(y) \subseteq p^{-1}(W) \subseteq V$; then $W\subseteq p(\overline{V}) \subseteq U$, and since $p(\overline{V})$ is closed, $y\in W\subseteq \overline{W} \subseteq p(\overline{V}) \subseteq U$.

Question: I don’t understand first line of proof, specifically $p^{-1}(y) \subseteq V$. In second line, $W\subseteq p(\overline{V})$.

Edit: let $\{y\}$ be a singleton set in $Y$. Since $p$ is surjective, $\exists x\in X$ such that $p(x)=y$. Since $X$ is $T_1$, $\{x\}$ is closed in $X$. $p$ is a closed map. So $p(\{x\})=\{y\}$ is closed in $Y$. Hence $Y$ is $T_1$.

Answer 1

$p^{-1}(y)$ is compact, $p^{-1}(U)$ is open and $p^{-1}(y)\subseteq p^{-1}(U)$. In regular space these properties imply the existence of an open set $V$ such that $p^{-1}(y)\subseteq V \subseteq \overline V \subseteq p^{-1}(U)$. [This is 1.5 (c) in the book].

Since $p$ is surjective, $p^{-1}(W) \subseteq V$ implies $W \subseteq p(V)\subseteq p(\overline V)$.

Answer 2

Here is a separate proof:-

Take a point $y\in Y$ and a closed set $C$ in $Y$. Then let $p^{-1}(y)=K$ and $p^{-1}(C)=F$ .

Then $K$ is a compact set and $F$ is a closed set such that $K$ and $F$ are disjoint.

We pick a point $x\in K$ and as $X$ is regular we can find disjoint open sets $U_{x}$ and $V_{x}$ such that $x\in U_{x}$ and $F\subset V_{x}$ .

Then we consider the cover of $K$ by open sets $\{U_{x}\}_{x\in K}$ .

As $K$ is compact there exist a finite subcover $U_{x_1},....,U_{x_n}$ of $K$.

and correspondingly open sets $V_{x_1},...,V_{x_n}$ such that $F\subset V_{x_i}$.

Then $\displaystyle\bigcup_{i=1}^{n}U_{x_i}$ is an open set containing $K$ and $\displaystyle\bigcap_{i=1}^{n}V_{x_i}$ is an open set containing $F$ .

We claim that they are disjoint.

if $x\in\displaystyle\bigcup_{i=1}^{n}U_{x_i}$ then $x\in U_{x_{i_{0}}}$ for some $i_{0}$ . Then it means that $x\notin V_{x_{i_{0}}}\implies x\notin \displaystyle\bigcap_{i=1}^{n}V_{x_i}$ .

Thus we have an open set $U=\displaystyle\bigcup_{i=1}^{n}U_{x_i}$ and an open set $V=\bigcap_{i=1}^{n}V_{x_i}$ .

Then $y\in p(U)$ and $C\subset p(V)$.

Now the set $X\setminus U$ is closed and hence $p(X\setminus U)$ is closed and henc e $Y\setminus p(X\setminus U)$ is open . and $y\notin p(X\setminus U)$ as $p^{-1}(y)\subset U$ . hence $y\in Y\setminus(p(X\setminus U))$.

And consider the open set $Y\setminus(p(X\setminus V))$ . Then we claim that $C\subseteq Y\setminus(p(X\setminus V))$ .

Let $z\in C$ and if possible let $z\in p(X\setminus V)$ . Then $z=p(x)$ for some $x\in X\setminus V$. But $p^{-1}(C)\subset V$ implies that $x\in V$ which is a contradiction.

Then we also claim that $\hat{U}= Y\setminus(p(X\setminus U))$ and $\hat{V}=Y\setminus(p(X\setminus V))$ are disjoint .

Let $w\in\hat{U}\implies w\notin p(X\setminus U)\implies w\in p(U)\subset p(X\setminus V)\implies w\notin\hat{V}$

Thus $\hat{U}$ and $\hat{V}$ are disjoint open sets such that $y\in\hat{U}$ and $C\subset\hat{V}$.

I should also mention that some books take the definition of Regular to be a space satisfying $T_{1}$ and $T_{3}$. In that case you have that since $p$ is a closed map and $\{x\}\subset X$ is closed , you have $\{p(x)\}$ is closed. It now follows that $Y$ is $T_{1}$ as $p$ is a surjection.