Let $p \colon X \rightarrow Y$ be a closed continuous surjective map such that $p^{-1}\big(\{ y \} \big)$ is compact for each $y \in Y$. (Such a map is called a perfect map.) (a) Show that if $X$ is Hausdorff, then so is $Y$.
My attempt: It’s easy to check, normal $\Rightarrow$ regular $ \Rightarrow$ $T_2$ $\Rightarrow$ $T_1$. By Exercise 6, Section 31 of Munkres’ Topology, $Y$ is $T_2$. Is this proof correct?
Edit: This solution is incorrect. Following are comments of Mr.Gandalf Sauron:
I guess you are trying to ask the following :- "Since Hausdorffness is preserved under a perfect map then is normality/regularity also preserved?" . The short answer is no. A map preserving weaker properties does not imply that it will preserve stronger properties. For example the map $f:\Bbb{Z}\to\Bbb{Z}$ given by $f(x)=2x$ is a group homomorphism but is not a ring homomorphism.
So although intuitively it might make sense to conclude as such but there is no logical pathway that lets you conclude that the preservation of Hausdorff property will imply preservation of Normality . Even if you look at the converse that "Since Normality is preserved, is Hausdorffness also automatically preserved?" then again the answer is that you have no logical pathway in connecting these two. You have to prove them separately .
The problem is that you are thinking as $N \subset R\subset H$ and as $p$ maps $H \to H$ does it imply that $R \to R$ and $N \to N$. That is true but it requires proof and is not true for general functions. Just because a morphism maps the class of all Hausdorff topological spaces $H \to H$. Does not mean that a particular subclass $R$ of $H$ need be mapped to $R$ itself and nor so for $N \to N$. Again conversely it is not true that as $N$ is mapped to $N$ that $R$ will be mapped to $R$ or that $H$ will be mapped to $H$. Here $N$ and $R$ are not subsets but are subclasses of the class of all Hausdorff topological spaces . $H$ denotes Hausdorff, $N$ is normal and $R$ is regular.
Best Answer
It's unclear to me why would you start with a condition that is not given in the question. However yes, if $X$ was regular then $Y$ is regular . This will allow you to conclude Hausdorffness if you go by Munkres' definition of Regular which is a $T_{1}$ and $T_{3}$ space.
For a proper proof you need to go like this:-
Let $y_{1},y_{2}\in Y$ . Then $K_{1}=p^{-1}(y_{1})$ and $K_{2}=p^{-1}(y_{2})$ are compact sets and since we are in a Hausdorff space they are also closed.
Let $x\in K_{1}$ and $x'\in K_{2}$ . There exists $U_{x,x'}$ and $V_{x,x'}$ such that they are disjoint open sets containing $x$ and $x'$ respectively.
Then consider the open cover of $K_{1}$ by $\{U_{x,x'}\}_{x\in K_{1}}$ . This has a finite subcover $\{U_{x_{1},x'},...,U_{x_{n},x'}\}$ . Correspondingly we have the collection $\{V_{x_{1},x'},...,V_{x_{n},x'}\}$ . Let $\displaystyle U_{x'}=\bigcup_{j=1}^{n}U_{x_{j},x'}$ and $\displaystyle V_{x'}=\bigcap_{j=1}^{n}V_{x_{j},x'}$.
Then for each $x'\in K_{2}$ we have open sets $U_{x'}$ and $V_{x'}$ such that $x\in V_{x'}$ , $K_{1}\subset U_{x'}$ and $U_{x'}\cap V_{x'}=\phi$ .
Thus we consider the cover of $K_{2}$ by $\{V_{x'}\}_{x'\in K_{2}}$ . This has a finite subcover $V_{1},...,V_{m}$ and correspondily open sets $U_{1},...,U_{m}$. Then if we define $\displaystyle V=\bigcup_{i=1}^{m}V_{i}$ and $\displaystyle U=\bigcap_{i=1}^{m}U_{i}$ . Then $U$ and $V$ are open sets such that $K_{1}\subset U$ and $K_{2}\subset V$ and $U\cap V=\phi $.
Now we take the open sets $ W=Y\setminus p(X\setminus U)$ and $W'=Y\setminus p(X\setminus V)$. Then $y_{1}\in W$ and $y_{2}\in W'$ and $W\cap W'=\phi$ .
This proves Hausdorffness of $Y$.