Solution check: Let $p:X\rightarrow Y$ be a closed continuous surjective map. Show that if $X$ is normal, then so is $Y$. Below is a question concerning Normal Spaces from Topology, James Munkres. Following that is my attempt at a solution, which I am not sure is correct and would appreciate if somebody could point out what (if anything) is wrong with it.
$\textbf{Question:}$ Let $p:X\rightarrow Y$ be a closed continuous surjective map. Show that if $X$ is normal, then so is $Y$.
$\textbf{Attempted solution:}$
Let $A$ and $B$ be any two closed sets of $Y$. Since $p$ is continuous, $C=p^{-1}(A)$ and $D=p^{-1}(B)$ are closed in $X$. Since $X$ is normal, there exist open sets $U_1$ and $U_2$ such that $C\subset U_1$ and $D\subset U_2$ and $U_1\cap U_2=\{\phi\}$.
Let $C_1=X-U_1$ and $C_2=X-U_2$. Then $C_1$ and $C_2$ are closed and since $p$ is a closed map, so are $p(C_1)$ and $p(C_2)$. Also $A\cap p(C_1)=\{\phi\}$ and $B\cap p(C_2)=\{\phi\}$ because if it wasn't so then $C_{1}\cap C$ and $C_{2}\cap B$ would be non-empty. Finally, let $V_1=Y-p(C_1)$ and $V_2=Y-p(C_2)$. Then $A\subset V_1$ and $B\subset V_2$ and $V_1$ are $V_2$ are disjoint because $V_1$ and $p(C_1)$ are disjoint and $V_2\subset p(U_2)\subset p(C_1)$.
 A: The argument seems correct, apart from a few slips:


*

*you should start with $A\cap B=\emptyset$;

*you seem to be confusing $\emptyset$ with $\{\emptyset\}$.


I'd make some passages clearer, in particular for showing where surjectivity is used.

Let $A$ and $B$ be disjoint closed subsets of $Y$. Then $p^{-1}(A)$ and $p^{-1}(B)$ are closed (by continuity of $p$) and disjoint (by general property of maps) subsets of $X$.
Since $X$ is normal, there are open sets $U$ and $V$ such that


*

*$p^{-1}(A)\subseteq U$

*$p^{-1}(B)\subseteq V$

*$U\cap V=\emptyset$


Since $p$ is closed, $p(X\setminus U)$ and $p(X\setminus V)$ are closed in $Y$. Let $U_1=Y\setminus p(X\setminus U)$ and $V_1=Y\setminus p(X\setminus V)$, which are open in $Y$.
Then
$$
U_1\cap V_1=Y\setminus(p(X\setminus V)\cup p(X\setminus U))
$$
Let's see that $p(X\setminus V)\cup p(X\setminus U)=Y$. If $y\in Y$, then $y=f(x)$ for some $x\in X$. Since $U\cap V=\emptyset$, we have either $x\in X\setminus U$ or $x\in X\setminus V$; so the thesis follows.
Therefore $U_1\cap V_1=\emptyset$.
Let $y\in A$. Suppose $y\notin U_1$; then $y\in p(X\setminus U)$, so $y=f(x)$ for some $x\in X\setminus U$. But this is impossible, because $x\in p^{-1}(A)\subseteq U$. Therefore $y\in U_1$.
Similarly, $B\subseteq V_1$.
A: Let $A$ be closed in $Y$ and $U$ be an open subset of $Y$ containing $A$. Then $f^{-1}(U)$ is an open set containing the closed set $f  ^{-1}(A)$. By normality of $X$, choose an open set $V$ in $X$, such that $A\subseteq f(V)$ and $f(\bar{V})\subseteq U$. Define $W=Y-f(X-V)\subseteq f(V)$, where $W$ is an open set containing $A$ and $\bar{W}\subseteq \overline{f(V)}=f(\bar{V})\subseteq U$, wherein equality follows since $f$ is closed and continuous. So, $Y$ is normal.
