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 $V_2\subset p(U_2)\subset p(C_1)$.

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    $\begingroup$ $A$ and $B$ should be disjoint. $\{\emptyset\}$ is different from $\emptyset$. Apart from this, it seems aa bit confused in the wording, but essentially correct. $\endgroup$ – egreg Jan 12 '15 at 0:10

The argument seems correct, apart from a few slips:

  1. you should start with $A\cap B=\emptyset$;
  2. 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

  1. $p^{-1}(A)\subseteq U$
  2. $p^{-1}(B)\subseteq V$
  3. $U\cap V=\emptyset$

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, $B\subseteq V_1$.

  • $\begingroup$ I do not understand last lines.. Let $y\in A$, as $f$ is surjective there exists $x\in X$ such that $y=p(x)$.. so, $x\in p^{-1}(y)\subset p^{-1}(A)\subset U$.. So, $x\in U$ which implies $x\in V^c$... so, $f(x)\in f(V^c)$ i.e., $y\in f(V^c)$... I do not understand how did you get $y\in U_1$ $\endgroup$ – user312648 Mar 12 '16 at 11:44
  • $\begingroup$ I must have been missing some simple thing... No idea what is that $\endgroup$ – user312648 Mar 12 '16 at 11:51
  • $\begingroup$ @cello I seem to have mixed up $U$ and $V$. It should be fixed, now. $\endgroup$ – egreg Mar 12 '16 at 12:11
  • $\begingroup$ To simplify a little more notice that $f(A\cup B)\subseteq f(A)\cup f(B)$. $\endgroup$ – Singh Mar 12 '16 at 13:50
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    $\begingroup$ @ZFR Since $X$ is normal, $\{x\}$ is closed; since $f$ is closed also $\{f(x)\}$ is closed. $\endgroup$ – egreg Dec 10 '18 at 21:35

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.


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