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If $X$ is normal and $A \subset X$ is closed, then the quotient space $X/A$ is normal.

I am trying to show this statement. The idea I have is using the fact that if $p:X \to Y$ is a closed continuous surjective map, then if $X$ is normal then so is $Y$ (Exercise 31.6 from Munkres).

Following is my proof. I would like verification.

Proof. By the above result, we need only show that the quotient map is closed. Now the equivalence class is $A$ if $x \in A$ and $\{x\}$ is $x \notin A$. Let $B$ be a closed subset of $X$. To show that $p(B)$ is closed in $X/A$ we can show that $p^{-1} p (B)$ is closed in $X$.

$p(B)=p(B\cap A) \cup p(B-A) = \{A\} \cup \bigcup_{x \in B-A} \{x\}$, which we may simply denote $A \cup (B-A)$. But clearly $p^{-1}(A)=A$ and $p^{-1}(B-A) = B-A$. Hence $p^{-1}(p(B)) = A \cup B$, which is closed since $A$ and $B$ are both closed subsets of $X$. Thus $p$ is a closed map.

nomadicmathematician
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1 Answers1

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Indeed, the quotient map from identifying a closed set to a point is a closed map, essentially for the reasons laid out. If $B \subseteq X$ is closed then $p^{-1}[p[B]] = B$ when $A \cap B = \emptyset$ and $p^{-1}[p[B]] = A \cup B$ when $A \cap B \neq \emptyset$, which in both cases are closed subsets of $X$, so the definition of the quotient topology tells us that in both cases $p[B]$ is closed in $X/{A}$ as required. Then you can indeed reduce the proof to the (hopefully already proved) exercise... You should make the case distinction I made because $p^{-1}[p[B]]$ does not always equal $A \cup B$, as you saw.

Henno Brandsma
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