Let $X$ be a topological space and $C \subseteq X$ closed. Give $X/C$ the quotient topology and suppose $X$ is normal. Then this implies that $X/C$ is normal.

Normal in my case is T1 and closed disjoint sets are contained in disjoint opens.

I think this should be quite easy to solve, however I am stuck on one part.

Here's my attempt: Suppose $D , F$ are closed disjoint sets in $X/C$. Then $\pi^{-1}(D)$ and $\pi^{-1}(F)$ are closed in $X$ since $\pi$ is continuous. Now, $X$ is normal so there exist disjoint opens $U$ and $V$ than contain $\pi^{-1}(D)$ and $\pi^{-1}(F)$ resp. Now does it hold that $\pi(U)$ and $\pi(V)$ are open in $X/C$, i.e. is $ \pi$ an open map? Does it also hold that $\pi(\pi^{-1}(D)) =D$ and $\pi(\pi^{-1}(F)) =F$? Then the rest of the proof is easy.

But I don't seem to find if $\pi$ is an open map. I think my last question should hold since $\pi$ is surjective, but I'm not confident that that is true.

  • 1
    $\begingroup$ Quotient maps may not be open. For instance, the quotient map of $[0,1]/\{0,1\}$ sends the open set $[0,1/2)$ to a non-open set. In order for $\pi(U)$ to be open, you need the open set $U$ to also be saturated with respect to the projection, i.e. that $\pi^{-1}(\pi(U))=U$. In this specific case, the condition is that $U\cap C=\varnothing$ or $U\supseteq C$. $\endgroup$ Jun 7 at 12:22
  • 1
    $\begingroup$ It's true that $\pi(\pi^{-1}(D))=D$: this is just the fact that $\pi$ is a surjective function. In general $f(f^{-1}(S))=S\cap\operatorname{im}f$ $\endgroup$ Jun 7 at 12:30
  • $\begingroup$ Does normal include Hausdorff? $\endgroup$
    – Paul Frost
    Jun 7 at 12:40
  • $\begingroup$ @PaulFrost yes. $\endgroup$
    – Mateo
    Jun 7 at 12:43
  • $\begingroup$ @Mateo The definition of normal is not standardized. See en.wikipedia.org/wiki/Separation_axiom and math.stackexchange.com/q/4459369. But let us assume it includes Hausdorff. $\endgroup$
    – Paul Frost
    Jun 7 at 13:13

2 Answers 2


Observe that the condition @Sassatelli mentioned, i.e. $\pi^{-1}(\pi(U_X))=U_X$ for $U_X\subset X$, is satisfied iff $U_X\cap C=\emptyset$ or $C\subset U_X$. Therefore I think the best way is to consider two cases.

Let's start with two closed sets $D,F$ in $X/C$ and let $D_X:=\pi^{-1}(D)$, $F_X:=\pi^{-1}(F)$. Of course $D_X,F_X$ are closed in $X$.

If neither of them contains $C$ then the sets $D_X$, $F_X$ and $C$ are closed and disjoint in $X$. Then there exist two disjoint sets $U_X,V_X$ that are open in $X$ that $D_X\subset U_X$, $F_X\subset V_X$ and moreover $U_X\cap C = V_X\cap C = \emptyset$ (important). Then $U:= \pi(U_X)$ and $V:= \pi(V_X)$ are open.

If one of them, say $D$, contains $C$ then we can find two disjoint sets $U_X,V_X$ that are open in $X$ that $D_X\subset U_X$, $F_X\subset V_X$. Then $U:= \pi(U_X)$ and $V:= \pi(V_X)$ are open.


Let us first show that all one-point subspaces of $X/C$ are closed.

Let $* \in X/C$ denote the point which is the common equaivalence class $[c]$ of the points of $C$. Then $\pi^{-1}(*) = C$ which shows that $\{*\}$ is closed. For any other point $[x] \in X/C$ we have $[x] = \{x\}$, thus $\pi^{-1}([x]) = \{x\}$, which shows that $\{[x] \}$ is closed.

Now let $D, F$ be closed disjoint subsets of $X/C$. We want to find disjoint open neigborhoods $U$ of $D$ and $V$ of $F$ in $X/C$. W.l.o.g. we may assume that one $D, F$ (say $D$) contains $*$: If none of $D, F$ contains $*$, then $D \cup \{*\}$ and $F$ are closed disjoint subsets of $X/C$, and if we can separate them by disjoint open neigborhoods, these of course also separate $D, F$.

$D' = \pi^{-1}(D)$ and $F' = \pi^{-1}(F)$ are disjoint closed subsets of $X$, where $C \subset D'$. There exist disjoint open neigborhoods $U'$ of $D'$ and $V'$ of $F'$ in $X$. Then $D \subset U = \pi(U')$ and $F \subset V = \pi(V')$. The sets $U,V$ are disjoint because $C \subset U'$ and $V' \cap C = \emptyset$. Moreover they are open because $\pi^{-1}(U) =U'$ and $\pi^{-1}(V) = V'$.


I assumed that normal includes Hausdorff. This allows to show that $X/C$ is $T_1$ which together with separation of closed disjoint subsets by disjoint open neigborhoods implies Hausdorff.

If you do not require that normal includes Hausdorff, the above proof still works because at least $\{*\}$ is closed (since $C$ is closed).


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