Here is Prob. 6, Sec. 31, in the book Topology by James R. Munkres, 2nd edition:

Let $p \colon X \to Y$ be a closed continuous surjective map. Show that if $X$ is normal, then so is $Y$. [Hint: If $U$ is an open set containing $p^{-1}(\{ y \} )$, show there is a neighborhood $W$ of $y$ such that $p^{-1}(W) \subset U$.]

My Attempt:

For any point $y\in Y$, we can find a point $x \in X$ such that $y = p(x)$, because $p$ is surjective. As $X$ is normal, so the set $\{ x \}$ is a closed set in $X$, and since $p$ is a closed map, the set $p(\{ x \}) = \{p(x) \} = \{y\}$ is a closed set in $Y$. Thus one-point sets are closed in $Y$.

Let $B$ be a closed set in $Y$, and let $U$ be an open set in $Y$ containing $B$. We need to find an open set $V$ in $Y$ such that $B \subset V$ and $\overline{V} \subset U$, by Lemma 31.1 in Munkres.

As $p \colon X \to Y$ is continuous and as sets $B$ and $U$ are, respectively, closed and open in $Y$, so the inverse images $p^{-1}(B)$ and $p^{-1} (U)$ are, respectively, closed and open in $X$.

And, as $B \subset U$, so we must also have $$ p^{-1}(B) \subset p^{-1}(U). \tag{0} $$

Thus $p^{-1}(B)$ is a closed set in $X$ and $p^{-1}(U)$ is an open set in $X$ such that $$p^{-1}(B) \subset p^{-1}(U).$$ So using Lemma 31.1 (b) in Munkres and normality of $X$, we can conclude that there exists an open set $W$ in $X$ such that $$ p^{-1}(B) \subset W \qquad \mbox{ and } \qquad \overline{W} \subset p^{-1}(U). \tag{A} $$

As $p \colon X \to Y$ is continuous, so we must have $$ p\left( \overline{W} \right) \subset \overline{ p(W) }, \tag{1} $$ by Theorem 18.1 (2) in Munkres.

On the other hand, as $W \subset \overline{W}$, so $$ p(W) \subset p \left( \overline{W} \right). $$ Moreover, as $p \colon X \to Y$ is a closed map and as $\overline{W}$ is a closed set in $X$, so the set $p \left( \overline{W} \right)$ is a closed set in $Y$. Thus $p \left( \overline{W} \right)$ is a closed set in $Y$ containing $p(W)$. So we must have $$ \overline{p(W)} \subset p \left( \overline{W} \right). \tag{2} $$

From (1) and (2) we obtain $$ p \left( \overline{W} \right) = \overline{p(W)}. \tag{3} $$

As $p^{-1}(B) \subset W$ and as $p$ is surjective, so $$ B = p\left( p^{-1}(B) \right) \subset p(W), $$ that is, $$ B \subset p(W). \tag{4} $$

As $\overline{W} \subset p^{-1}(U)$ and as $p$ is surjective, so $$ p \left( \overline{W} \right) \subset p \left( p^{-1}(U) \right) = U, $$ that is, $$ p \left( \overline{W} \right) \subset U. \tag{5} $$

Thus from (3), (4), and (5) we obtain $$ B \subset p(W) \subset p \left( \overline{W} \right) = \overline{p(W)} \subset U. \tag{6} $$

Now suppose that $y \in Y \setminus p(X \setminus W)$. Then $y \in Y$ and $y \not\in p(X \setminus W)$, and since $p \colon X \to Y$ is surjective, $y = p(x)$ for some point $x \in X$ such that $x \not\in X \setminus W$ and hence $x \in W$, which implies that $y \in p(W)$. Therefore $$ Y \setminus p(X \setminus W) \subset p(W). \tag{7a} $$

So what next? How to proceed from here? If $p$ were bijective or if $p$ were also an open map, then the proof would go through easily. But here $p$ is only surjective, and $p$ is closed.

How to use the hint given by Munkres?


Now as $p^{-1}(B) \subset W$ from (A) above and as $W \subset X$, so $$ p^{-1}(B) \cap (X\setminus W) = \emptyset. $$ So if $y \in B \cap p(X \setminus W)$, then $y = p(x)$ for some element $x \in X \setminus W$, and then $x \in p^{-1}(B) \cap (X \setminus W)$, which contradicts the fact that $p^{-1}(B)$ and $X \setminus W$ are disjoint. Therefore $$ B \cap p(X \setminus W) = \emptyset, $$ and this, together with the fact that $B \subset Y$, implies that $$ B \subset Y \setminus p(X \setminus W). \tag{7b}$$

From (7a) and (7b) abbove we obtain $$ B \subset Y \setminus p(X \setminus W) \subset p(W). \tag{7} $$

Now as $W$ is an open set in $X$, so $X \setminus W$ is closed in $X$, and since $p \colon X \to Y$ is a closed map, the image set $p(X \setminus W)$ is a closed set in $Y$. Therefore the set $Y \setminus p(X \setminus W)$ is an open set in $Y$. Let us put $$ V \colon= Y \setminus p(X \setminus W). $$ Then $V$ is an open set in $Y$, and (7) becomes $$ B \subset V \subset p(W), $$ and this together with (6) yields $$ B \subset V \subset p(W) \subset \overline{p(W)} \subset U. \tag{8} $$

Now since $V \subset p(W)$, therefore we have $$ \overline{V} \subset \overline{p(W)},$$ and this together with (8) gives $$ B \subset V \subset \overline{V} \subset U.$$

Thus we have shown that

(1) one-point sets in $Y$ are closed, and

(2) for any closed set $B$ in $Y$ and for any open set $U$ in $Y$ such that $B \subset U$, there exists an open set $V$ in $Y$ such that $B \subset V$ and $\overline{V} \subset U$.

Hence by Lemma 31.1 (b) in Munkres $Y$ is normal.

Is this proof sound enough now? If so, is my reasoning clear enough too? Or are there still any lacks and gaps left?


The most straightforward way is just to take the bull by the horns and go direct:

That $Y$ is $T_1$ is, as you noticed, simply a consequence of closedness plus ontoness: if $p(x)=y$ then $\{y\}= p[\{x\}]$, the image of a closed set in $X$.

If $C$ and $D$ are closed and disjoint in $Y$, $C'=f^{-1}[C]$ and $D'=f^{-1}[D]$ are closed (by continuity) and disjoint in $X$ so have disjoint open neighbourhoods $C'\subseteq U'$ and $D' \subseteq V'$. Then by closedness of $f$, $U= Y\setminus f[X\setminus U']$ and $V= Y\setminus f[X\setminus V']$ are open in $Y$ and simple set theory shows that $U \cap V=\emptyset$ and $C \subseteq U$ and $D \subseteq V$, as required.

Cf. my answer here or my proof of Munkres' hint here for inspiration.

IMHO you make it way too complicated...

Added Some more explanation of the arguments involved:

  • $U \cap V=\emptyset$: suppose not, then we have $y \in U \cap V$. As $f$ is surjective we write $y=f(x)$ for some $x \in X$. If $x \in X\setminus U'$ this would imply $f(x) \in f[X\setminus U']$ so $f(x) \notin U$, contradiction. So $x \notin X\setminus U'$ or $x \in U'$. Entirely similar is the argument that $x \in V'$, but then we contradict $U'\cap V'=\emptyset$. This final contradiction shows that $U$ and $V$ are indeed disjoint. We could also applied de Morgan and ontoness for a more "algebraic" proof:

$$U \cap V = \left(Y\setminus f[X\setminus U']\right) \cap \left(Y\setminus f[X\setminus V']\right) = \\ Y\setminus \left(f[X\setminus U'] \cup f[X\setminus V']\right)= Y\setminus f[(X\setminus U') \cup (X\setminus V')]=\\ Y\setminus f[X\setminus (U' \cap V')] = Y\setminus f[X\setminus \emptyset]= \emptyset$$

  • $C \subseteq U$: Take $y \in C$ and assume $y \notin U$. This means that $y \in f[X\setminus U']$, so that $y=f(x)$ for some $x \in X\setminus U']$, or $x \notin U'$. But as $y=f(x) \in C$, $x \in C' = f^{-1}[C]$ and so $x$ would contradict the inclusion $C'\subseteq U'$. So $y \in U$ as required. $D \subseteq V$ is entirely similar again.

Hope this makes the total argument clearer.

  • $\begingroup$ your proof is just beautiful, and I admit that my approach is a bit too convoluted. But can you please also expand your post by including an explicit demonstration of the inclusion and disjointness assertions toward the end? $\endgroup$ – Saaqib Mahmood Jun 19 at 4:38
  • $\begingroup$ @SaaqibMahmood later today, time permitting. $\endgroup$ – Henno Brandsma Jun 19 at 4:39
  • $\begingroup$ thanks. I've just edited your original post. Do you approve of my doing so? $\endgroup$ – Saaqib Mahmood Jun 19 at 4:40
  • $\begingroup$ @SaaqibMahmood it’s fine by me. I typically use the stand out paragraph format for theorems or lemmas, but if you like this better, fine. $\endgroup$ – Henno Brandsma Jun 19 at 4:55
  • $\begingroup$ @SaaqibMahmood I've added some explanations. $\endgroup$ – Henno Brandsma Jun 20 at 22:13

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