# Prob. 6, Sec. 31, in Munkres' TOPOLOGY, 2nd ed: The image of a normal space under a closed continuous map is also normal

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?

PS:

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?

• – Basanta R Pahari Jun 18 at 18:41
• Without the hint (but it sort of uses the idea: this answer) e.g. – Henno Brandsma Jun 18 at 21:24
• @BasantaRPahari. Yes. Much simpler proof. – DanielWainfleet Jun 18 at 21:27
• The hint I show here, in case you're interested. – Henno Brandsma Jun 18 at 21:28

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.

• 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? – Saaqib Mahmood Jun 19 at 4:38
• @SaaqibMahmood later today, time permitting. – Henno Brandsma Jun 19 at 4:39
• thanks. I've just edited your original post. Do you approve of my doing so? – Saaqib Mahmood Jun 19 at 4:40
• @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. – Henno Brandsma Jun 19 at 4:55
• @SaaqibMahmood I've added some explanations. – Henno Brandsma Jun 20 at 22:13