# Exercise 12, Section 26 of Munkres’ Topology

Let $$p:X\to Y$$ be a closed continuous surjective map such that $$p^{-1}(\{y\})$$ is compact, for each $$y\in Y$$. (Such a map is called a perfect map) Show if $$Y$$ is compact, then $$X$$ is compact. [Hint: If $$U$$ is open set containing $$p^{-1}(\{y\})$$, there is a neighborhood $$W$$ of $$y$$ such that $$p^{-1}(W)$$ is contained in $$U$$]

My attempt: Inspired by James Dugundji, we first generalize hint:

Let $$p: X\to Y$$ be a closed map. If $$S\subseteq Y$$ and $$U\in \mathcal{N}_{p^{-1}(S)}$$, then $$\exists V\in \mathcal{N}_S$$ such that $$p^{-1}(V)\subseteq U$$.

Proof: $$p^{-1}(S)\subseteq U\in \mathcal{T}_X$$. $$X-U$$ is closed in $$X$$. Since $$p$$ is closed map, $$p(X-U)$$ is closed in $$Y$$. So, $$Y-p(X-U)\in \mathcal{T}_Y$$. Let $$V= Y-p(X-U)$$. We claim $$S\subseteq V$$. Assume towards contradiction, $$\exists s\in S$$ such that $$s\notin V=Y-p(X-U)$$. So $$s\in p(X-U)$$. $$\exists x\in X-U$$ such that $$p(x)=s\in S$$. Which implies $$x\in p^{-1}(S)=\{z \in X|p(z)\in S\} \subseteq U$$. $$x\in U \cap (X-U)$$. Thus we reach contradiction. Hence $$S\subseteq V$$. $$p^{-1}(V)=p^{-1}(Y-p(X-U))=X- p^{-1}(p(X-U))\subseteq U$$. Hence $$\exists V\in \mathcal{N}_S$$ such that $$p^{-1}(V) \subseteq U$$. This is precisely chapter 3 theorem 11.2 of Dugundji topology.

Let $$U=\{ U_\alpha| \alpha \in J\}$$ be an open cover of $$X$$. $$p^{-1}(y)$$ is compact, $$\forall y\in Y$$. $$U$$ is an open cover of $$p^{-1}(y)$$, $$\forall y\in Y$$. So $$\exists \{U_{y, 1}, …,U_{y,n_y}\}$$ finite subcover of $$U$$ in $$p^{-1}(y)$$. Let $$U_y =\bigcup_{i=1}^{n_y} U_{y,i}$$. By hint/chapter 3 theorem 11.2 of Dugundji topology, $$\forall y\in Y$$, $$\exists V_y \in \mathcal{N}_y$$ such that $$p^{-1}(V_y)\subseteq U_y$$. So $$\{V_y| y\in Y\}$$ is an open cover of $$Y$$. Since $$Y$$ is compact, $$\exists \{ V_{y_1},…,V_{y_n}\}$$ finite subcover of $$Y$$. $$\bigcup_{i=1}^n V_{y_i}=Y$$. So $$p^{-1}(\bigcup_{i=1}^n V_{y_i})=p^{-1}(Y)=X =\bigcup_{i=1}^n p^{-1}(V_{y_i}) \subseteq \bigcup_{i=1}^n U_{y_i}$$. Thus $$X=\bigcup_{i=1}^n U_{y_i}= \bigcup_{y\in \{y_1,..,y_n\}, i\in J_{n_y}} U_{y,i}$$. Hence $$\{ U_{y, i}| y\in \{y_1,..,y_n\}, i\in J_{n_y}\}$$ is finite subcover of $$U$$. Another way to write it, $$\{ U_{y_k ,i}| i\in J_{n_{y_k}}, k\in J_n\}$$. Is my proof correct?

Note: We only used $$p^{-1}(y)$$ is compact and $$p$$ is closed map conditions.

• I wasn’t able to solve this problem at the time of doing section 26. While I was doing problems on perfect maps(exercise 7, section 32), I recognize a pattern, which is using $p^{-1}(y)$ is compact and $p$ is closed(generalize hint) condition. May 14 at 13:34
• Here is slight variation of above proof. May 14 at 13:39
• I tried to answer :) , if you don’t know the fip property tell me and I will be more clear. May 14 at 13:56

The idea I think is the following:

$$X$$ is compact if and only if any family $$\{C_\alpha\}$$ of closed sets of $$X$$ with the f.i.p. Property admits $$\cap_\alpha C_\alpha\neq \emptyset$$.

Take a family of closed sets $$\{C_\alpha\}$$ with the f.I.p. Property on $$X$$. Then $$\{p(\cap_{\alpha\in I}C_\alpha)\}_ {|I|<\infty }$$ is a family of closed sets of $$Y$$ with the f.I.p. Property, so that $$\cap_{|I|<\infty}p(\cap_{\alpha\in I} C_\alpha)$$ is non-empty. Take a point $$y$$ in the intersection. Then $$\{\cap_{\alpha\in I}C_\alpha\cap p^{-1}(y)\}_{|I|<\infty}$$ has the f.I.p. Property over $$p^{-1}(y)$$.

In fact $$C_\alpha$$ is closed and so $$\cap_{|I|<\infty} C_\alpha$$ is closed too in $$X$$. By definition of the topology induced to a subset, we get that $$\cap_{|I|<\infty} C_\alpha \cap p^{-1}(y)$$ is closed in $$p^{-1}(y)$$. Thus $$\{\cap_{\alpha\in I}C_\alpha\cap p^{-1}(y)\}_{|I|<\infty}$$ is a family of closed sets of $$p^{-1}(y)$$. Now we talk about the fip property. Take a finite numbers of closed sets in the family $$(\cap_{\alpha\in I_1}C_\alpha)\cap p^{-1}(y), \dots, (\cap_{\alpha\in I_s}C_\alpha)\cap p^{-1}(y)$$. Then

$$\cap_{j=1}^s (\cap_{\alpha\in I_j}C_\alpha)\cap p^{-1}(y)= (\cap_{\alpha\in I_1\cup \dots \cup I_s}C_\alpha)\cap p^{-1}(y)$$

that is non empty because $$y\in \cap_{|I|<\infty}p(\cap_{\alpha\in I} C_\alpha)$$ and so there exists an $$x\in \cap_{\alpha\in I_1\cup \dots \cup I_s}C_\alpha$$ such that $$p(x)=y$$.

( I want to point out that the choice of $$\{p(C_\alpha)\}$$ doesn’t work because $$\{C_\alpha\cap p^{-1}(y)\}$$ does not admit the fip property).

Using compactness of $$p^{-1}(y)$$ we get

$$(\cap_\alpha C_\alpha)\cap p^{-1}(y)\neq \emptyset$$

And so

$$\cap_\alpha C_\alpha\neq \emptyset$$

• How did I forget this approach! Yes, I known this characterization of compactness. After seeing your first line of post, I “almost” solved this problem. I was able to show $\bigcap_{\alpha \in J} p(C_\alpha )\neq \emptyset$. If $f$ were injective, then $\emptyset \neq p^{-1}(\bigcap_{\alpha \in J} p(C_\alpha ))=\bigcap_{\alpha \in J} p^{-1}(p(C_\alpha))= \bigcap_{\alpha \in J} C_\alpha$. Our desired result. But $p$ is not injective, in general. Then I gave up(saw your complete post/solution). I have two question, how to show $C_\alpha \cap p^{-1}(y)$ is closed in $p^{-1}(y)$ and FIP. May 14 at 15:01
• In the 2nd sentence of your 2nd paragraph, why not just take $\{p(C_\alpha)\}$?
– Ruy
May 14 at 15:32
• @Ruy because that family does not admit fip property in general 😊 May 14 at 15:55
• Why $\{ p(C_\alpha)\}_{\alpha \in J}$ don’t admit FIP? $\cap_{\alpha \in I} C_\alpha \neq \emptyset$, where $I$ is finite subset of $J$ $\Rightarrow$ $\emptyset \neq p(\cap_{\alpha \in I} C_\alpha)\subseteq \cap_{\alpha \in I}p( C_\alpha)$. May 14 at 16:56
• @user264745 I added where is exactly the obstruction, but if you want I can be more precise 😀 May 14 at 17:30