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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.

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  • $\begingroup$ 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. $\endgroup$
    – user264745
    May 14 at 13:34
  • $\begingroup$ Here is slight variation of above proof. $\endgroup$
    – user264745
    May 14 at 13:39
  • $\begingroup$ I tried to answer :) , if you don’t know the fip property tell me and I will be more clear. $\endgroup$ May 14 at 13:56

1 Answer 1

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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

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

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    $\begingroup$ 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. $\endgroup$
    – user264745
    May 14 at 15:01
  • $\begingroup$ In the 2nd sentence of your 2nd paragraph, why not just take $\{p(C_\alpha)\}$? $\endgroup$
    – Ruy
    May 14 at 15:32
  • $\begingroup$ @Ruy because that family does not admit fip property in general 😊 $\endgroup$ May 14 at 15:55
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    $\begingroup$ 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 $J$ $\Rightarrow$ $\emptyset \neq p(\cap_{\alpha \in I} C_\alpha)\subseteq \cap_{\alpha \in I}p( C_\alpha)$. $\endgroup$
    – user264745
    May 14 at 16:56
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    $\begingroup$ @user264745 I added where is exactly the obstruction, but if you want I can be more precise 😀 $\endgroup$ May 14 at 17:30

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