6

I've been stuck on this problem for a very long time:

Problem: Let $X,Y$ be topological spaces, $Y$ compact, and let $q:X\to Y$ be a continuous and closed map. Assume that $q^{-1}(\{y\})$ is compact for every $y\in Y$. Show that $X$ is compact.

What I have so far:

$q:X\to q(X)$ is also continuous and closed, and $q(X)$ is compact (as a closed subset of a compact space). Thus, we could assume that $q(X)=Y$, or that $q$ is onto $Y$.

If $q^{-1}(A)\subset X$ is open, Then $(q^{-1}(A))^c=q^{-1}(A^c)$ is closed, and so ($q$ is onto) $A^c=q(q^{-1}(A^c))$ is closed and $A$ is open. Thus $q$ is a quotient map.

Thus, the quotient space $X^*=\{q^{-1}(\{y\})|y\in Y\}$ is compact, as it's homeomorphic to $Y$.

How do I carry on from here?

Ohad
  • 932
  • How can you show it is open? A continuous surjective closed map need not be open – TomGrubb Sep 30 '17 at 17:23
  • Oops, I got mixed up :) I'll fix that. – Ohad Sep 30 '17 at 17:26
  • I believe that the trick would be to start with an open cover for $X$. Then push the complements of those open sets through your closed map and work with the compactness of Y to get a finite cover that you can pull back. – Robert Thingum Sep 30 '17 at 17:48
  • Also a duplicate of this, and appears as exercise 12, page 172 in Munkres, Topology, 2nd ed., Prentice Hall, 2000. – fred goodman Oct 01 '17 at 15:50

1 Answers1

1

I found this proof in some notes of mine. I'm sure it's not original but I don't know the source.

Let $\mathcal A$ be a family of closed subsets of $X$ with the finite intersection property. We need to show that $\bigcap_{A \in \mathcal A} A \ne \emptyset$. We can suppose wlog that this family is closed under finite intersections, because augmenting $\mathcal A$ with all finite intersections of its elements preserves both the FIP and $\bigcap_{A \in \mathcal A} A $.

Consider $\{q(A)\}_{A \in \mathcal A}$; this is a family of closed subsets of $Y$ since $q$ is a closed map. Show that this family has the finite intersection property. Since $Y$ is compact there exists a point $y \in \bigcap_{A \in \mathcal A} q(A)$. Equivalently, $q^{-1}(y) \cap A \ne \emptyset$ for all $A \in \mathcal A$. The family $\{q^{-1}(y) \cap A\}_{A \in \mathcal A}$ has the FIP because for $A_i \in \mathcal A$, $$ (q^{-1}(y) \cap A_1) \cap \cdots \cap (q^{-1}(y) \cap A_k) = q^{-1}(y) \cap (A_1 \cap \cdots \cap A_k), $$ $A_1 \cap \cdots \cap A_k \in \mathcal A$ by the assumption that $\mathcal A$ is closed under finite intersections, and we have just shown that $q^{-1}(y) \cap A \ne \emptyset$ for all $A \in \mathcal A$. But the family $\{q^{-1}(y) \cap A\}_{A \in \mathcal A}$ is a family of closed subsets of the compact set $q^{-1}(y)$. Hence $\bigcap_{A \in \mathcal A} (q^{-1}(y) \cap A) \ne \emptyset$.

Notes: Continuity of $q$ is not used in the proof. However, the result is a lemma for the following proposition. Call a map $f: X \to Y$ between topological spaces \emph{proper} if the inverse image of every compact subset of $Y$ is compact in $X$.

Proposition: Let $X$ and $Y$ be topological spaces with $Y$ Hausdorff. Let $f: X \to Y$ be continuous and closed with each fiber $f^{-1}(y)$ compact in $X$. Then $f$ is proper.

Proof: Let $K$ be compact in $Y$. Then $K$ is closed since $Y$ is Hausdorff, and $f^{-1}(K)$ is closed in $X$ since $f$ is continuous. Let $q$ denote the restriction of $f$ to $f^{-1}(K)$. The hypotheses of the original question apply to $q$, and hence $f^{-1}(K)$ is compact.

fred goodman
  • 4,443
  • Do we need the continuity of $q$? – Julien Sep 30 '17 at 23:16
  • Why should ${ q^{-1}(y) \cap A_\alpha }$ have the FIP just because $\mathcal{A}$ does? – Mr. Chip Oct 01 '17 at 00:31
  • 1
    Mr. Chip, we assumed at the outset that $\mathcal A$ is closed under finite intersections. Hence $(q^{-1}(y)\cap A_{\alpha_1}) \cap \cdots \cap (q^{-1}(y)\cap A_{\alpha_k}) = q^{-1}(y) \cap ( A_{\alpha_1} \cap \cdots \cap A_{\alpha_k}) \ne \emptyset$, because $A_{\alpha_1} \cap \cdots \cap A_{\alpha_k} \in \mathcal A$. – fred goodman Oct 01 '17 at 02:58