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Dugundji's Definition: A map $f:X\to Y$ between topological spaces is called perfect (or proper), if it is a closed continuous surjection such that each fiber $f^{-1}(\{y\})$, $y\in Y$, is compact.

Show that a continuous surjection $f:X\to Y$ between topological spaces is proper if, and only if, $f^{-1}(K)\subset X$ is compact, whenever $K\subset Y$ is compact.

My try:

$(\Rightarrow)$ We need just to show that $K\subset Y$ compact $\Rightarrow f^{-1}(K)\subset X$ compact. Let $K\subset Y$ be a compact set and let $\scr U$ be an open convering of $f^{-1}(K)$. Then $\scr F$$=\{F_U\,:\, U\in \scr U\}$, $F_U:=X-U$, is a family of closed sets. Since $f$ is closed, $\scr V$$=\{Y-f(F_U)\,:\,U\in \scr U\}$ is a family of open subsets of $Y$. But... what then?

$(\Leftarrow)$ It is obvious that each fiber $f^{-1}(\{y\})$ is compact, since every $\{y\}\subset Y$ is compact. Let $F\subset X$ be closed. How can I show that $f(F)\subset Y$ is closed? (in order to show that $f:X\to Y$ is closed)

Derso
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1 Answers1

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Let's distinguish the definitions:

A continuous map is perfect if it is a closed surjection with compact fibers.

A continuous map is proper if the preimage of each compact set is compact.

The forward direction, perfect => proper, holds in full generality. Going with your notation, we can assume $\mathcal U$ is closed under finite unions. This means each fiber is covered by some $U\in\mathcal U,$ and hence each $k\in K$ is covered by some open set in $\mathcal V.$ By compactness of $Y$ some finite union of sets in $\mathcal V$ covers $Y,$ which can only come from a finite cover of $f^{-1}(K)$ by sets in $\mathcal U.$

The backward direction does not hold in general, for example the constant map from the one-point space (or any compact space) to the open point of the Sierpinski space is proper but not closed. It does hold when $Y$ is compactly generated Hausdorff: Continuous proper map into compactly generated Hausdorff space is closed.

Dap
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