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A perfect map $f$ is a closed continuous surjective function such that the preimage of every point is compact. One property of perfect maps is that if $f \, \colon \, X \to Y$ is perfect, and $Y$ is compact, then $X$ is compact too.

My question (rephrased): if $f$ is a continuous surjective function such that the preimage of every point is compact, and $Y$ is compact, does it follow that $X$ is compact?

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I think that such a map is automatically closed if $X$ is Hausdorff. –  Giuseppe Negro Dec 12 '11 at 22:06
    
i think it's closed if Y is Hausdorff. –  Alex J. Dec 12 '11 at 22:09
    
yes, i rephrased my question, hopefully it's clear now. –  Alex J. Dec 12 '11 at 22:17
    
Thanks for the clarification! –  Dylan Moreland Dec 12 '11 at 22:17

1 Answer 1

No.

Let $Y$ be the closed unit interval with the usual Euclidean topology, and let $X$ be the closed unit interval with the discrete topology. Let $f:X\to Y$ be the identity map. Clearly $f$ is continuous and surjective and has compact fibres, and $Y$ is compact, but $X$ is not. Note that $X$ and $Y$ are even metrizable, so they’re about as nice as one could hope for.

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what is a compact fiber/fibre? –  Alex J. Dec 12 '11 at 22:52
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@Alex: The fibres of a map are the preimages of points under that map. –  Brian M. Scott Dec 12 '11 at 23:02

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