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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.2011-12-12
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    i think it's closed if Y is Hausdorff.2011-12-12
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    yes, i rephrased my question, hopefully it's clear now.2011-12-12

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