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I'm stuck on this statement. Could anyone please help me out?

Let $X$ be a compact manifold, every map $f: X \longrightarrow Y$ is proper.

The definition of proper: a smooth map between manifolds is called proper if inverse images of compact subsets are compact.

I know that continuous maps map compact sets to compact sets. But this seems to be the converse of that... Is there anything that I'm missing here? Thanks!

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up vote 6 down vote accepted

In $Y$, compact sets are closed (assuming $Y$ is Hausdorff). $f$ is continuous, so the inverse image of a closed set is closed. But a closed subset of a compact (Hausdorff) space is compact. So the inverse image of a compact set is compact.

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Thanks for the help! – Evariste Jun 9 '13 at 6:03

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