Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
add comment

1 Answer

up vote 5 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.

share|improve this answer
    
Thanks for the help! –  Evariste Jun 9 '13 at 6:03
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.