# Does noncompact manifold or orbifold have the homotopy type,of CW complex?

I forget for a while, we don't need the compactness condition here right?

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Relevant comments and references can be found on MO also mathoverflow.net/questions/36838/… even though the question is not exactly the same. –  yasmar Apr 14 '12 at 21:14

According to The Topology of CW-Complexes by Lundell and Weingram (Van Nostrand Reinhold, 1969) the answer is yes for (separable) manifolds.

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And the proof for smooth manifolds is relatively nice -- on a smooth manifold there is a proper non-negative smooth function $f : M \to \mathbb R$ so $f^{-1}([0,a])$ is a smooth submanifold of $M$ for $a$ a regular value of $f$, and these manifolds exhaust $M$. –  Ryan Budney Sep 17 '10 at 13:34