# 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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– yasmar Apr 14 '12 at 21:14

## 2 Answers

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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1
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

For smooth manifolds the following holds. By the existence of Morse functions one can deduce a handle-body decomposition of a manifold. This decomposition then yields a CW structure on a space homotopy equivalent to the manifold. I don't think that compactness is needed in any of the above arguments.

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The Morse function does not give a CW structure on the manifold.  Moreover, the results you're quoting (without reference) do require compactness. – Ryan Budney Apr 14 '12 at 20:29
– mland Apr 14 '12 at 20:55
The .pdf file you link to does not support your claims, at least not as stated. The CW structure is not on the manifold, there is a homotopy-equivalence to an induced CW-complex.  In particular, the Morse function they use on a non-compact manifold is a proper Morse function, meaning the level-sets are compact.  This is not the same thing as a Morse function. – Ryan Budney Apr 14 '12 at 21:27
Yes the argument was to brief. But I still think for smooth manifolds the existence of such a "proper" morse function suffices to see that the manifold has the homotopy type of a CW complex. Of course the CW structure is only given on a space homotopy equivalent to the manifold we started with. – mland Apr 14 '12 at 21:30
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