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

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  • $\begingroup$ Relevant comments and references can be found on MO also mathoverflow.net/questions/36838/… even though the question is not exactly the same. $\endgroup$
    – yasmar
    Commented Apr 14, 2012 at 21:14

2 Answers 2

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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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    $\begingroup$ 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 $M$. $\endgroup$ Commented Sep 17, 2010 at 13:34
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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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    $\begingroup$ The Morse function does not give a CW structure on the manifold. Moreover, the results you're quoting (without reference) do require compactness. $\endgroup$ Commented Apr 14, 2012 at 20:29
  • $\begingroup$ Let me maybe make this more precise. Any manifold admits a Morse function. Moreover such a morse function induces a handlebody decomposition which in turn induces a CW structure. (collapsing handles to their cocores). See also imsc.res.in/~jniyer/handlebody.pdf and Milnor's book on Morse theory. $\endgroup$
    – mland
    Commented Apr 14, 2012 at 20:55
  • $\begingroup$ 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. $\endgroup$ Commented Apr 14, 2012 at 21:27
  • $\begingroup$ 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. $\endgroup$
    – mland
    Commented Apr 14, 2012 at 21:30

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