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Can every topological (not necessarily smooth or PL) manifold be given the structure of a CW complex?

I'm pretty sure that the answer is yes. However, I have not managed to find a reference for this.

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    $\begingroup$ @algori : I thought you had posted an (important sounding) comment? Why did you delete it? $\endgroup$ Aug 27, 2010 at 4:48
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    $\begingroup$ It turns out that my first comment was a bit wrong. Here are the slides of A. Ranicki's talk in Orsay. www.maths.ed.ac.uk/~aar/slides/orsay.pdf It says on p. 5 there that a compact manifold of dimension other than 4 is a CW complex. There is a related conjecture that says that each closed manifold of dimension $\geq 5$ is homeomorphic to a polyhedron (there are 4-manifolds for which this is false). See arxiv.org/pdf/math/0212297. I'm not sure what if anything is known about the noncompact case. $\endgroup$
    – algori
    Aug 27, 2010 at 4:50
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    $\begingroup$ Update: recent work of Davis, Fowler, and Lafont front.math.ucdavis.edu/1304.3730 shows that in every dimension ≥6 there exists a closed aspherical manifold that is not homeomorphic to a simplicial complex. $\endgroup$
    – Lee Mosher
    May 1, 2013 at 16:10
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    $\begingroup$ Hatcher's Algebraic Topology p. 529 has a paragraph answering this question very clearly for compact manifolds (not including results in 2013 of course). However his references are to two long dense books, without page specification. $\endgroup$
    – hsp
    Sep 3, 2013 at 15:47

2 Answers 2

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Kirby and Siebenmann's paper "On the triangulation of manifolds and the Hauptvermutung" Bull AMS 75 (1969) is the standard reference for this, I believe.

The result is that compact topological manifolds have the homotopy-type of CW-complexes, to be precise.

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  • $\begingroup$ I think the fact that they have the homotopy type of a CW complex is due to Milnor (it is in his paper about spaces homotopy equivalent to CW complexes). Do Kirby-Siebenmann just prove this, or do they prove that all compact manifolds are homeomorphic to CW complexes? Also, how about the noncompact case? $\endgroup$ Aug 27, 2010 at 4:08
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    $\begingroup$ But I thought the question was whether each has the "homeomorphism type" of a CW complex. $\endgroup$
    – Dev Sinha
    Aug 27, 2010 at 4:22
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    $\begingroup$ @Ryan, the open problem is not whether any compact manifold is homeomorphic to a CW complex (this was proved by Kirby-Siebenmann). The open problem is whether it has a (non-combinatorial) triangulation. @grad student, whatever is known in the noncompact case must be Kirby-Siebenmann's book. $\endgroup$ Aug 27, 2010 at 13:15
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    $\begingroup$ @IgorBelegradek That doesn't sound right: The paper by Kirby-Siebenmann claims that they have the homotopy type of a CW complex, not that they are homeomorphic. Furthermore, the answers to this MSE post seem to provide references that the case $n=4$ is still open. I don't know much (if anything) about these topics, but could you provide a reference to back up that claim? $\endgroup$
    – Danu
    Aug 9, 2016 at 20:39
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    $\begingroup$ @Danu: Kirby-Siebenmann's book p.107: every closed topological manifold of dimension $\ge 6$ is homeomorphic to a CW complex. In the comment above I was of course talking about higher dimensions. The statement that every manifold (compact or not) is homotopy equivalent to a CW complex is much easier. For compact manifolds this is proved e.g. in Hatcher's textbook (in appendix on CW complexes). $\endgroup$ Aug 9, 2016 at 21:35
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See http://arxiv.org/abs/math/0609665

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    $\begingroup$ That manifold isn't 2nd countable. Like most mathematicians, I only care about manifolds that are Hausdorff and 2nd countable. $\endgroup$ Aug 27, 2010 at 3:56
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    $\begingroup$ I hope that the fact that you only care about those does not preclude you from enjoying learning about the rest. $\endgroup$ Aug 27, 2010 at 4:12

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