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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.

A grad student
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    @algori : I thought you had posted an (important sounding) comment? Why did you delete it? – A grad student Aug 27 '10 at 04:48
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    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.

     – algori Aug 27 '10 at 04:50
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    Update: recent work of Davis, Fowler, and Lafont http://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. – Lee Mosher May 01 '13 at 16:10
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    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. – hsp Sep 03 '13 at 15:47

2 Answers2

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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.

Ryan Budney
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  • 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? – A grad student Aug 27 '10 at 04:08
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    But I thought the question was whether each has the "homeomorphism type" of a CW complex. – Dev Sinha Aug 27 '10 at 04:22
  • It's been a while since I've looked at that Milnor paper -- I suspect maybe he's arguing that manifolds have the homotopy-type of countable CWs, while Kirby-Siebenmann deal with compact manifolds and finite CWs. ? – Ryan Budney Aug 27 '10 at 04:23
  • @Ryan : Yes, I think that is what Milnor proved (it's also been a long time since I looked at it). – A grad student Aug 27 '10 at 04:27
  • @Dev, yes, that sounds right. As algori mentions that's apparently an open problem. As far as I know that problem hasn't attracted a whole lot of attention. – Ryan Budney Aug 27 '10 at 04:30
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    @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. – Igor Belegradek Aug 27 '10 at 13:15
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    @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? – Danu Aug 09 '16 at 20:39
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    @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). – Igor Belegradek Aug 09 '16 at 21:35
  • @IgorBelegradek Thank you for the clarifications. – Danu Aug 09 '16 at 22:00
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See http://arxiv.org/abs/math/0609665

Mariano Suárez-Álvarez
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