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It is well know that a topological manifold $M$ is homotopy equivalent to a CW-complex $X$. In addition, if $M$ is compact, one even has $M \simeq X$ for some finite CW-complex.

In a Mathoverflow thread, it was discussed if one can take an $n$-dimensional CW-complex $X$ if $M$ is of dimension $n$. However, I have not found a clear reference in this thread, but I think most people believe it to be true.

My question is: Can we at least say for sure (with a good reference/reason) that we can find a finite dimensional CW-complex $X$ with $M \simeq X$ for every topological manifold $M$?

It seems as if Milnor in On spaces having the homotopy type of a CW-complex proves this claim, because the manifold gets embedded into a finite dimensional simplicial complex. But I have some trouble reading this article. Is there a better reference or resource for this fact? It seems quite useful to me.

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abenthy
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    If you just want an (unreadable) reference, you can argue as follows. 1. Kirby and Siebenmann proved (in their book) that every topological $n$-manifold (assuming $n\ne 4$) $M$ has a handle decomposition, in particular, it is homotopy-equivalent to an $n$-dimensional CW complex. To deal with the case $n=4$, replace $M$ with $M\times [0,1]$ and then apply Kirby-Siebenmann. As for a readable reference, I am not sure, will think about one. – Moishe Kohan Dec 11 '16 at 07:05
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    Take a look at the book Lundell and Weingram "The Topology of CW-Complexes", it might be better than Milnor's exposition. – Moishe Kohan Dec 11 '16 at 07:07

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