3
$\begingroup$

When constructing a classifying space $BPL$ for piecewise linear microbundles, one would like it to be a polyhedron, i.e. a locally finite simplicial complex. Milnor solved this by showing that the isomorphism classes of pl microbundles $mb_{PL}(S^n)$ over each sphere are countable. Then some special version of Browns representablility theorem yields a countable CW complex $BPL$ and simplicial approximation gives a countable simplicial complex version of this space.

Now, one can easily show that being (locally) countable, locally finite and locally compact is the same thing for a CW or simplicial complex up to homotopy equivalence. (Whitehead does something like this (including simplicial approximation) in his "combinatorial homotopy" article. There is also a summary on the first page of Milnor's "on spaces having the homotopy type of a CW complex". There he also says that this is that same as being homotopy equivalent to an absolute neighbourhood retract.)

Knowing this I was wondering about the statement in the title:

Given a (connected) CW complex $X$ with countable homotopy groups, the functor $[.,X]$ satisfies the conditions of Brown's theorem and moreover it takes countable values on the spheres. So there should, by a similar argument as above, be a countable CW complex $Y$ representing this functor and by the usual arguments (Yoneda, Whitehead) we can see that it is homotopy equivalent to $X$.

The converse holds as the homotopy groups of a finite CW complex are finitely generated and taking the limit over a countable index set preserves countablility.

Did I make some stupid mistakes in the argument above and if not: Does anyone know a good reference for this "countable version" of Brown's theorem or even directly for this classification of CW complexes of "countable type" in terms of homotopy groups?

Thank you in advance!

$\endgroup$
3
$\begingroup$

If $X$ is simply-connected, then the homotopy groups will be countable iff the homology groups are countable; and then one can build $X$ by a homology resolution using Moore spaces for countable groups, resulting in countably many cells.

$\endgroup$
  • 1
    $\begingroup$ Thank you a lot; do you any reference, where I can look up homology resolutions ? I think I got the idea, but I'd like to see how you do this formally. In addition I'd prefer not to require simply-connected, do you think it can be handled by taking the universal cover? $\endgroup$ – J. Steinebrunner Aug 21 '16 at 15:41
  • 1
    $\begingroup$ What if we take Eilenberg-Mac Lane-spaces instead of Moore spaces? Then, the analogue should be a Posnikov system and we wouldn't need the simply-connected. I can prove that $K(G,n)$ is of countable type for $G$ countable, but I do not know whether there are problems, because this is only the homotopy fibre. $\endgroup$ – J. Steinebrunner Aug 21 '16 at 15:47
  • 3
    $\begingroup$ I think going to the universal cover works, so Jeff's arguement resolves your question also in the non-simply-connected case. Note however that the analogous statement for "countable" replaced by "finitely generated" is not true, see e.g. this paper: jstor.org/stable/… paper $\endgroup$ – Jens Reinhold Aug 22 '16 at 7:34

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.