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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!

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    $\begingroup$ The claim in this question that the homotopy groups of a finite CW complex are finitely generated is incorrect. See (counter-)example 4.27 in Hatcher. $\endgroup$
    – Communicative Algebra
    Commented Sep 30, 2019 at 14:29

2 Answers 2

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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.

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    $\begingroup$ 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? $\endgroup$
    – Jan Steinebrunner
    Commented Aug 21, 2016 at 15:41
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    $\begingroup$ 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. $\endgroup$
    – Jan Steinebrunner
    Commented Aug 21, 2016 at 15:47
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    $\begingroup$ 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
    Commented Aug 22, 2016 at 7:34
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I couldn't quite figure out the proof for non-simply connected spaces sketched in the comments to Jeff Strom's answer, so I'm posting a more detailed answer in which direction ($\Leftarrow$) of the claim is instead proved using CW approximation.

Claim: Cosider a topological space that is homotopy equivalent to a connected CW complex. The space is homotopy equivalent to a countable CW complex if and only if all its homotopy groups are countable.

Proof: ($\Rightarrow$) Suppose $X$ is a homotopy equivalent to a connected countable CW complex. We may as well assume that $X$ itself is a CW complex with these properties. By Theorem C of Wall, or by a direct argument, $\pi_1(X)$ is countable. It follows that the universal cover $\tilde X$ of $X$ with the canonical CW structure (see Proposition 2.3.9 in [Fritsch & Piccinini]) is also countable. Applying Theorem C of Wall once again, we find that all homology groups of the simply connected CW complex $\tilde X$ are countable. As countable abelian groups form a Serre class within the category of all abelian groups, this implies that all homotopy groups of $\tilde X$ are also countable. As the higher homotopy groups of $X$ agree with those of $\tilde X$, this shows that all homotopy groups of $X$ are countable, as claimed.

($\Leftarrow$) Suppose conversely that $X$ is homotopy equivalent to a connected CW complex with countable homotopy groups. Then the usual construction of a CW approximation $X'\to X$ as in, say, § 10.5 of May, yields a countable CW complex $X'$ which is, by Whitehead's Theorem, homotopy equivalent to $X$. To see that $X'$ is countable, recall that $X'$ is constructed inductively as a colimit of spaces $X_k$. The cells of $X_1$ are indexed by elements of the homotopy groups $\pi_i(X)$, which are countable by assumption, so $X_1$ is countable. The additional cells added to obtain $X_{k+1}$ from $X_k$ are indexed by pairs of elements of the group $\pi_k(X_k)$. Assuming by induction that $X_k$ is countable, we find from the previous part of the proof that $\pi_k(X_k)$ is countable, and hence that $X_{k+1}$ is countable. Thus, each subcomplex $X_k$ is countable, and hence so is $X'$. $\square$

Wall: Finiteness Conditions for CW-Complexes
Fritsch & Piccinini: Cellular Structures in Topology
May: A Concise Course in Algebraic Topology

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