# A CW is of countable type, iff all its homotopy groups are countable? (References?)

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?

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.
• 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. – J. Steinebrunner Aug 21 '16 at 15:47