Show that a path-connected space \(X\) has a CW approximation with countably many cells if \(\pi_n(X)\) is countable for all \(n\).
Solution
We use the standard inductive construction of a CW approximation, keeping track of cardinalities.
Start with one \(0\)-cell and attach \(1\)-cells representing a generating set for \(\pi_1(X)\). Since \(\pi_1(X)\) is countable, only countably many \(1\)-cells are needed. Attach countably many \(2\)-cells to kill the kernel of the resulting map on \(\pi_1\): the kernel is countable because the fundamental group of a countable CW complex is countable.
Inductively suppose a countable CW complex \(K_n\) and a map \[ f_n:K_n\to X \] have been constructed inducing isomorphisms on \(\pi_i\) for \(i<n\) and a surjection on \(\pi_n\). The group \(\pi_n(K_n)\) is countable: every map \(S^n\to K_n\) has compact image, hence lies in a finite subcomplex, and there are only countably many finite subcomplexes and countably many homotopy classes of maps into each. Thus the kernel of \[ (f_n)_*:\pi_n(K_n)\to\pi_n(X) \] is countable. Attach one \((n+1)\)-cell for each element of a countable generating set of this kernel. Then attach countably many further \((n+1)\)-cells, if necessary, to represent the countable group \(\pi_{n+1}(X)\) and make the next homotopy map surjective.
At every stage only countably many cells are added. Taking the union \[ K=\bigcup_n K_n \] produces a CW complex with countably many cells and a map \(K\to X\) inducing isomorphisms on every homotopy group. Hence \(K\to X\) is a CW approximation, and \[ \boxed{X\text{ has a CW approximation with countably many cells}.} \]