Degree-wise free chain complexes over $Z$ and CW-complexes

Question

Given a Degree-wise free chain complexes (of finite type), say $C_*$, is there a CW-complex $X$ such that $C_*$ is the cellular chain complex of $X$?

Answer 1

The comment by jgon is unfortunately nonsense as $\vee_\beta S_\beta^{n-1}$ is a quotient of $X_{n-1}$ not a subcomplex. This approach will not work in general. For instance, suppose you have built up a CW complex $X_{n-1}$ which gives the correct chain complex up to degree $(n-1)$. Suppose that it happens to be $X_2$ which you have built and it agrees with the 2-torus $T^2$. But suppose the chain complex you want to construct has $C_3 = C_2 = C_0 = \Bbb Z$, where the differential is zero except on $d: C_3 \to C_2$.

Instead you should make your CW complex out of model examples. It is not hard to prove that every free finitely generated chain complex over $\Bbb Z$ is a sum of copies of shifts of $\Bbb Z$ and $\Bbb Z \xrightarrow{d} \Bbb Z$. You can add the first term by wedging on a sphere. You can add the second term by wedging on a suspension of the mapping cone $\text{Cone}(z^d)$, where $z^d: S^1 \to S^1$ is the degree $d$ map. That is, I am suggesting you wedge together a sum of CW complexes, where each is either a sphere $S^k$ or the suspension $\Sigma^{k-2} Cone(z^d)$.

Some special care should be taken to show that there are no issues at $H_0$ but this is inconsequential.