Are chain complexes chain equivalent to free ones?

Question

Given a chain complex $A_\bullet\in\mathrm{Ch(\mathbf{Ab})}$, are there exist some chain complex $A'_\bullet\in\mathrm{Ch(\mathbf{Ab})}$ which is chain equivalent to $A_\bullet$ such that $A'_p$ are all free abelian groups?

Answer 1

I'm answering my own question.

No. A counter example is $A_\bullet := A_0\to A_1\to A_2 := 0\to\mathbf{Z}/2\mathbf{Z}\to 0$.

Assume there exists a chain complex $A'_\bullet$ of free abelian group with $\varphi:A_\bullet\to A'_\bullet, \psi:A'_\bullet\to A_\bullet$ and a homotopy $\alpha$ between $\psi\circ\varphi$ and $\mathrm{id}_{A_\bullet}$.

It lead to a contradiction Because

• $\phi_1:\mathbf{Z}/2\mathbf{Z}\to B$ is a null map because $B$ is free and
• the only $\alpha_0$ and $\alpha_1$ are null maps.