Concerning homology groups of the universal covering space of a finite polyhedron

Question

Suppose that $P$ is a finite polyhedron. I know that $H_i (\tilde{P})$ is a finitely generated $\mathbb{Z}[\pi_1 (P)]$-module for all $i$.

My question is that:
If $\pi_1 (P)$ is a free group, then is $H_i (\tilde{P})$ a projective $\mathbb{Z}[\pi_1 (P)]$-module?

Thanks in advance.

Answer 1

If you take $P$ to be the suspension of $\mathbb{R}\mathrm{P}^2$ (which I am not sure counts as a finite polyhedron for you), then $\pi_1(P)$ is the trivial free group, but $H_2(\tilde P)=H_2(P)=H_1(\mathbb{R}\mathrm{P}^2)=\mathbb{Z}/2\mathbb{Z}$ is not a projective $\mathbb{Z}$-module.