Relation between $\mathbb{Z}\pi_1 (X)$-module $\pi_n (X)$ and $\mathbb{Z}$-module $H_n (X)$ or $\mathbb{Z}\pi_1 (X)$-module $H_n (\tilde{X})$

Question

Let $X$ be a finite CW-complex of $n$.

1. For $i\geq 2$, $\pi_i (X)$ is a $\mathbb{Z}\pi_1 (X)$-module.

2. for $i\geq 2$, $H_i (\tilde{X})$ is a finitely generated $\mathbb{Z}\pi_1 (X)$-module, where $\tilde{X}$ is the universal covering of $X$.

I know that the connection between homotopy groups and homology groups is given by the Hurewicz-Theorem. But my question is that:

My question: Can we express $\pi_i (X)$ (as a $\mathbb{Z}\pi_1 (X)$-module) in terms of $H_i (X)$'s (as $\mathbb{Z}$-module) or $H_i (\tilde{X})$ (as $\mathbb{Z}\pi_1 (X)$-module) for $2\leq i\leq n$? (particularly when they are free).

Answer 1

Not sure if you're already aware of an exact sequence which implies Hurewicz's theorem, but if $X$ is a space with $\pi_i(X)=0$ for $1<i<n$ then we have $$H_{n+1}(X)\to H_{n+1}(\pi)\to \pi_n(X)_{\pi}\to H_n(X)\stackrel{\psi}{\to} H_n(\pi)\to0$$ where $\pi=\pi_1(X)$ is the fundamental group and $H_*(\pi)$ is discrete group cohomology, and $\pi_n(X)_{\pi}$ denotes the coinvariants. See e.g. Ken Brown "Cohomology of Groups" exercise VII.7.6.

Generally, without assumptions on the homotopy groups, that map $\psi$ exists (in all degrees) and is canonical: You take a projective resolution of $\mathbb Z$ over $\mathbb Z\pi$, and you take the complex of free $\mathbb Z\pi$-modules $C_*(\text{universal cover of }X)$, to build a canonical chain map between those resolutions that induces $\psi$. This is also explained in the above reference somewhere in Chapter II.