Let $R$ be an integral domain with unit. In Burde-Zieschang's Knots it is claimed (page 219) that for a pair of spaces $(X,Y)$ $$H_i(X,Y;R) \cong H_i(X,Y;\mathbb{Z}) \otimes_{\mathbb{Z}} R.$$
Is it the case that $R$, as a $\mathbb{Z}$-module, is projective or free (so that $\mathrm{Tor}^{\mathbb{Z}}_1(-, R)$ always vanishes)? Or otherwise why does the $\mathrm{Tor}$ term never appear?
It's not true in general, for instance with $R= \mathbb Z/p, p$ prime it fails badly.
e.g. $p=2, X=\mathbb RP^2, Y=\emptyset$, then $H_2(\mathbb RP^2;\mathbb Z) = 0$ but $H_2(\mathbb RP^2;\mathbb Z/2) \cong \mathbb Z/2$
Maybe the author meant characteristic $0$ integral domain ? (in which case the underlying abelian group is torsion-free and hence, since $\mathbb Z$ is principal, flat)