Cup product pairing is nonsingular for closed R orientable manifolds when R = Z and torsion in $H^∗(M;Z)$ is factored out?

I was reading Hatcher on the cup product being nondegenerate. I could follow the proof in the case when $$R$$ is a field, but I am not sure how to interpret

For field coefficients or for integer coefficients with torsion factored out, $$h$$ is an isomorphism.

Any help on what does he mean to factor the torsion out? Certainly $$h$$ is not the original $$h$$ defined on $$H^{n-k}(M;R)$$. Is $$h$$ in this case the induced homomorphism on the quotient group? If so, why is $$h$$ well-defined? Also, why is $$h$$ an isomorphism in this case? Lastly, why doesn't quotienting out the torsion part affect the equation $$\phi\to \psi([M]\cap \phi)=(\phi\cup\psi)[M]$$?

PS: I attached Hatcher's proof on page 250 below.

$$\newcommand{\Hom}{\operatorname{Hom}}$$
This is a fair question. If $$A$$ is an abelian group, let me write $$T(A)$$ for the torsion subgroup of $$A$$. In the following, I will omit both the space and the coefficients from notation. Note that $$\mathbb{Z}$$ is torsion-free, so the cup product pairing $$H^{n-k}\times H^k\rightarrow\mathbb{Z}$$ factors through a pairing $$H^{n-k}/T(H^{n-k})\times H^k\rightarrow\mathbb{Z}$$. This is adjoint to a homomorphism $$H^{n-k}/T(H^{n-k})\rightarrow\Hom(H^k,\mathbb{Z})$$. On the other hand, the cup product pairing $$H^{n-k}\times H^k\rightarrow\mathbb{Z}$$ is adjoint to a homomrphism $$H^{n-k}\rightarrow\Hom(H^k,\mathbb{Z})$$. Since $$\Hom(H^k,\mathbb{Z})$$ is torsion-free, this factors through a homomorphism $$H^{n-k}/T(H^{n-k})\rightarrow\Hom(H^k,\mathbb{Z})$$. These two maps just constructed are actually one and the same, as they do the same thing on representatives.
More precisely, note that this map is really just the composite of the factorization $$H^{n-k}/T(H^{n-k})\rightarrow\Hom(H_{n-k},\mathbb{Z})$$ (exists, since $$\Hom(H_{n-k},\mathbb{Z})$$ is torsion-free) and the regular old isomorphism $$\Hom(H_{n-k},\mathbb{Z})\rightarrow\Hom(H^k,\mathbb{Z})$$. The first map is actually an isomorphism too, by the Universal Coefficient Theorem. The kernel of the Kronecker map is the canonical image of $$\mathrm{Ext}(H_{n-k-1},\mathbb{Z})$$, but this is a torsion group, so its image is contained in $$T(H^{n-k})$$. Conversely, each torsion element lies in the kernel, as already observed. The reason the $$\mathrm{Ext}$$-term is a torsion group is that it's isomorphic to the torsion subgroup of $$H_{n-k-1}$$, since $$H_{n-k-1}$$ is finitely generated. The fact that the homology groups of a compact manifold are finitely generated is proven by Hatcher in the appendix. Together, this shows the pairing $$H^{n-k}/T(H^{n-k})\times H^k\rightarrow\mathbb{Z}$$ is non-degenerate on the left. Again since $$\mathbb{Z}$$ is torsion-free, it descends to a pairing $$H^{n-k}/T(H^{n-k})\times H^k/T(H^k)\rightarrow\mathbb{Z}$$, which is necessarily still non-degenerate on the left. Non-degeneracy on the right follows again by commutativity.
Alternatively (and even quicker), this can be deduced from the case with field coefficients, by noting that $$H^{n-k}(M;\mathbb{Z})/T(H^{n-k}(M;\mathbb{Z}))$$ injects into $$H^{n-k}(M;\mathbb{Q})\cong H^{n-k}(M;\mathbb{Z})\otimes_{\mathbb{Z}}\mathbb{Q}\cong H^{n-k}(M;\mathbb{Z})/T(H^{n-k}(M;\mathbb{Z}))\otimes_{\mathbb{Z}}\mathbb{Q}$$. Then, prove the general fact that if $$A,B$$ are torsion-free abelian groups (which is equivalent to the canonical maps $$A\rightarrow A\otimes\mathbb{Q}$$ and $$B\rightarrow B\otimes\mathbb{Q}$$ being injections) and $$(A\otimes\mathbb{Q})\times(B\otimes\mathbb{Q})\rightarrow\mathbb{Q}$$ is a non-degenerate pairing restricting to a pairing $$A\times B\rightarrow\mathbb{Z}$$, this restricted pairing is non-degenerate as well (this applies to the cup product pairings as cup products and the Kronecker pairing are both natural with respect to change of coefficients).