Integral cohomology detects adjacent homology torsion

A homomorphism $\mathbb{Z}\to \mathbb{Z}$ is uniquely determined by the arbitrary integer image of $1$, so its Hom group is $\mathbb{Z}$. A homomorphism $\mathbb{Z}\mathrm{/}m\to \mathbb{Z}$ sends $1$ to an integer $a$ with $ma=0$, which implies $a=0$ when $m>0$; its Hom group is zero. Hom from a finite direct sum is the direct sum of the Hom groups: restriction to each summand and summing their values are inverse homomorphisms. Therefore $\mathrm{Hom}(H_n,\mathbb{Z})\cong {\mathbb{Z}}^r$.

For $\mathbb{Z}$, use the resolution $0\to 0\to \mathbb{Z}\stackrel{1}{\to }\mathbb{Z}\to 0$; its degree-one Hom group is zero, hence its Ext is zero. For $\mathbb{Z}\mathrm{/}m$ with $m>0$, use $0\to \mathbb{Z}\stackrel{m}{\to }\mathbb{Z}\to \mathbb{Z}\mathrm{/}m\to 0$. Multiplication by $m$ is injective and its image is precisely the quotient kernel. Applying Hom into $\mathbb{Z}$ yields $\mathbb{Z}\stackrel{m}{\to }\mathbb{Z}$ in degrees zero and one, because evaluation at $1$ takes precomposition to multiplication by $m$. Thus Ext in degree one is $\mathbb{Z}\mathrm{/}m$. Taking the finite direct sum of these resolutions gives an exact free resolution of $H_{n-1}$: each kernel and image is computed coordinatewise. Hom and then cohomology also split coordinatewise for this finite sum, giving ${\mathrm{Ext}}^1(H_{n-1},\mathbb{Z})\cong {⨁}_{j=1}^t\mathbb{Z}\mathrm{/}{m}_j$. The comparison in [F3] identifies this calculation with the UCT term.

Apply the splitting in [F1] and substitute steps 1.1 and 1.2 to obtain $H^n(X;\mathbb{Z})\cong {\mathbb{Z}}^r\oplus {⨁}_{j=1}^t\mathbb{Z}\mathrm{/}{m}_j$. In this direct sum a finite-order element has zero free coordinate, since a nonzero integer vector has infinite order. Conversely every element with zero free coordinate is killed by the product of the finitely many $m_j$ (or by $1$ if there are none). The torsion subgroup is therefore exactly the displayed finite summand, abstractly the torsion subgroup of $H_{n-1}$, and the free rank is $r$.

At $n=0$, $H_{-1}=0$ gives $s=t=0$, so $H^0$ is free of rank $r$ under the stated finite-generation hypothesis. Empty $X$ gives $r=s=t=0$. Empty torsion data in either input is allowed; a single cyclic summand contributes exactly one $\mathbb{Z}\mathrm{/}m$ in the next cohomology degree. The same resolution computation for $m=1$ gives a zero cyclic group and zero Ext, so omitted trivial summands do not change the formula; $m=0$ is not treated as torsion and belongs to the separate free case. The isomorphism of torsion groups uses chosen decompositions and the UCT splitting; no canonical duality for finite groups is claimed. AC is inherited from [F1] and the comparison in [F3]; the finite cyclic calculations add no choice requirement.