Poincaré duality gives a nonsingular cup pairing

For each singular $n$-simplex $\sigma$, the formulas [F3] give $\beta (\alpha \cap \sigma )=\alpha (\sigma [0,\dots ,p])\beta (\sigma [p,\dots ,n])=(\alpha ⌣\beta )(\sigma )\mathrm{.}$ Linearity gives $\beta (\alpha \cap z)=(\alpha ⌣\beta )(z)$. Evaluation on a cycle is unchanged if a cocycle changes by $\delta h$, because $(\delta h)(c)=h(\mathrm{\partial }c)=0$; it is unchanged if the cycle changes by $\mathrm{\partial }d$, because a cocycle vanishes on such a boundary. Thus the identity descends, using [F1], to $⟨a⌣b,[M]⟩=⟨b,D_M(a)⟩\mathrm{.}$ It is bilinear in both classes and is independent of all three representatives.

Suppose $R=F$ is a field. It is a PID: every nonzero ideal contains a nonzero $u$, hence contains $u^{-1}u=1$ and is the whole ring, while the zero ideal is principal. Apply [F5]'s general PID-complex lemma to the free singular complex $C_{\ast }(M;F)$ and the coefficient module $F$, with the cochain identification [F6]. By [F2] the groups $H_j(M;F)$ are finitely generated. A finite spanning list over a field can be reduced to a basis: whenever it is dependent, solve a nonzero dependence coefficient for one vector and delete that vector without changing the span; the list shortens, so the procedure terminates with a finite independent spanning list. Thus these groups are finite-dimensional and finite free. A finite free module has a length-zero free resolution, so the degree-one Hom cohomology computing its Ext is zero. The comparison assertion in [F5] consequently kills the Ext term in every degree, including $H_{-1}=0$. Thus evaluation is an isomorphism $h_j:H^j(M;F)\stackrel{\sim }{\to }{\mathrm{Hom}}_F(H_j(M;F),F)\mathrm{.}$ It is this field-complex application, rather than an unjustified replacement of integral homology by field homology in the topological UCT statement, that gives the required map.

Suppose $R=\mathbb{Z}$. By [F2] and [F7], write $H_{j-1}(M;\mathbb{Z})\cong {\mathbb{Z}}^r\oplus {⨁}_{i=1}^t\mathbb{Z}\mathrm{/}(d_i)$ with $d_i>1$. Use the free resolution whose degree-zero term is ${\mathbb{Z}}^{r+t}$, whose degree-one term is ${\mathbb{Z}}^t$, and whose differential sends its $i$th basis vector to $d_i$ times the $(r+i)$th basis vector. Its cokernel is the displayed group and its differential is injective. Applying Hom into $\mathbb{Z}$ gives degree-one cokernel ${⨁}_i\mathbb{Z}\mathrm{/}(d_i)$, a finite group. By comparison in [F5] this computes the Ext term in the integral UCT. Therefore the kernel of evaluation $h_j:H^j(M;\mathbb{Z})\to {\mathrm{Hom}}_{\mathbb{Z}}(H_j(M;\mathbb{Z}),\mathbb{Z})$ is finite, and hence consists of torsion elements. Conversely every torsion element maps to zero: the Hom target is torsion-free, since an integer multiple of a homomorphism is zero only when each of its integer values is zero. Thus $\ker h_j=\mathrm{Tor}{H}^j(M;\mathbb{Z})$. Every integer homomorphism from $H_j$ kills its torsion, so UCT surjectivity gives an induced isomorphism ${\bar{h}}_j:H^j(M;\mathbb{Z})\mathrm{/}\mathrm{Tor}\stackrel{\sim }{\to }{\mathrm{Hom}}_{\mathbb{Z}}(H_j(M;\mathbb{Z})\mathrm{/}\mathrm{Tor},\mathbb{Z})\mathrm{.}$ These torsion-free quotients are finite free by [F2] and [F7]. At $j=0$ take $r=t=0$ for $H_{-1}=0$.

Over the field, the adjoint in the $b$ variable of the pairing in step 1.1 is the composite $H^q(M;F)\stackrel{h_q}{\to }{\mathrm{Hom}}_F(H_q(M;F),F)\stackrel{D_M^{\ast }}{\to }{\mathrm{Hom}}_F(H^p(M;F),F)\mathrm{.}$ Its first map is the evaluation isomorphism of step 1.2, and its second map is precomposition with the isomorphism [F1]; its inverse is precomposition with $D_M^{-1}$. Thus this adjoint is an isomorphism. Over $\mathbb{Z}$, the pairing vanishes on torsion in either variable, since its values are integers and it is bilinear. Duality [F1] sends torsion onto torsion, since it and its inverse commute with integer multiplication, so it induces an isomorphism on the free quotients. The same composite with ${\bar{h}}_q$ of step 1.3 gives an isomorphism from the second free quotient onto the integer dual of the first.

Repeat step 2.1 with $p,q$ interchanged. It says that the map sending $a$ to the functional $b\mapsto ⟨b⌣a,[M]⟩$ is an isomorphism, over the field or on the integer free quotients. By [F4], the other adjoint of our original pairing is this map multiplied by $(-1{)}^{pq}$. Multiplication by that unit is its own inverse, so this adjoint is an isomorphism too. This proves perfectness and integral unimodularity with both arguments in the stipulated order; no identification with an infinite-dimensional double dual has been assumed.

When $M$ is empty all groups are zero and both adjoints are isomorphisms of zero modules. For a point with orientation unit $u$, $n=p=q=0$ and the pairing is $(a,b)\mapsto uab$, whose adjoints multiply by the unit $u$. The same proof includes disconnected closed manifolds and degree endpoints $p=0,n$. If $p$ is outside $0,\dots ,n$, both factors vanish by [F2] and the negative-degree conventions, so perfectness holds for the zero modules. A field and $\mathbb{Z}$ are nonzero, so the zero ring is outside the hypotheses. Degenerate simplices satisfy step 1.1 without alteration. All AC use is inherited from [F1] and [F5] as specified in [F8]; finite cyclic resolutions and the two adjoint compositions add no infinite selection.