[proofplan]
We show that both adjoint maps of the cup product pairing are isomorphisms by factoring them through Poincaré duality and the universal coefficients theorem. The adjoint map $H^k(M; R) \to \operatorname{Hom}_R(H^{d-k}(M; R), R)$ is the composition of the UCT isomorphism $H^k(M; R) \cong \operatorname{Hom}_R(H_k(M; R), R)$ with the map induced by the Poincaré duality isomorphism $D_M: H^{d-k}(M; R) \xrightarrow{\sim} H_k(M; R)$. Since $H_*(M; R)$ is free over $R$, both factors are isomorphisms, and the same argument applies to the other adjoint.
[/proofplan]
[step:Define the cup product pairing and its adjoint maps]
The cup product pairing is the bilinear map
\begin{align*}
\langle \cdot, \cdot \rangle: H^k(M; R) \otimes_R H^{d-k}(M; R) &\to R \\
\alpha \otimes \beta &\mapsto (\alpha \smile \beta)[M],
\end{align*}
where $[M] \in H_d(M; R)$ is the fundamental class. The two adjoint maps are:
\begin{align*}
\Phi_L: H^k(M; R) &\to \operatorname{Hom}_R(H^{d-k}(M; R), R), &\quad \Phi_L(\alpha)(\beta) &= (\alpha \smile \beta)[M], \\
\Phi_R: H^{d-k}(M; R) &\to \operatorname{Hom}_R(H^k(M; R), R), &\quad \Phi_R(\beta)(\alpha) &= (\alpha \smile \beta)[M].
\end{align*}
We must show both $\Phi_L$ and $\Phi_R$ are isomorphisms.
[/step]
[step:Factor the left adjoint through Poincaré duality and universal coefficients]
The cap-cup adjunction formula states that for $\sigma \in H_d(M; R)$, $\varphi \in H^k(M; R)$, and $\psi \in H^{d-k}(M; R)$:
\begin{align*}
\psi(\sigma \frown \varphi) = (\varphi \smile \psi)(\sigma).
\end{align*}
Applying this with $\sigma = [M]$ and writing $D_M(\varphi) = [M] \frown \varphi \in H_{d-k}(M; R)$:
\begin{align*}
\psi(D_M(\varphi)) = (\varphi \smile \psi)[M] = \langle \varphi, \psi \rangle.
\end{align*}
Therefore the left adjoint factors as
\begin{align*}
\Phi_L: H^k(M; R) \xrightarrow{D_M} H_{d-k}(M; R) \xrightarrow{\mathrm{ev}} \operatorname{Hom}_R(H^{d-k}(M; R), R),
\end{align*}
where $\mathrm{ev}$ sends $c \in H_{d-k}(M; R)$ to the functional $\psi \mapsto \psi(c)$.
We now identify $\mathrm{ev}$ with the universal coefficients map. The universal coefficients theorem provides a natural short exact sequence
\begin{align*}
0 \to \operatorname{Ext}^1_R(H_{d-k-1}(M; R), R) \to H^{d-k}(M; R) \xrightarrow{h} \operatorname{Hom}_R(H_{d-k}(M; R), R) \to 0.
\end{align*}
Since $H_*(M; R)$ is free over $R$ by hypothesis, $H_{d-k-1}(M; R)$ is free, so $\operatorname{Ext}^1_R(H_{d-k-1}(M; R), R) = 0$. Thus $h$ is an isomorphism, and $\mathrm{ev}$ coincides with the inverse $h^{-1}$. Since both $D_M$ (by [Poincaré Duality](/theorems/2291)) and $h^{-1}$ are isomorphisms, $\Phi_L = h^{-1} \circ D_M$ is not quite the right decomposition — let us be more precise.
Rewriting: $\Phi_L(\varphi) = \psi \mapsto \psi(D_M(\varphi))$. The evaluation map $\mathrm{ev}: H_{d-k}(M; R) \to \operatorname{Hom}_R(H^{d-k}(M; R), R)$ is defined by $\mathrm{ev}(c)(\psi) = \psi(c)$. Since $h: H^{d-k}(M; R) \to \operatorname{Hom}_R(H_{d-k}(M; R), R)$ is an isomorphism (UCT with vanishing Ext), and both $H_{d-k}(M; R)$ and $H^{d-k}(M; R)$ are free of the same rank, the natural evaluation $\mathrm{ev}$ is an isomorphism. Hence $\Phi_L = \mathrm{ev} \circ D_M$ is a composition of two isomorphisms, and therefore an isomorphism.
[guided]
The argument connects three pieces: the cap-cup adjunction formula, Poincaré duality, and the universal coefficients theorem.
The cap-cup formula $\psi(\sigma \frown \varphi) = (\varphi \smile \psi)(\sigma)$ tells us that evaluating the cup product on $[M]$ is the same as evaluating $\psi$ on the cap product $D_M(\varphi) = [M] \frown \varphi$. This means the left adjoint map $\Phi_L(\varphi)$ is simply "evaluate on $D_M(\varphi)$" — it factors through the Poincaré duality map.
The remaining question is whether the evaluation map $\mathrm{ev}: H_{d-k}(M; R) \to \operatorname{Hom}_R(H^{d-k}(M; R), R)$ is an isomorphism. This is where the freeness hypothesis enters. The universal coefficients theorem gives $H^{d-k}(M; R) \cong \operatorname{Hom}_R(H_{d-k}(M; R), R)$ (the Ext term vanishes because $H_{d-k-1}(M; R)$ is free). Since $H_{d-k}(M; R)$ is a finitely generated free $R$-module, the natural map from a free module to its double dual is an isomorphism, making $\mathrm{ev}$ an isomorphism as well. The composition $\Phi_L = \mathrm{ev} \circ D_M$ is therefore an isomorphism.
[/guided]
[/step]
[step:Show the right adjoint is also an isomorphism by the same argument]
The right adjoint $\Phi_R: H^{d-k}(M; R) \to \operatorname{Hom}_R(H^k(M; R), R)$ sends $\beta$ to $\alpha \mapsto (\alpha \smile \beta)[M]$. By the same cap-cup formula with roles exchanged (using graded commutativity of the cup product: $\alpha \smile \beta = (-1)^{k(d-k)} \beta \smile \alpha$), we have
\begin{align*}
\Phi_R(\beta)(\alpha) = (\alpha \smile \beta)[M] = (-1)^{k(d-k)} (\beta \smile \alpha)[M].
\end{align*}
Define $D'_M: H^{d-k}(M; R) \to H_k(M; R)$ by $D'_M(\beta) = [M] \frown \beta$, which is an isomorphism by [Poincaré Duality](/theorems/2291). Then $\Phi_R(\beta)(\alpha) = (-1)^{k(d-k)} \alpha(D'_M(\beta))$, so $\Phi_R = (-1)^{k(d-k)} \mathrm{ev}' \circ D'_M$ where $\mathrm{ev}': H_k(M; R) \to \operatorname{Hom}_R(H^k(M; R), R)$ is the evaluation map. Since $H_k(M; R)$ is free over $R$, the same UCT argument shows $\mathrm{ev}'$ is an isomorphism. The sign $(-1)^{k(d-k)}$ is a unit in $R$, so $\Phi_R$ is an isomorphism.
[/step]
