Ordinary cohomology does not give noncompact Poincaré duality

For any $x,y\in X$, the straight path $\gamma (t)=(1-t)x+ty$ joins them. By [F4], a degree-zero cocycle $\phi$ must satisfy $\phi (y)-\phi (x)=0$, so it is constant. Conversely a constant function has zero endpoint difference on every path, so is a cocycle. There are no degree-zero coboundaries by [F3]. Evaluation at $0$ and the assignment of the constant function with value $a\in R$ are inverse $R$-linear maps between $H^0(X;R)$ and $R$. In particular the constant cocycle $1$ represents a nonzero class because $1\mathrm{\ne }0$.

The homotopy $H(x,t)=(1-t)x$ contracts nonempty $X$ to its origin, so [F2] identifies its homology with that of a point. At a point there is one singular simplex in each degree, and its boundary in degree $k>0$ is multiplication by ${\sum }_{j=0}^k(-1{)}^j$, equal to $1$ for even $k$ and $0$ for odd $k$. Therefore every positive-degree kernel equals the next boundary image, and $H_k(\ast ;R)=0$ for $k>0$, while $H_0(\ast ;R)=R$. Since $n>0$, this proves $H_n(X;R)=0$ and $H_0(X;R)=R$, the latter with the origin's point class as generator.

An isomorphism $H^0(X;R)\to H_n(X;R)$ cannot exist: by step 1.1 its domain contains the nonzero constant class $1$, while by step 1.2 every element of its codomain is zero, so every homomorphism has that nonzero class in its kernel. This is precisely the degree-zero failure of the proposed replacement of $H_c^p$ by $H^p$ in dimension $n$. The straight-path and point-chain computations used no AC.

Assume now [F5]. The space $X$ is the oriented Euclidean coordinate ball of [F1], which therefore supplies $H_c^p(X;R)=0$ for $p\mathrm{\ne }n$ and the stated normalized cap isomorphism in degree $n$. In particular $H_c^0(X;R)=0$ since $n>0$, so the correct degree-zero duality is $0\to 0$, consistent with step 1.2. In degree $n$, evaluation equal to $1$ on the oriented relative class maps to the origin's generator of $H_0$ by [F1]. The AC use is the freeness/projection and free-comparison-lift selections in that lemma's local UCT proof; it is not needed for step 2.1.

The hypotheses exclude both $n=0$ and $R=0$ for a reason: at $n=0$ the space is a point and both ordinary degree-zero groups equal $R$; over the zero ring both groups in the claimed mismatch are zero. The chosen witness space is nonempty, with the specified origin, and $n=1$ is already a counterexample with the same contraction. The contraction has endpoints the identity and the constant map; the straight paths have their stated endpoints. The point-chain calculation includes degenerate simplices in every positive degree. No infinite selection of points, paths or cocycle representatives is used: all paths and the constant cocycles are given by formulas.