Cohomology over a field is dual to homology over that field

A field is a PID: any nonzero ideal contains a nonzero $a$, hence $1=a^{-1}a$, and is the whole ring; the zero ideal is generated by zero. The complex $C$ is nonnegative and free on the singular simplex sets. Thus [F2] applies, and [F1] identifies its middle group with $H^n(X;k)$.

Put $B=B_{n-1}C\subseteq Z=Z_{n-1}C$. Apply [F3] first to the empty independent set in $B$ to obtain a basis $L$ of $B$, then to $L\subseteq Z$ to extend it to a basis $T$ of $Z$. For any linear $\psi :B\to k$, define a map on $T$ by its existing values $\psi (l)$ on $L$ and zero on $T\setminus L$. Every vector of $Z$ has a unique finite basis expansion, so summing its coefficients against these values defines a linear $g:Z\to k$ with $g{\mathrm{\mid }}_B=\psi$. Consequently restriction ${\mathrm{Hom}}_k(Z,k)\to {\mathrm{Hom}}_k(B,k)$ is onto, and the cokernel Ext term in [F2] is zero.

Exactness from step 1.1 and vanishing from step 1.2 make evaluation both injective and surjective. Its formula is $[\phi ]\mapsto ([c]\mapsto \phi (c))$, and its naturality in $X$ is exactly the chain-map naturality in [F2]. The auxiliary bases used to prove Ext vanishing do not occur in this map. Arbitrary functionals are admitted: step 1.2 uses finite expansions of individual vectors, with no finite support requirement on the functional or finite bound on the basis.

For $n=0$, $B_{-1}=Z_{-1}=0$ and the same result follows with empty bases. For empty $X$, both sides are zero. On a point in degree zero, evaluation identifies $k$ with ${\mathrm{Hom}}_k(k,k)$ via multiplication by the value at the constant zero-simplex; the value $1$ yields the identity. If $H_n(X;k)=0$, the isomorphism forces $H^n(X;k)=0$. AC is inherited in [F2] and used explicitly in both basis applications of step 1.2. No identification of an infinite-dimensional vector space with its double dual is made.