Dualizing real chain complexes requires an exactness argument

If $f:C_n\to \mathbb{R}$ is a cocycle, then $f(d_{n+1}c)=0$ for all $c$, so $f$ vanishes on $B_n$. Its restriction to $Z_n$ therefore descends to $H_n$. If $f$ is replaced by $f+g\circ d_n$, its values on cycles are unchanged. Likewise replacing $z$ by $z+d_{n+1}c$ leaves $f(z)$ unchanged. Hence ${\epsilon }_C^n$ is well-defined and linear. This construction and these two representative checks use no choice.

Assume AC. Given $h:H_n\to \mathbb{R}$, compose it with the quotient $Z_n\to H_n$ to obtain a functional on $Z_n$. By [F1] extend this to $f:C_n\to \mathbb{R}$. It vanishes on $B_n$, since its restriction to $Z_n$ does, and hence $f$ is a cocycle. Its image under ${\epsilon }_C^n$ is $h$. This proves surjectivity. The sole selection in this step is the functional extension furnished under AC.

If a cocycle class is sent to zero, its representative $f$ vanishes on $Z_n$. Define $b:B_{n-1}\to \mathbb{R}$ by $b(d_{n}c)=f(c)$. If $d_{n}c=d_n{c}^{\mathrm{\prime }}$, then $c-c^{\mathrm{\prime }}\in Z_n$, so $f(c)=f(c^{\mathrm{\prime }})$; thus b is well-defined. Applying this rule to sums and scalar multiples of any preimages proves linearity, without selecting a family of preimages. Extend $b$ to $g:C_{n-1}\to \mathbb{R}$ using [F1]. Then $f=g\circ d_n={\delta }^{n-1}g$, so its class is zero. Conversely every coboundary vanishes on cycles, as already checked in step 1.1. This proves injectivity and both directions of the zero-class criterion.

Let $u:C_{\bullet }\to D_{\bullet }$ be a real-linear chain map. Precomposition defines $u^{\ast }:D^n\to C^n$ and commutes with coboundary, because $u_n{d}_{n+1}^C=d_{n+1}^D{u}_{n+1}$. For a cocycle $f$ on D and a cycle $z$ on C, ${\epsilon }_C^n([f\circ u_n])([z])=f(u_{n}z)=({\epsilon }_D^n([f])\circ H_n(u))([z])\mathrm{.}$ This proves naturality. If $H_n(u)$ is an isomorphism, precomposition by it is an isomorphism of real duals, with inverse precomposition by its inverse. Steps 2.1 and 2.2 and this commuting identity show that $H^n(u^{\ast })$ is an isomorphism. No bases are chosen to define the canonical evaluation map or the naturality square.

In contrast to the AC conclusion of step 3.1, now assume only the DC and Baire-property hypotheses of the second branch. Put $C_2=E$, $C_1=P$, $C_0=P\mathrm{/}E$, with $d_2$ the inclusion, $d_1$ the quotient and every other group and differential zero. The composite $d_1{d}_2$ is zero. The inclusion has zero kernel, the quotient has kernel E, and the quotient is surjective. It follows directly that $H_2(C)=H_1(C)=H_0(C)=0$, and all remaining homology groups vanish because their chain groups are zero. Thus the unique chain map $C\to 0$ is a quasi-isomorphism in every degree.

The dual complex in degrees $0,1,2$ is $(P\mathrm{/}E{)}^{\ast }\stackrel{q^{\ast }}{\to }{P}^{\ast }\stackrel{i^{\ast }}{\to }{E}^{\ast },$ and its differential out of degree 2 is zero. Therefore $H^2(C^{\bullet })=E^{\ast }\mathrm{/}\mathrm{im}(P^{\ast }\to E^{\ast })\mathrm{.}$ By the second clause of [F1], the explicit functional $\mathrm{\ell }$ is omitted from that image; its class in this quotient is nonzero. The dual of $C\to 0$ is $0\to C^{\bullet }$, whose induced degree-2 map from zero cannot be surjective. This is a conditional real-vector-space example, with no change of coefficient field and no model-existence assertion.

The zero complex satisfies the positive assertion with the unique isomorphism $0\to 0$ in every degree. For a nonnegative complex, at degree zero $Z_0=C_0$; the kernel case of step 2.2 gives $f=0$ directly, and no negative-degree extension is required. For a complex supported at one degree with group $\mathbb{R}$ and zero differential, evaluation is the usual map ${\mathbb{R}}^{\ast }\to {\mathbb{R}}^{\ast }$ and is the identity. The proof does not assume injective differentials or nonzero chain groups: all zero and repeated maps are governed by the displayed square-zero identity. The abstract conditional complex of step 4.1 is not asserted to be a singular chain complex of any space.