The kronecker pairing is independent of cocycle and cycle representatives

Replacing $\phi$ by $\phi +\delta \psi$ changes its value on $c$ by $(\delta \psi )(c)=\psi (\mathrm{\partial }c)=\psi (0)=0$. Replacing $c$ by $c+\mathrm{\partial }b$ changes its value under $\phi$ by $\phi (\mathrm{\partial }b)=(\delta \phi )(b)=0$. The replacement cocycle is still closed and the replacement cycle still a cycle by their defining quotient subgroups. Applying the two equalities successively therefore allows both representatives to change at once. This proves descent through both quotients.

On representatives $(\phi +\psi )(c)=\phi (c)+\psi (c)$ and $\phi (c+c^{\mathrm{\prime }})=\phi (c)+\phi (c^{\mathrm{\prime }})$. Zero and negatives obey the same evaluation rules. In the ring version, $\phi (rc)=r\phi (c)$ and $(r\phi )(c)=r\phi (c)$ by $R$-linearity; the two vanishing calculations of step 1.1 use the same positive differential and remain valid for $R$-linear maps.

For representatives of $\alpha ,z$, the left spatial pairing is $(\phi f_{\mathrm{\#}})(c)=\phi (f_{\mathrm{\#}}c)$, exactly the right pairing by [F3]. Its representatives are valid cycles and cocycles because the induced maps preserve them. Likewise $(u\phi )(c)=u(\phi (c))$ proves coefficient naturality. Step 1.1 makes these representative identities identities on quotient classes.

Step 1.1 proves representative independence, step 2.1 descends to biadditivity and the stated $R$-bilinearity, and step 2.2 proves naturality. In degree zero there are no negative cochains to change the cocycle, but changing a zero-cycle by a one-boundary is still covered by the second calculation. Negative-degree groups and empty-space or zero-coefficient groups pair to zero. On a point, the degree-zero chain $m[x]$ evaluates to $m\phi (x)$, including $m=0$ and $m=1$; no generators in higher unnormalized chain degrees are discarded. The proof compares arbitrary representatives without choosing a representative function, so it uses no AC.