An additive coefficient group does not determine a cup multiplication

Both operations are bilinear and commutative. The first is associative coordinatewise and has unit $(1,1)$. For the second, if $z=(e,f)$, both $(x\ast y)\ast z$ and $x\ast (y\ast z)$ equal $(ace,acf+ade+bce)$; its unit is $(1,0)$. Thus these define commutative unital rings on the same additive group. They are respectively $\mathbb{Z}\times \mathbb{Z}$ and $\mathbb{Z}[ϵ]\mathrm{/}({ϵ}^2)$, with $(a,b)$ corresponding to $a+bϵ$ in the latter.

More strongly, let $\mu :G\times G\to G$ be bilinear and invariant under every additive automorphism, meaning $f(\mu (x,y))=\mu (f(x),f(y))$. With $f=-\mathrm{i}\mathrm{d}$, bilinearity gives $-\mu (x,y)=\mu (-x,-y)=\mu (x,y)$. Thus $2\mu (x,y)=0$, and torsion-freeness of ${\mathbb{Z}}^2$ forces $\mu =0$. A zero multiplication cannot have a unit on a nonzero group, since $\mu (u,(1,0))=0\mathrm{\ne }(1,0)$. Consequently no unital multiplication can be recovered in a manner invariant under all additive automorphisms. The zero bilinear pairing is indeed canonical; the refuted claim concerns a unital multiplication, not the existence of any pairing.

In the first ring the idempotents are exactly $(0,0),(1,0),(0,1),(1,1)$, since an integer satisfies $a^2=a$ exactly when $a=0$ or $1$. In the second ring, an idempotent satisfies $a^2=a$ and $(2a-1)b=0$. For $a=0$ or $1$, the second equation forces $b=0$. There are exactly two idempotents. Any ring isomorphism bijects idempotents, so these two rings are not isomorphic.

At the point space, degree-zero cochains with values in $G$ are just $G$, and [F1]'s formula multiplies their values. The two displayed ring structures therefore give different cup operations already there, and step 1.2 rules out a natural unital choice from additive data alone. For example $(1,0)\cdot (0,1)=0$, while $(1,0)\ast (0,1)=(0,1)$. This fixed nonempty, nonzero, degree-zero example has no endpoint or higher-simplex qualification. The zero coefficient group would not witness failure; neither would empty-space cochains. All operations, automorphism and idempotents used here are explicit, with no AC.