Cochain cup product is not strictly graded commutative

Define the integral one-cochain $a$ to have value $1$ on the singular simplex $e_{01}$ and value $0$ on every other singular one-simplex; define $b$ similarly with support $\{e_{12}\}$. Each extends uniquely to a homomorphism on the free group of finite singular one-chains. The two edges are distinct maps (their initial vertices differ), so $a(e_{01})=b(e_{12})=1,b(e_{01})=a(e_{12})=0.$ No choice of a basis is involved: singular simplex maps are the specified generators.

Formula [F1] gives $(a⌣b)(s)=a(e_{01})b(e_{12})=1,(b⌣a)(s)=b(e_{01})a(e_{12})=0.$ Since $p=q=1$, graded commutativity would require the first value to be the negative of the second. But $1\mathrm{\ne }-0$ in $\mathbb{Z}$. Thus these are unequal cochains, even with the required sign.

Here $\mathrm{\partial }s=e_{12}-e_{02}+e_{01}$, so $\delta a(s)=1$ and $\delta b(s)=1$. Neither cochain is a cocycle, and [F2] does not assert the refuted identity for them. This calculation uses two nondegenerate one-faces of a single nondegenerate two-simplex. Mixed degree-zero and positive-degree cochains can also witness failure when the zero-cochain takes different values at the two endpoints of an edge; the present example instead keeps both cochains in degree one. Empty spaces and the zero coefficient ring cannot furnish this witness. All faces include their endpoints, and all other simplex values, including degenerate ones, were explicitly set to zero. No AC is used.