The singular coboundary squares to zero

If $n\ge 0$, let $\phi \in C^n(X;G)$ and $c\in C_{n+2}(X;\mathbb{Z})$. Associativity of composition and [F1]–[F2] give $(({\delta }^{n+1}{\delta }^n)\phi )(c)=(\phi {\mathrm{\partial }}_{n+1}){\mathrm{\partial }}_{n+2}(c)=\phi (0)=0.$ This holds for every chain, hence the cochain is zero, for every $\phi$. It includes $n=0$, where the boundary composite is ${\mathrm{\partial }}_1{\mathrm{\partial }}_2$.

If $n<0$, the domain $C^n$ is zero and ${\delta }^n=0$ by [F1], so the composite is zero. In particular $n=-1$ has zero first map even though the later groups may be nonzero. For empty $X$ or $G=0$ all groups are zero and both calculations remain valid.

Steps 1.1 and 1.2 cover every integer $n$. Thus the graded cochains and coboundary form a complex, with no topological restrictions on $X$ or freeness/injectivity restriction on $G$. A point still has singular simplices in every nonnegative degree; step 1.1 applies to that unnormalized complex without discarding them. No representatives, bases or primitives were selected, so no AC is used.