= Two-coboundary
{title2=$\delta b(g,h)=g\cdot b(h)-b(gh)+b(g)$}
For an additive left $G$-<module> $M$, a two-coboundary is a degree-two <group cochain> obtained from a degree-one cochain $b:G\to M$ by
$$
(\delta b)(g,h)=g\cdot b(h)-b(gh)+b(g).
$$
It is automatically a <two-cocycle> because $\delta^2=0$. Expanding the four terms in $g\cdot(\delta b)(h,k)-(\delta b)(gh,k)+(\delta b)(g,hk)-(\delta b)(g,h)$ makes every contribution cancel.
For trivial action and $G=\mathbb Z$, take $M=\mathbb Z$ and $b(n)=n^2$. Then $(\delta b)(m,n)=n^2-(m+n)^2+m^2=-2mn$, a nonzero cochain representing the zero class in $H^2(G,M)$. Being cohomologically trivial does not require being pointwise zero.
Changing a section in an abelian-kernel <group extension> changes its extension cocycle by a two-coboundary. This is the mechanism behind quotienting two-cocycles by two-coboundaries when defining second <group cohomology>.
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