A C4 coboundary changes the pentagonator and is gauged away

The cochain $\beta$ is normalized because it equals $1$ whenever any input is $0$. At $(g,h,k,l)=(1,2,3,1)$ in $C_4$, the five factors in [F1] are respectively $\beta (2,3,1)=1$, $\beta (3,3,1{)}^{-1}=1$, $\beta (1,1,1)=-1$, $\beta (1,2,0{)}^{-1}=1$, and $\beta (1,2,3)=1$. Thus $\omega (1,2,3,1)=-1$, so the resulting 4-cochain is not the constant cochain.

If $g=0$, the first two factors in $\delta \beta$ are equal with opposite exponents and the remaining three have an input $0$, so all remaining normalized values are $1$. If $h=0$, the second and third factors cancel and the other factors are normalized; if $k=0$, the third and fourth factors cancel and the other factors are normalized; if $l=0$, the fourth and fifth factors cancel and the other factors are normalized. Thus $\omega =\delta \beta$ is normalized. Applying the degree-four coboundary to this expression gives a product of the 30 composites $d_j{d}_i\beta$ from the six degree-four faces and five degree-three faces. For each $0\le i<j\le 5$, the simplicial face identity $d_j{d}_i=d_i{d}_{j-1}$ pairs two occurrences with opposite exponents; the interior face identities use associativity in $C_4$, and the endpoint identities delete the same first or last entry. All 15 pairs cancel, so $\delta \omega =1$ and $\omega$ is a 4-cocycle.

Theorem [F1] applies to $\omega$ and ${\omega }^{\mathrm{\prime }}=1$ with cochain ${\beta }^{-1}$ (whose coboundary is ${\omega }^{-1}$), and gives a monoidal 2-equivalence $2{\mathrm{V}\mathrm{e}\mathrm{c}\mathrm{t}}_{C_4}^{\omega }\simeq 2{\mathrm{V}\mathrm{e}\mathrm{c}\mathrm{t}}_{C_4}$. The example checks a nonconstant representative and its coherent untwisting; it does not assert a nonzero class in $H^4(C_4;{\mathbb{C}}^{\times })$.