The nonequivariance defect of an infinitesimal moment map is a constant Lie-algebra two-cocycle

Fix $\xi ,\eta \in \mathfrak{g}$. By [F2] and [F3], $d\{{\mu }^{\xi },{\mu }^{\eta }\}=-{\iota }_{[X_{{\mu }^{\xi }},X_{{\mu }^{\eta }}]}\omega =-{\iota }_{[-{\xi }_M,-{\eta }_M]}\omega =-{\iota }_{[{\xi }_M,{\eta }_M]}\omega ,$ and [F4] rewrites this as $-{\iota }_{[\xi ,\eta {]}_M}\omega$, which equals $d{\mu }^{[\xi ,\eta ]}$ by the moment equation for $[\xi ,\eta ]$. Hence $dc(\xi ,\eta )=0$.

The defect is alternating and bilinear in $(\xi ,\eta )$: it is a difference of the Poisson bracket of two functions depending linearly on the parameters and of the function ${\mu }^{[\xi ,\eta ]}$, which is bilinear in $(\xi ,\eta )$ by multilinearity of the bracket and linearity of the components; skew-symmetry of the Poisson bracket and of the Lie bracket give $c(\eta ,\xi )=-c(\xi ,\eta )$ and $c(\xi ,\xi )=0$.

By step 1.1 the smooth function $c(\xi ,\eta )$ has zero differential, so it is locally constant by [F8]; since $M$ is connected, it is constant on $M$.

Jacobi for the Poisson bracket applied to ${\mu }^{\xi },{\mu }^{\eta },{\mu }^{\zeta }$ reads $0=\{\{{\mu }^{\xi },{\mu }^{\eta }\},{\mu }^{\zeta }\}+\{\{{\mu }^{\eta },{\mu }^{\zeta }\},{\mu }^{\xi }\}+\{\{{\mu }^{\zeta },{\mu }^{\xi }\},{\mu }^{\eta }\}\mathrm{.}$ Replacing each inner bracket by ${\mu }^{[\cdot ,\cdot ]}+c(\cdot ,\cdot )$ and using that a constant Poisson-commutes with every function, the three $\mu$-terms combine into ${\mu }^{[[\xi ,\eta ],\zeta ]+[[\eta ,\zeta ],\xi ]+[[\zeta ,\xi ],\eta ]}=0$ by the Jacobi identity in $\mathfrak{g}$, and the three defect terms give exactly $c([\xi ,\eta ],\zeta )+c([\eta ,\zeta ],\xi )+c([\zeta ,\xi ],\eta )$. Hence this cyclic sum vanishes. By alternation it is the negative of the zero-based differential displayed in [F7], so it vanishes if and only if $dc=0$.

Since $c$ is constant on the connected manifold $M$, the bracket identity of [F9] holds if and only if $c=0$, and it suffices to test $c=0$ at a single point of $M$. By [F9] that bracket identity is equivalent to equivariance under the identity component $G^0$, and hence to coadjoint equivariance of $\mu$ when $G$ is connected; for a general $G$, equivariance under all of $G$ additionally requires equivariance under one representative of each coset of $G\mathrm{/}{G}^0$.