Right-invariant fields carry the opposite Lie bracket

Let $m:G\times G\to G$ be multiplication and let $j_1(g)=(g,e)$, $j_2(g)=(e,g)$. Under [F7], $d(j_1{)}_e(A)=(A,0)$ and $d(j_2{)}_e(B)=(0,B)$. Since $m\circ j_1=m\circ j_2={\mathrm{id}}_G$, the chain rule [F6] and linearity [F8] give $d{m}_{(e,e)}(A,B)=A+B$.

The map $g\mapsto m(g,\mathrm{inv}(g))$ is constant at $e$. By [F8], its differential annihilates every tangent vector because derivations annihilate constant germs. Applying [F6] and step 1.1 gives $0=d{m}_{(e,e)}(Z,d(\mathrm{inv}{)}_{e}Z)=Z+d(\mathrm{inv}{)}_{e}Z$. Hence $d(\mathrm{inv}{)}_e=-{\mathrm{id}}_{\mathfrak{g}}$.

For $h\in G$, the identity $\mathrm{inv}\circ L_h=R_{h^{-1}}\circ \mathrm{inv}$ and [F6] give $d(\mathrm{inv}{)}_{h}d(L_h{)}_{e}Z=d(R_{h^{-1}}{)}_{e}d(\mathrm{inv}{)}_{e}Z=-d(R_{h^{-1}}{)}_{e}Z$. By [F4], as $h$ varies this says ${\mathrm{inv}}_{\ast }(Z^L)=(-Z{)}^R=-Z^R$.

Inversion is a diffeomorphism by [F2]. Apply [F9], step 3.1, and [F5]: $[X^R,Y^R]=[-X^R,-Y^R]=[{\mathrm{inv}}_{\ast }{X}^L,{\mathrm{inv}}_{\ast }{Y}^L]={\mathrm{inv}}_{\ast }[X^L,Y^L]={\mathrm{inv}}_{\ast }([X,Y{]}_G^L)=-[X,Y{]}_G^R\mathrm{.}$ The first equality uses the bilinearity visible directly in the commutator formula [F10].

A Lie group is nonempty. If $\dim G=0$, all fields and brackets vanish; in dimension one the tangent bracket vanishes by alternation, so the displayed identity again reads zero equals zero. No metric or nondegeneracy condition occurs, and the group is boundaryless by convention. The stated ${\mathrm{A}\mathrm{C}}_{\omega }$ is inherited through [F3], [F4], and [F5]; the product differential, inversion, and bracket calculation are canonical and add no choice. The proposition is a one-way sign identity, not a biconditional.