Adjoint exponential identity

Put $A(t)={\mathrm{Ad}}_{{\exp }_G(tX)}$. By [F2]--[F3], $A$ is smooth, $A(0)=I$, and $A(t+s)=A(t)A(s)=A(s)A(t)$. The chain rule and [F4] give $A^{\mathrm{\prime }}(0)=d(\mathrm{Ad}{)}_{e}X={\mathrm{ad}}_X$.

Fix $t$ and differentiate $A(t+s)=A(s)A(t)$ with respect to $s$ at zero. Using step 1.1 gives $A^{\mathrm{\prime }}(t)={\mathrm{ad}}_X\circ A(t)$ and $A(0)=I$.

In one fixed basis of $\mathfrak{g}$, [F5] gives a unique solution of $E^{\mathrm{\prime }}(t)={\mathrm{ad}}_{X}E(t)$, $E(0)=I$, on every compact interval containing zero. Solutions on overlapping intervals agree by uniqueness, so they define the global linear-ODE exponential $E(t)=e^{t{\mathrm{ad}}_X}$ without any arbitrary selection. Step 2.1 and uniqueness give $A(t)=E(t)$ on every such interval. Evaluating at $t=1$ yields ${\mathrm{Ad}}_{{\exp }_{G}X}=e^{{\mathrm{ad}}_X}$.

A Lie group is nonempty and boundaryless. If $\dim G=0$, both sides are the unique endomorphism of the zero space; in dimension one, ${\mathrm{ad}}_X=0$ and the ODE gives both sides equal to $I$. Degenerate adjoint endomorphisms are allowed. The compact-interval solutions glue globally, so there is no endpoint issue, and no metric occurs. The only choice use is the stated ${\mathrm{A}\mathrm{C}}_{\omega }$ inherited through [F3]--[F4]; one finite basis and uniquely determined ODE solutions add no choice. No biconditional is asserted.