Left-invariant vector fields evaluate isomorphically at the identity

Pointwise addition and scalar multiplication preserve smooth vector fields. Because each differential $d(L_g{)}_h$ is linear, they also preserve left invariance; the same holds on the right. Thus ${\mathfrak{X}}_L(G)$ and ${\mathfrak{X}}_R(G)$ are real vector spaces, and evaluation at $e$ is linear on each.

Fix $v\in T_{e}G$. The map $g\mapsto (g,v)$ is a smooth section of the product bundle $G\times T_{e}G$. Composing it with the left trivialization in [F3] shows that $v_g^L=d(L_g{)}_e(v)$ is a smooth vector field. Its identity-value formula makes it left invariant by [F2], and $v_e^L=d(L_e{)}_e(v)=v$.

Conversely, [F2] forces every $X\in {\mathfrak{X}}_L(G)$ to satisfy $X_g=d(L_g{)}_e(X_e)$ at every $g$. Hence $X=(X_e{)}^L$, so $v\mapsto v^L$ and ${\mathrm{ev}}_e^L$ are mutually inverse linear maps.

Replacing the left trivialization by the right trivialization in [F3] gives a smooth field $v_g^R=d(R_g{)}_e(v)$. The right identity-value characterization in [F2] proves invariance and uniqueness, while $R_e={\mathrm{id}}_G$ gives $v_e^R=v$. Thus $v\mapsto v^R$ is the inverse of ${\mathrm{ev}}_e^R$.

A Lie group is nonempty. If $\dim G=0$, then $T_{e}G=0$ and both invariant-field spaces contain only the zero field, so both evaluation maps are the unique zero-dimensional isomorphisms; dimension one needs no change. No metric or nondegeneracy condition occurs, and the group is boundaryless by convention. The stated ${\mathrm{A}\mathrm{C}}_{\omega }$ is inherited through [F2] and [F3]; fixing the supplied vector $v$ and performing pointwise linear operations adds no choice. The theorem asserts two explicit isomorphisms, not a biconditional.