Translations are diffeomorphisms and their differentials trivialize the tangent bundle

The group laws give $L_g\circ L_{g^{-1}}=L_{g^{-1}}\circ L_g={\mathrm{id}}_G$ and $R_g\circ R_{g^{-1}}=R_{g^{-1}}\circ R_g={\mathrm{id}}_G$. All four translations are smooth by [F2], so $L_g$ and $R_g$ are diffeomorphisms with the asserted inverses.

Let $m:G\times G\to G$ be multiplication. Near an arbitrary $g_0$, choose product coordinates in which $m$ is represented by a smooth map $\mu (a,b)$. The local matrix of $d(L_g{)}_e$ is the second-variable Jacobian $D_2\mu (a,b_e)$, whose entries are smooth in $a$. Equivalently, this is the restriction of the smooth global differential $dm$ supplied by [F3]. Therefore ${\mathrm{\Phi }}_L$ is a smooth bundle map. The same calculation with the variables reversed gives smoothness of ${\mathrm{\Phi }}_R$.

By [F4] and step 1.1, each fibre map $d(L_g{)}_e:T_{e}G\to T_{g}G$ and $d(R_g{)}_e:T_{e}G\to T_{g}G$ is a linear isomorphism. Hence ${\mathrm{\Phi }}_L$ and ${\mathrm{\Phi }}_R$ are fibrewise linear bijections over ${\mathrm{id}}_G$.

Apply [F5] to the smooth fibrewise bijections from steps 2.1 and 1.2 over ${\mathrm{id}}_G$. Both ${\mathrm{\Phi }}_L$ and ${\mathrm{\Phi }}_R$ are vector-bundle isomorphisms. Fibrewise, their inverses are $(g,V)\mapsto (g,d(L_{g^{-1}}{)}_{g}V)$ and $(g,V)\mapsto (g,d(R_{g^{-1}}{)}_{g}V)$, respectively.

A Lie group is nonempty. If $\dim G=0$, all tangent fibres are zero spaces and the displayed maps are the unique fibre maps; if $\dim G=1$, the same proof applies. No metric or nondegeneracy condition occurs, and the group is boundaryless by the page convention. The only choice assumption is the stated ${\mathrm{A}\mathrm{C}}_{\omega }$, used through [F3] for the canonical smooth tangent bundles/global differential; all group operations and local computations are supplied or pointwise and add no choice. The item asserts explicit inverse identities but no biconditional.