Given a Lie Group, we have that the left multiplication $$L_g:G \to G$$ such that$L_g(x)=g x$ induces the map $$dL_g(1): T_1G \to T_gG$$
If $v \in T_1G$, I have to prove that $dL_g(1)(v)$ is a smooth section of $T_gG$.
I know that a vector field on G is a smooth section of $T_gG$ and I tried to use this in order to solve the exercise, but a vector filed is map $F:M \to TM$, such that $F \circ \pi=Id_G $, where $M$ is a smooth manifold, $TM$ its tangent bundle and $\pi$ the fiber map. Conversely $dL_g(1)(v)$ does not go from $G$ to $T_gG$ so I don't know how to proceed.
Have you any suggestions?