Proving that on a Lie group $G$ the space of left-invariant vector fields is isomorphic to $T_e G$ Let $G$ be a Lie Group. Given a vector $v\in T_e G$, we define the left-invariant vector field $L^v$ on $G$ by 
$$L^v(g)=L^v|_g=(dL_g)_e v.$$
I want to show that $v\mapsto L^v$ is a linear isomorphism between $T_e G$ and $\mathfrak{X}^L(G)$ (The set of all left invariant vector fields on $G$). 
I am comfortable showing the map is linear and 1-1. To show surjectivity, let $X\in \mathfrak{X}^L(G)$ be arbitrary. We want to show there exists $v\in T_e G$ such that $L^v|_g = X|_g$ for all $g\in G$. Define the vector $v$ as
$v:=(dL_{g^{-1}})_g X|_g$. Then, 
$\begin{align*}
L^v|_g&=(dL_g)_e\left[(dL_{g^{-1}})_g X|_g\right]\\
\\
&=(dL_g)_e[X|_e]\\
\\
&=X|_g,
\end{align*}$ 
where the last two lines are because $X$ is a left-invariant vector field.
Is this argument sufficient?

This is sufficient, though it's perhaps slightly awkward in that it defines $v$ to be the image of different vectors of different maps $(dL_{g^{-1}})_g$, all of which happen to coincide. One may as well take $v$ to be the image under just one of these maps, say, the one for $g = e$. Explicitly, this is just the map $\mathfrak{X}^L(G) \to T_e G$ given by $$X \mapsto X|_e,$$
which, by regarding $X$ as a map $G \to TG$ and $X|_e$ as a map $\{e\} \to T_e G$, we can think of as just the restriction of $X$ to $\{e\}$.
