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Let $G$ be a Lie group, and let $v: G \rightarrow T(G)$ be a left invariant vector field. Then $v$ is uniquely determined by its value at $e = 1_G$, so $v \mapsto v(e)$ defines an injection of left invariant vector fields into $T_e(G) = \mathfrak g$. I'm trying to show this correspondence is surjective.

Given $X \in \mathfrak g$, the corresponding $v$ should be defined by the formula $v(x) = T_e(\ell_x)(X)$, where $\ell_x: G \rightarrow G$ denotes left translation.

The first thing I'm trying to check is that $v$ is smooth. Should this be obvious? I think I have a way of showing arguing this (in my answer below), but I would be interested in knowing whether there is a simpler way.

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Let $x_0 \in G$, and let $U$ be a chart of $G$ containing $x_0$.

Let $U'$ be a chart of $G$ containing $e$ and centering $e$ at the zero vector, and assume $U$ and $U'$ are chosen small enough so that the product set $U.U'$ is contained in some chart $V$ of $G$. To the chart $V$ containing $x_0$, we can associate a chart $V \times \mathbb R^n$ of $T(G)$, and the map $x \mapsto T_e(\ell_x)(X)$ is given in local coordinates by a map $U \rightarrow V \times \mathbb R^n$, which we want to show is smooth.

The product map $\mu: G \times G \rightarrow G$ restricts to a map $\mu: U \times U' \rightarrow V$, which is given in coordinates as $(\mu_1, ... , \mu_n)$. Each $\mu$ and $\mu_i$ is a function of $2n$ variables $x_1, ... , x_n, x_1', ... , x_n'$, or just $(x,x')$.

Now for each $x = (x_1, ... , x_n) \in U$, the derivative of $\ell_x: U' \rightarrow V$ at $x'$ is given by the Jacobian $(\frac{\partial \mu_i}{\partial x_j'}(x,x'))$.

Then $v(x)$ is given in local coordinates by $x \mapsto (x_1, ... , x_n, a_1, ... , a_n)$, where $^T(a_1, ... , a_n)$ the vector given by the product of the matrix $(\frac{\partial \mu_i}{\partial x_j'}(x,0))$ and the column vector $X$, where $X \in \mathfrak g$ is given in local coordinates with respect to the chart $U'$ and the point $e$.

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