Question on Left-Invariant Vector Fields Let $G$ be a Lie group, and $\xi \in T_{e}G$ a tangent vector at the identity. Given a function $f \in C^{\infty}(G)$, verify that $ g \rightarrow ((\ell_{g})_{*}\xi)f$ is a $C^{\infty}$ function on $G$, and deduce that $X_{\xi}: f \rightarrow (g \rightarrow ((\ell_{g})_{*}\xi)f)$ defines a vector field on $G$. 
Ultimately, all this is saying is that the Lie algebra $\mathfrak{g}$, i.e. the set of left-invariant vector fields on $G$, defines a natural linear map $\mathfrak{g} \rightarrow T_{e}G$, for which we have $\mathfrak{g} \ni X \rightarrow X_{e} \in T_{e}G$ such that this is a linear isomorphism. 
Proving injectivity is easy. We just take any $g \in G$. Then $X_{g}=\ell_{g*}X_{e}$, so our map is injective. Now we need only prove surjectivity. 
First, is my interpretation of the problem correct? Second, how would one show surjectivity? Thanks very much!
 A: Curiously your proof of injectivity 

We just take any g∈G. Then $X_g=ℓ_{g^∗}X_e$

also provides the proof of surjectivity. I think we need to be a bit more verbose to see the difference between the two uses of this statement.
Injectivity:
Suppose that $X$ and $Y$ are left invariant vector fields with $X_e = Y_e$. In order to show that $X = Y$ we just take any g∈G. Then $X_g=ℓ_{g^∗}X_e$ and similarly $Y_g=ℓ_{g^∗}Y_e = ℓ_{g^∗}X_e$, hence $X_g = Y_g$.
Surjectivity:
Suppose that $\xi \in T_eG$. We want to show that there exists a left invariant vector field  $X$ with $X_e = \xi$. We just take any g∈G. Then $X_g=ℓ_{g^∗}X_e = ℓ_{g^∗}\xi$ defines the desired $X$. (Of course this is just the same thing you say at the beginning of your post.)
Actually what I think causes the confusion is that in the text at the beginning of your post you construct vector field out of a tangent vector and then you interpret this construction as a map going the other way (i.e. making a tangent vector out of a vector field). So instead of showing injectivity and surjectivity you might also consider checking that the two constructions are indeed eachother's inverses.
