# Lie algebra and tangent space at the identity are isomorphic

Define the Lie algebra $\mathfrak{g}$ of a Lie group G to be the set of all left-invariant vector fields of $G$.
I want to prove that

$$f: \mathfrak{g} \rightarrow T_eG \\ X \mapsto X(e)$$ is a linear isomorphism.

The only thing that I have left to prove is surjectivity.Letting $v\in T_eG$, we set $X\in \mathcal{X}(G)$ such that $\forall x\in G,\,X(x)=L_{x_{*,e}}(v)$ (I still have to show that such an $X$ is indeed a vector field on $G$). We get that $$f(X)=X(e)=L_{e_{*,e}}(v)=v$$ where the last equality holds because $$L_{e_{*,e}}(v)f = v(f \circ L_e) = v(f \circ id_G) = v(f)$$ for any $f\in C^\infty(G)$. (I am thinking of the tangent vectors as derivations).

Now all we have left to do is prove that such an $X$ is left-invariant, and it is therefore an element of $\mathfrak{g}$, but I do not know how. I tried to prove that $$\forall f\in C^\infty(G),\, L_{{x}_*,y}(X(y))f=X(L_x(y))f$$ but with no success.

To show that $X$ is left-invariant, you need to show that $$L_q^{*}(X(x)) = X(qx)$$ for $q \in G$, right? (Here the "star" denotes the tangent map -- the derivative of left multiplication, which taked $T_x(G)$ to $T_{gx}(G)$.)
Now $X(x) = L_x^{*} X(e)$, by definition (if I'm reading your notation right). So what you want to prove is that $$L_q^{*}( L_x^{*} X(e)) = X(qx).$$ The left hand side is, by the chain rule, just $$(L_q \circ L_x)^{*} (X(e)) = L_{qx}^{*} (X(e)).$$ Now recalling that $X(x) = L_x^{*}X(e)$ for any $x$, we apply this to $qx$, getting that $$X(qx) = L_{qx}^{*} (X(e))$$ and we're done.