Evaluating a differential on the tangent space of the product manifold I'm trying to solve a problem on Lie groups (more precisely, exercise 11 on the third chapter of Warner's Foundations of differentiable manifolds and Lie groups; or problem 7-2 in Lee's Introduction to smooth manifolds), which states that if I have $\gamma_1,\gamma_2\colon\mathbb{R}\to G$ two smooth curves on a Lie group $G$ such that $\gamma_1(0) = \gamma_2(0) = e$, then $\gamma(t) = \gamma_1(t)\gamma_2(t)$ is such that
$$\gamma'(0) = \gamma_1'(0) + \gamma_2'(0)$$
the hint that the books give is to consider the differential of the product map $m\colon G\times G\to G$ $(dm)_{(e,e)}\colon T_{(e,e)}G\times G\to T_eG$, with the usual identification $T_{(e,e)}G\times G \simeq T_eG\oplus T_eG$, and to prove that $(dm)_{(e,e)}(X_e, Y_e) = X_e + Y_e$ by proving $(dm)_{(e,e)}(X_e,0) = X_e$ and $(dm)_{(e,e)}(0, Y_e) = Y_e$.
I'm aware how, from this, the exercise would follow, but I'm having problems in proving that $(dm)_{(e,e)}(X_e,0) = X_e$. I start by considering an $f\in C^\infty(G)$ and evaluating
$$(dm)_{(e,e)}(X_e,0)(f) = (X_e,0)(f\circ m) = \cdots$$
but then I don't know how to procede: how do you evaluate tangent vectors on the tangent space of a product manifold?, how can I get to $X_e(f)$?
Moreover, on the answer of this question it is said that you can perform the product rule on $(\gamma_1\gamma_2)'(0)$ by identifying the Lie group $G$ with that of matrices, but I feel as if said argument is illegal and it's just begging the question. Am I mistaken?, can I just identify every single Lie group with a Lie group of matrices?
 A: Here's a different approach towards showing that $dm_{(e,e)}(X_e,0)=X_e$.  Start with a smooth curve $\psi\colon\mathbb{R}\to G$ so that $\psi(0)=e$ and $\psi'(0)=X_e$.  Define $\tilde{\psi}\colon\mathbb{R}\to G\times G$ by $\tilde{\psi}(t)=(\psi(t),e)$, so that $\tilde{\psi}(0)=(e,e)$ and $\tilde{\psi}'(0)=(X_e,0)$.  Then
\begin{equation}
dm_{(e,e)}(X_e,0)=\frac{d}{dt}((m\circ\tilde{\psi})(t))|_{t=0} = \frac{d}{dt}(\psi(t))|_{t=0} = X_e,
\end{equation}
because $(m\circ\tilde{\psi})(t)=\psi(t)\cdot e = \psi(t)$.  The same trick will also show that $dm_{(e,e)}(0,Y_e)=Y_e$.
A: For sake of completeness, I attach a full proof of the exercise inspired in the accepted answer:
Let $\gamma_1'(0) = X_e$ and $\gamma_2'(0) = Y_e$. Consider the smooth curves
$$\widetilde{\gamma_1}(t) = (\gamma_1(t),e)$$
$$\widetilde{\gamma_2}(t) = (e,\gamma_2(t))$$
these are such that 
$$\widetilde{\gamma_1}'(0) = (\gamma_1'(0),0) = (X_e,0)$$
$$\widetilde{\gamma_2}'(0) = (0,\gamma_2'(0)) = (0,Y_e)$$
taking this into account, we see that
$$(dm)_{(e,e)}(X_e,0) = (m\circ\widetilde{\gamma_1})'(0) = \gamma_1'(0) = X_e$$
and $(dm)_{(e,e)}(0,Y_e) = Y_e$; thus, by identifying $T_{(e,e)}G\times G$ with $T_eG\oplus T_eG$, we see that
$$(\gamma_1\gamma_2)'(0) = (m\circ (\gamma_1, \gamma_2))'(0) = (dm)_{(e,e)}(X_e,Y_e) = (dm)_{(e,e)}(X_e,0) + (dm)_{(e,e)}(0,Y_e) = X_e + Y_e = \gamma_1'(0) + \gamma_2'(0)$$
