A homework problem about Tangent space of Lie groups at the identity element This is one of our homework problem.

Let $G$ be a Lie group be defined as a manifold with group structure such that the map $F:G\times G \mapsto G, F(a, b)=ab^{-1}$ is smooth. Show that $$dF_e(a,b)=a-b,$$ where $e$ is the identity.

I assume the general form of $dF_e(a,b)=Aa+bB$ where $A$ and $B$ are matrices. Then use composition of $F_e$ with other functions I show that $dF_e$ can only be $a-b$. But it doesn't seem to be the rigorous solution so I'm wondering what should be the most normal way of proving it.
 A: One way to go about proving this identity is to prove the following : 
$$d_e\iota=-\mathrm{id}_{\mathfrak g}$$
where $\iota:G\to G,g\mapsto g^{-1}$ is the inverse,  $\mathfrak g=T_eG$ is the tangent space at the identity of $G$, and 
$$d_{(e,e)}\mu(X,Y)=X+Y$$
for all $X, Y\in\mathfrak{g}$, where $\mu:G\times G\to G, (g,h)\mapsto gh$ is the product map, and compose the two.
In order to prove these two identities, just use the fact that the differential of a map $f:M\to N$ evaluated on a tangent vector $X\in T_mM$ equals the derivative of $f\circ c$ at $0$ for any smooth path $c$ with $c'(0)=X$. There are on a Lie group natural paths for the question at hand given by the exponential of said Lie group, or by integrating left (or right) invariant vector fields. These paths are one-parameter subgroups of $G$ (at least those that start at the identity of $G$) and thus behave well with respect to passing to inverses.

For instance with the inverse, observe that for all $X\in\mathfrak g$, $\iota(\exp_G(X))=\exp_G(-X)$. Now since $d_0\exp_G=\mathrm{id}_{\mathfrak g}$ it follows that 
$$d_e\iota=-\mathrm{id}_{\mathfrak g}$$

For the product map, use the natural (inverse) isomorphisms
$$(d_{(m,n)}\pi_M,d_{(m,n)}\pi_N):T_{(m,n)} (M\times N)\stackrel{\sim}{\leftrightarrow} T_mM\times T_nN:(d_mi_n,d_ni_m)$$
where $\pi_M,\pi_N$ are the natural projections and $i_n: M\to M\times N$ and $i_m: N\to M\times N$ are the natural injections.
Since $i_e:G\to G\times G, g\mapsto (g,e)$ composed with $\mu$ gives $\mathrm{id}_G$ (and similarly for $j_e:G\to G\times G, g\mapsto (e,g)$) it follows that 
$$\mathfrak g\times \mathfrak g\xrightarrow{(d_ei_e,d_ej_e)}T_e(G\times G)\xrightarrow{d_e\mu}\mathfrak g$$
sends $(X,Y)$ to $X+Y$.

Now combine the two and you'll find (using the same isomorphism to identify $T_e(G\times G)$ with $\mathfrak g\times \mathfrak g$ as earlier) that the differential of $(g,h)\mapsto gh^{-1}$ at $(e,e)$ sends the pair $(X,Y)$ to $X-Y$.
