$d(exp)_{0}:T_0 \frak{g} \to$ $T_eG$ is the identity map I'm learning about Lie groups, and do not have a thorough background in differentiable manifolds.
I have the following definition:
For a map $F:M \to N$ between manifolds and $a \in M$, the differential is given by $dF_a:T_aM \to T_{F(a)}N$ is given by $dF_a(X_a)(f) = X_a(f\circ F)$ where $f:M \to R$ is $C^{\infty}$.
How do you derive the statment:

$d(exp)_{0}:T_0 \frak{g} \to$ $T_eG$ is the identity map

I understand that because $\frak{g}$ is a vector space, it is equivalent to its tangent space, so I see why the statement is well defined.
I have seen two explanations for the above, neither of which I get:
1.
Fix $X \in T_eG$. Then for $s \in \mathbb{R}$ $exp(sX) = \gamma^{X}(s)$, where $\gamma^X$ is the associated 1-parameter group to $X$. (So far so good).
Then they say: $d(exp)_0(X) \underset{(1)}{=} \frac{d}{ds}(exp(0 + sX))|_{s=0}  \underset{(2)}{=} X$.
Equality $(2)$ I understand. Question 1: Where does $(1)$ come from?
2.
$\sigma: t: \mapsto tX$ is a curve in $\frak{g}$. It has $X$ as a tangent vector at $t = 0$ (Question 2: why is that the tangent vector at $t = 0$? How do I get that $d\sigma(\frac{d}{dr}|_0) = X$?)
Next $t \mapsto exp(tX)$ is a curve in $G$, which has $X_e$ as a tangent vector at $t = 0$ (this I get, it is by definition of the map $exp$).
Question 3: How does this show the statement?

Thanks for any help, I will accept an answer that allows me to understand all three questions, since I see them as crucial to my understanding of the subject. Also please try to be as rigorous as possible, otherwise I have trouble following the (very confusing )notation of differentiable manifolds. 
 A: Let $f:M\rightarrow N$ be a differentiable map between manifolds and $x\in M,u\in T_xM$, $df_x(u))={d\over{dt}}_{t=0}f\circ c$ where $c:[0,1]\rightarrow M$ is a differentiable curve such that $c'(0)=u$. To see this just apply the chain rule $(f\circ c)'=df_{c(t)}.c'(t)$, if $t=0$ you have the result.
Here consider $M={\cal G}$ the Lie algebra and $N=G$, $exp:U\subset {\cal G}\rightarrow G$ and $c(s)=sX$ is a curve in ${\cal G}$ such that $c'(0)=X$.
Here ${\cal G}$ has the structure of a vector space thus for every $u\in {\cal G}, T_u{\cal G}={\cal G}$.
A: Q1. Let $\gamma\colon \mathbb R\to M$ be a smooth curve and define $\gamma'(t_0) = d\gamma_{t_0}(\left.\frac d{dt}\right|_{t_0}).$ If $F\colon M\to N$ is a smooth map, then $$dF_{\gamma(t_0)}(\gamma'(t_0)) = dF_{\gamma(t_0)}(d\gamma_{t_0}(\left.\frac d{dt}\right|_{t_0})) = d(F\circ\gamma)_{t_0}(\left.\frac d{dt}\right|_{t_0}) = (F\circ\gamma)'(t_0) = \left.\frac d{dt}(F\circ\gamma)(t)\right|_{t=t_0}.$$
To get your expression, let $F = \mathrm{exp}\colon\mathfrak g\to G$ and $\gamma(t) = tX$. Notice that $\gamma'(0) = X$ (which you ask in Q2 why is true).
Q2. I'll use the same notation I've used above. Note that $\gamma'(0)\colon T_0\mathbb R\to T_0\mathfrak g\cong\mathfrak g.$ The key here is to understand that we identify vector space with its tangent space by assigning directional derivative to a vector. Thus, $$\gamma'(0)f = d\gamma_0(\left.\frac d{dt}\right|_{0})f = (\left.\frac d{dt}\right|_{0})(f\circ\gamma) = \left.\frac d{dt}f(tX)\right|_{t=0} \!\!\!\!= \lim_{t\to 0}\frac{f(0+tX)-f(0)}{t} = D_Xf(0) = Xf,$$ where the last equality is identification of $T_0\mathfrak g$ and $\mathfrak g$.
Q3. By the above, since $\gamma(0) = 0$ and $\gamma'(0) = X,$ we have
$$d(\mathrm{exp})_0(X) = d(\mathrm{exp})_0(\gamma'(0)) = \left.\frac d{dt}\mathrm{exp}(tX)\right|_{t = 0} = X.$$ Thus, $d(\mathrm{exp})_0$ is identity (if we identify $T_0\mathfrak g$ and $\mathfrak g$).
