Correspondence between one-parameter subgroups of G and TeG I am reading the proof of this theorem from Andreas Arvanitoyeorgos and I cannot get some points in it, highlighted below.
Theorem. The map $\phi \to d\phi_0(1)$ defines a one-to-one correspondence between one-parameter subgroups of $G$ and $T_eG$.
Proof:
Let $v \in T_eG$ and $X^v_g=(dL_g)_e(v)$ be (the value of) the corresponding left-invariant vector field. 
We need to find a smooth homomorphism $\phi_v: \mathbb R \to G$. 
Let $\phi: I \to G$ be the unique integral curve of $X^v$ such that $\phi(0)=e$ and $d\phi_t=X^v_{\phi(t)}$. 
This curve is a homomorphism because if we fix an $s \in I$ such that $s+t \in I$ for all $t \in I$ then the curves
$t \to \phi(s+t)$ 
and 
$t \to \phi(s)\phi(t)$
satisfy the previous equation (the second curve by the left-invariance of $X^v$), and take the common value $\phi(s)$ when $t=0$. By uniqueness of the integral curve then 
$\phi(s+t)=\phi(s)\phi(t)$ 
$(s,t \in I)$
Extend to $\mathbb R$ and define $\phi_v(t)=\phi(t/n)^n$. 
The map $v \to \phi_v$ is the inverse of $\phi \to d\phi_0(1)$ and this completes the proof. 
So my questions are these:
1- We need to find a smooth homomorphism $\phi_v: \mathbb R \to G$. (Why?) 
In particular this part is not clear to me: The map $v \to \phi_v$ is the inverse of $\phi \to d\phi_0(1)$ and this completes the proof. 
2- The map $t \to \phi(s)\phi(t)$ satisfies the integral equation of the vector field by the left-invariance of $X^v$ (Why?)
I see that given the left-invariant vector field we can have 
$dL_{\phi(s)}X^v_{\phi(t)}=X^v_{\phi(s)\phi(t)}$
and $d(\phi(s)\phi(t))=Y_{\phi(s)\phi(t)}$
But I cannot get it that $Y_{\phi(s)\phi(t)}=X^v_{\phi(s)\phi(t)}$ so that we have 
$d(\phi(s)\phi(t))=X^v_{\phi(s)\phi(t)}$ to say that $\phi(s)\phi(t)$ is the same integral curve.
3-(What 1 in $d\phi_0(1)$ here refer to? Is it n=1?)
 A: $1.$  The correspondence is that there is a bijection from one parameter subgroups and elements of $T_eG$.  A one parameter subgroup is a homomorphism $\mathbb{R}\rightarrow G$.  The correspondence takes a one parameter subgroup, $\phi$, such that $\phi(0)=e$ and computes the derivative $d\phi_0(1)\in T_eG$.  The object is to find a $\phi$ such that $d\phi_0(1)=v$ for some given $v\in T_eG$.
$2.$  Call $\psi(t)=\phi(s)\phi(t)$.  Then
$$
d\psi_t=\phi(s)d\phi_t=\phi(s)X^v_{\phi(t)}=X^v_{\phi(s)\phi(t)}=X^v_{\psi(t)}
$$
where the second equality follows by definition of $\phi$.  For the third equality $\phi(s)\in G$ and $X^v$ is invariant under multiplication from group elements, i.e. $gX^v_h=X^v_{gh}$.
$3.$  The number $1$ is just the time you are plugging into the derivative.  Something like if $f(x)=x^2$, then $f'(1)=2$.  Here $\phi$ is a function from $\mathbb{R}$ to $G$.  So, it has a derivative at $t=0$.  That derivative is a function from $\mathbb{R}$ to $T_eG$.  So, $\phi_0(1)$ is that derivative evaluated at $x=1$.  In the above $\phi_0(1)=v$.
