Correspondence between one-parameter subgroups of $G$ and $T_eG$ 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 
$$\phi(s+t)=\phi(s)\phi(t)$$
where $s,t \in I$.
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:


*

*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. 

*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.

*(What 1 in $d\phi_0(1)$ here refer to? Is it $n=1$?)
 A: 1) One way to show that a map constitutes a 1-1 correspondence is by showing that it has an inverse. This is what is done here: elements of $T_eG$ are tangent vectors to $G$ at the unit element, and one-parameter subgroups of $G$ are smooth homomorphisms $\mathbb R\to G$. Note that this is a notion more general than a closed 1-dimensional subgroup, for which the theorem wouldn't hold.
The inverse should map elements of $T_eG$ to 1-parameter subgroups.
2) An integral curve is a curve that is everywhere tangent to a given vector field. From existence and unicity theorems of ordinary differential equations we get that through every point there is a unique integral curve. $\phi$ is the unique integral curve that passes through $e$ at $t = 0$, so $\phi(0) = e$ and $\dot\phi(0) \equiv d\phi_0 = X^v_e$. If instead we denote this curve $\Phi^e$, and more generally $\Phi^g$ for the unique integral curve that passes through $g$ at $t = 0$, then clearly we have $\Phi^{\phi(s)}(t) = \phi(s + t)$. The translated curve $g\phi$ passing through $g$ at $t = 0$ is an integral curve of the translated vector field, which is the same vector field by translation invariance, so it must be the unique integral curve passing through $g$. Now take $g = \phi(s)$.
3) In physics this would often just be written $\dot\phi(0)$, but strictly speaking, since $\phi$ is a smooth map $\mathbb R\to G$, its derivative in 0 is a map from the tangent space of $\mathbb R$ at 0, which is $\mathbb R$ again, to the tangent bundle of $G$ at $\phi(0) = e$, i.e. $T_eG$. Evaluating this in $1\in\mathbb R$ (or if you prefer $1\in T_0\mathbb R$) gives an element of $T_eG$. You could take any other fixed non-zero element, if you adapt your construction of $\phi_v$ accordingly.
