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 onetoone correspondence between oneparameter 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 leftinvariant 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 leftinvariance 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 leftinvariance of $X^v$ (Why?)
I see that given the leftinvariant 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 $n=1$?)
This post imported from StackExchange Physics at 20141011 09:48 (UTC), posted by SEuser Victor Vahidi Motti