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?)
$1.$ 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 $\phi_0(1)=v$.