I'm trying to understand one line I saw in a proof that any element of the Galilean group (i.e. any Galilean transformation) on the canonical ($\mathbb R \times \mathbb R ^3$) Galilean space can be written uniquely as $g_1 \circ g_2 \circ g_3 $, where
$g_3(\mathbf G)$ : Rotation $(t,\mathbf x ) \rightarrow (t,\mathbf x)$
$g_2(s,\mathbf s)$ : Translation $(t,\mathbf x) \rightarrow (t+s,\mathbf {x+s})$
$g_1(\mathbf v)$ : Uniform-velocity motion $(t,\mathbf x) \rightarrow (t,\mathbf x+t\mathbf v) $
The proof first shows uniqueness, by demonstrating that if $\phi$ is in the Galilean group, the $s,\mathbf G, \mathbf s,\mathbf v$ parameters above will be determined by the values of $\phi(0,\mathbf 0), \phi(1,\mathbf 0)$ and $\phi(0,\mathbf e_i)$ for $i=1,2,3$.
Then to show existence, it says :
Given $\phi$ in the Galilean group, we can construct $s,\mathbf G, \mathbf s,\mathbf v$ as above and then let $\psi=g_1 \circ g_2 \circ g_3$. Now, does $\phi(s,\mathbf x)=\psi(s,\mathbf x)$ for all $s,\mathbf x$? In other words, if $\phi$ and $\psi$ are Galilean transformations that agree on $\phi(0,\mathbf 0), \phi(1,\mathbf 0)$ and $\phi(0,\mathbf e_i)$ for $i=1,2,3$, then we need to show that $\phi=\psi$ agree on all $(t,\mathbf x)$. Equivalently, if $\chi$ is a Galilean transformation that leaves fixed $\phi(0,\mathbf 0), \phi(1,\mathbf 0)$ and $\phi(0,\mathbf e_i)$ for $i=1,2,3$, we need to show it is the identity on all $(t,\mathbf x)$.
and goes on to show that $\chi(t,\mathbf x)=(t,\mathbf x)$ for all $(t,\mathbf x)$. My question is how is the line I italicized above "equivalent" to the line before? A transformation that leaves $\phi(0,\mathbf 0), \phi(1,\mathbf 0)$, $\phi(0,\mathbf e_i)$ fixed is only one specific case of a Galilean transformation. How is showing that it leaves all the other points fixed too the same as showing that an arbitrary Galiliean transformation $\phi$ agreeing on $\phi(0,\mathbf 0), \phi(1,\mathbf 0)$, $\phi(0,\mathbf e_i)$ with some transformation $\psi$ will also agree on all other points?