Affine geometry and simply transitive action I'm learning affine geometry, specifically the notion of action, and need help to understand the following example : 
Let us take $\mathbb{A} = \{(x, y) \in \mathbb{R^2} : y > 0\}$ and, as the $\mathbb{R}$-vector space, $E = \mathbb{R^2}$. As the action of the vector space on the set we take  $$\mathbb{A} \times E \to \mathbb{A} \\ (x, y), (u_1, u_2) \mapsto (x + u_1, e^{u_2}y).$$
Note that, for all $u_2 \in \mathbb{R}, e^{u_2}y > 0$, and hence $(x + u_1, e^{u_2}y) \in \mathbb{A}$.
(?) It is relatively easy to prove that this action is simply transitive. 

I don't understand the last sentence. What does it mean for this specific example to say that the action is simply transitive and how do I prove it? 
 A: A group action $*: X \times G \to X,\, (x,g) \mapsto x*g$ is transitive if for any $x,y\in X$ there is a $g \in G$ such that $x*g=y.$
A group action $*: X \times G \to X,\, (x,g) \mapsto x*g$ is simply transitive if for any $x,y\in X$ there is a unique $g \in G$ such that $x*g=y.$
In your particular example $G=E$ as an additive abelian group $(E,+)$. Let's show first that the action is transitive.
Let $(x', y'),(x,y) \in \mathbb{A}.$ We want to show that there is a $(g,h) \in E$ such that $(x',y')*(g,h)=(x' + g,e^h y')=(x,y).$
So, choose $g=(x-x')$ and $h=\ln(\frac{y}{y'})$. Since $y,y' >0, \ln(\frac{y}{y'})\in \mathbb{R}$ is well defined and $(g,h)\in E$. Then
$$(x',y')*(g,h)=(x' + (x-x'), e^{\ln(\frac{y}{y'})}y')=(x,\frac{y}{y'}y')=(x,y).$$
Hence, the action is transitive. To show that $(g,h)\in E$ is unique, suppose $(g',h')\in E$ also satisfies $(x',y')*(g',h')=(x,y).$ Then
$$(x'+ g, e^hy')=(x,y)=(x' +g',e^{h'}y'),$$
so
$$x'+g=x'+g' \, \text{ and } \, e^hy' = e^{h'}y', $$
so
$$g =g' \, \text{ and } \, y'>0 \implies e^h = e^{h'} \implies h=\ln(e^h)=\ln(e^{h'})=h'.$$
Therefore $(g,h)=(g',h')$ and so the action is simply transitive.
