Answer
$G$ is one-to-one.
(a) the image of $G$ are points in the first quadrant excluding the coordinate axes in the $xy$-plane.
(b)
- the image of the vertical lines $u=c$ are the vertical lines $x = {{\rm{e}}^c}$ in the first quadrant of the $xy$-plane.
- the image of the horizontal lines $v=c$ are the lines $y = x{{\rm{e}}^c}$ in the first quadrant excluding the origin in the $xy$-plane.
Work Step by Step
(a) Since the functions ${{\rm{e}}^u}$ and ${{\rm{e}}^{u + v}}$ are one-to-one, so $G\left( {u,v} \right) = \left( {{{\rm{e}}^u},{{\rm{e}}^{u + v}}} \right)$ is one-to-one.
From the map we get $x = {{\rm{e}}^u}$ and $y = {{\rm{e}}^{u + v}}$. Notice that ${{\rm{e}}^u} > 0$ and ${{\rm{e}}^{u + v}} > 0$, so $x,y > 0$.
Since $u,v \in \mathbb{R}$ and $x,y > 0$, so the image of $G$ are points in the first quadrant excluding the coordinate axes in the $xy$-plane.
(b)
1. Consider the vertical lines $u=c$. In this case, $v \in \mathbb{R}$.
Using the map $G\left( {u,v} \right) = \left( {{{\rm{e}}^u},{{\rm{e}}^{u + v}}} \right)$ we get
$G\left( {c,v} \right) = \left( {{{\rm{e}}^c},{{\rm{e}}^{c + v}}} \right)$
So, $x = {{\rm{e}}^c}$ and $y = {{\rm{e}}^{c + v}} = {{\rm{e}}^c}{{\rm{e}}^v}$. Since $x,y > 0$, the image of the vertical lines $u=c$ are the vertical lines $x = {{\rm{e}}^c}$ in the first quadrant of the $xy$-plane.
2. Consider the horizontal lines $v=c$. In this case, $u \in \mathbb{R}$.
Using the map $G\left( {u,v} \right) = \left( {{{\rm{e}}^u},{{\rm{e}}^{u + v}}} \right)$ we get
$G\left( {u,c} \right) = \left( {{{\rm{e}}^u},{{\rm{e}}^{u + c}}} \right)$
So, $x = {{\rm{e}}^u}$ and $y = {{\rm{e}}^{u + c}}$.
Substituting $x = {{\rm{e}}^u}$ in $y = {{\rm{e}}^{u + c}}$, we get $y = x{{\rm{e}}^c}$.
Since $x,y > 0$, the image of the horizontal lines $v=c$ are the lines $y = x{{\rm{e}}^c}$ in the first quadrant excluding the origin in the $xy$-plane.
